Part A

Metric and Topological Spaces

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Lecture A1Introduction to the Course

This course is an introduction to the basic ideas of topology and metric spacetheory for first-year graduate students. Topology studies the idea of “continuity”in the most general possible context. As a separate subject, it takes its originfrom the pioneering papers of Poincare at the end of the 19th century, and itsdevelopment is one of the central stories of 20th century pure mathematics. (Thehistory by Dieudonne, A History of Algebraic and Differential Topology, 1900 -1960, is available free to Penn State members online. — but that history beginsroughly where this course leaves off.) But the idea of continuity is so centralthat topological methods crop up all over mathematics — in analysis of course(including its “computational” twin, numerical analysis); in geometry; in purealgebra; even in number theory. So, if you’re a first-year grad student, you mostlikely need to take this course (and maybe even Math 528 as well).

In this lecture we’ll just describe the course protocol and prerequisites — we’llget started properly next time.

There is a recommended course textbook — Introduction to Topology by Gamelinand Greene, which is available in a cheap Dover reprint for 15 bucks. However,more important than that are the online course materials, which are available intwo ways:

• Through the course website, which ishttp://www.math.psu.edu/roe/527-FA10/index.html

• Through the course page on ANGEL, Penn State’s course management sys-tem, at http://cms.psu.edu

Registered students (i.e., those taking this course for a grade) will want to usethe ANGEL site which will contain homework assignments, quizzes and support-ing material, which all need to be taken on time in order to receive credit. The AN-GEL page also contains the detailed course syllabus, which contains University-required information about exam schedules, academic integrity requirements, of-fice hours, and suchlike. Please be sure to read this information carefully.

Detailed lecture notes for each lecture will be posted 24 hours in advance onthese sites. To succeed in the course you need to read and study the notes beforethe lecture to which they refer. This is very important!

Most lectures will contain one or more in-class exercises. You are expectedto attempt these exercises before the next class session. Students may be called

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on to present their solutions in class. You grade your own exercises using theANGEL system, and this self-assessment will form a (small) component of yourfinal in-class grade.

Exercise A1.1. Here is your first in-class exercise. Find the corresponding as-sessment item on ANGEL (it’s called “inclassAug23”), and enter your response(“I fully understand this”). You must complete this assignment before tomorrowevening, when it will expire.

Now for some information about the prerequisites for the course, that is, thingsthat I will assume that you know. There are four groups of these.

1. Number systems and their properties: specifically the natural numbers N,integers Z, rational numbers Q, real numbers R, and complex numbers C. Coursesin analysis or foundations of mathematics often contain an account of proceduresfor “constructing” all or some of these, but we’ll just take them as given. The mostimportant of these number systems for us is the system R of real numbers. Thebasic properties of the reals are summarized by

Proposition A1.2. R is an archimedean complete ordered field.

What does this mean?

(a) A field — basic properties of additional, multiplication, subtraction, divi-sion — commutative, associative, distributive laws.

(b) ordered. There is an order relations < with the expected properties. Inparticular x2 > 0 for all x.

(c) Archimedean — no ‘infinite’ or ‘infinitesimal’ real numbers. There is aunique (injective) homomorphism of rings Z → R taking 1 to 1. (Proof?)The archimedean property is that every element of R is smaller than (theimage of) some element of Z.

(d) complete — no ‘holes’, contrast Q.

There are various ways of expressing completeness. The standard way is theLeast Upper Bound Axiom. Let S be any nonempty set of real numbers. A numbera ∈ R is an upper bound for S if, for all x ∈ S, x 6 a. Let US be the set of allupper bounds for S. The least upper bound axiom says that, if S and US are bothnonempty, then US has a least member. This is called the least upper bound for Sand written supS.

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Example A1.3. Let S = x ∈ R : x2 < 2. Then US = a ∈ R : a > 0, a2 > 2.US has a least member, namely supS =

√2. The corresponding statement with

R replaced everywhere by Q would be false.

Exercise A1.4. (Cantor’s nested interval theorem) Let [an, bn], n = 1, 2, . . . be asequence of closed intervals in R that are nested in the sense that [an+1, bn+1] ⊆[an, bn] and whose lengths bn − an tend to zero. Show that the intersection⋂n[an, bn] contains one and only one real number c.

Exercise A1.5. Use the previous exercise to show that the real numbers are un-countable, i.e. cannot be listed as c1, c2, . . .. Hint: Suppose such a listing ispossible. Pick an interval [a1, b1] of length at most 1 that doesn’t contain c1, thena subinterval [a2, b2] of length at most 1

2that doesn’t contain c2, and so on. Apply

the nested interval theorem. What can you say about the limit point c?

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Lecture A2Metric Spaces

We continue with our list of prerequisites for the course (we discussed number1 last time.)

2. Set-theoretic language such as unions (A ∪ B), intersections (A ∩ B),complements (A\B), subsets (A ⊆ B) and so on. In topology it is often necessaryto deal with infinite unions and intersections. Let F be a family of sets (that is, aset whose members are subsets of some other set). The intersection of the family,written

⋂F∈F F or just

⋂F , is the set of objects that belong to every member of

the family F : in symbols

x ∈⋂

F ⇔ ∀F ∈ F , x ∈ F.

Similarly the union of the family, written⋃F∈F F or just

⋃F , is the set of objects

that belong to some member of the family F : in symbols

x ∈⋃

F ⇔ ∃F ∈ F , x ∈ F.

De Morgan’s Laws extend to this context: if F is a family of subsets of some setX , then

X \⋂F∈F

F =⋃F∈F

X \ F

and the same with intersection and union reversed.

3. Quantified statements and their proofs. We already met some quantifiersin the previous section: things like ∀ and ∃. Throughout pure mathematics, butespecially analysis and topology, one meets concepts that are built up of layers ofquantifiers nested together in complicated ways. For instance, a function f : R→R is continuous at a given point x0 if

∀ε > 0 ∃δ > 0∀x ∈ R |x− x0| < δ ⇒ |f(x)− f(x0)| < ε.

It is necessary that you can grasp what such a complicated statement says, andthat you can understand how the form of the statement dictates the shape of itsproof. For instance the statement above begins “∀ε > 0” so its proof has to begin“Let ε > 0 be arbitrary”, followed by an argument that establishes the rest of thestatement for a given arbitrary value of ε. You could call a simple pattern like thisa ”proof skeleton” corresponding to the “for all” quantifier.

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Exercise A2.1. Give some examples of other proof skeletons corresponding toother parts of a statement, e.g. ∃,⇒,⇔ and so on.

Exercise A2.2. Write down in symbols (as above) what it is for the function f tobe not continuous at x0.

4. Axiomatic method This is the “standard operating procedure” of modernmathematics. We understand specific examples by fitting them into a generaltheory. The general theory consists of a collection of axioms, such as those for agroup or a vector space. There are many examples which realize the axioms, andwe develop a theory that applies to all of them.

We will study topology from this point of view. We’ll begin with the axiomsfor metric spaces.

Definition A2.3. A metric space is a setX equipped with a function d : X×X →R (called a metric or distance function) such that:

(i) d(x, x′) > 0 for all x, x′; moreover, d(x, x′) = 0 if and only if x = x′.

(ii) d(x, x′) = d(x′, x) for all x, x′ (symmetry).

(iii) d(x, x′′) 6 d(x, x′) + d(x′, x′′) for all x, x′, x′′ (triangle inequality).

The motivating example is the metric d(z, w) = |z − w| on C or R.

Definition A2.4. Let (X, dX) and (Y, dY ) be metric spaces. An isometry from Xto Y is a bijection f : X → Y such that

dY (f(x1), f(x2)) = dX(x1, x2)

for all x1, x2 ∈ X .

For the usual metric on the plane, the isometries are just the congruences ofEuclidean geometry. Two metric spaces that are related by an isometry are equiv-alent from the point of view of metric space theory.

Here are some more examples of metric spaces.

Example A2.5. The vector space Rn can be made into a metric space by definingthe distance between points x = (x1, . . . , xn) and y = (y1, . . . , yn) to be

d(x, y) =(|x1 − y1|2 + · · ·+ |xn − yn|2

)1/2.

(The proof that this does indeed satisfy the triangle inequality is a standard ex-ercise.) The same formula also makes Cn into a metric space. These are calledEuclidean spaces.

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Example A2.6. One can also define metrics on Rn by

d′(x, y) =(|x1 − y1|+ · · ·+ |xn − yn|

)and

d′′(x, y) = sup|x1 − y1|, . . . , |xn − yn|

.

These are different metrics from the standard one (though, as we’ll see later, theyare in a certain sense “equivalent”.)

Example A2.7. Let X be any set. The discrete metric on X is defined by

d(x, y) =

0 (x = y)

1 (x 6= y)

Example A2.8. Let A be a finite set (the alphabet) and consider the set An ofn-tuples of elements of A, which we think of as n-letter words in the alphabet A.Define a distance on An by

d(x, y) = #i : 1 6 i 6 n, xi 6= yi;

in other words, the number of positions in which the two words differ. This Ham-ming distance was introduced in 1950 to give a technical foundation to the theoryof error-correcting codes (Hamming, Error-detecting and error-correcting codes,Bell System Technical Journal 29(1950), 147–160.)

Example A2.9. Let X be any metric space and let Y be a subset of X . Then thedistance function on X , restricted to Y , makes Y into a metric space; this metricstructure is called the subspace metric on Y .

Example A2.10. Here is an infinite-dimensional example. Consider the collectionC[0, 1] of all continuous functions [0, 1]→ C. We can define a metric by

d(f, g) = sup|f(t)− g(t)| : t ∈ [0, 1].

Convergence of a sequence of functions in this metric is called uniform conver-gence.

Definition A2.11. A subset U of a metric space X is open if for every x ∈ Uthere is ε > 0 such that the entire ball

B(x; ε) := x′ ∈ X : d(x, x′) < ε

is contained in U .

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The triangle inequality shows that every ball is open, so there are plenty ofopen sets. A set F whose complement X \ F is open is called closed. Notecarefully that ‘closed’ does not mean the same as ‘not open’. Many sets are neitheropen nor closed, and some may be both.

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Lecture A3Open and Closed Sets

We finished last time with an important definition: A subset U of a metricspace X is open if for every x ∈ U there is ε > 0 such that the entire ball

B(x; ε) := x′ ∈ X : d(x, x′) < ε

is contained in U .

Example A3.1. In a discrete metric space (Example A2.7), the open ball B(x, 12)

is just the set x. Consequently, in a discrete space, every subset is open.

Example A3.2. Consider the three metrics on Rn defined in examples A2.5 and A2.6.Each open ball in any one of these metrics contains an open ball (with the samecenter but possibly different radius) in any one of the other metrics. Consequently,these three metrics all have exactly the same open sets. That is what is meant bysaying that they are equivalent.

Lemma A3.3. The union of any collection of open sets is open. The intersectionof a finite collection of open sets is open. The empty set ∅, and the entire metricspace X , are open.

Proof. Let F be a collection of open subsets of a metric space X and let U =⋃F be the union of the family. If x ∈ U , then there is some V ∈ F such that

x ∈ V . Since V is open, there is ε > 0 such that B(x; ε) ⊆ V . Since V ⊆ U ,we also have B(x; ε) ⊆ U . Thus for any x ∈ U there exists ε > 0 such thatB(x; ε) ⊆ U ; which is to say that U is open.

Now let F = U1, . . . , Un be a finite collection of open sets and let U =⋂F = U1 ∩ · · · ∩ Un. If x ∈ U , then for each i = 1, . . . , n there is εi > 0 such

that B(x; εi) ⊆ Ui. Let ε = minε1, . . . , εn > 0. Then B(x; ε) ⊆ Ui for all i, andthus B(x; ε) ⊆ U . Thus for any x ∈ U there exists ε > 0 such that B(x; ε) ⊆ U ;which is to say that U is open.

Equivalently, using de Morgan’s laws, we have

Lemma A3.4. The intersection of any collection of closed sets is closed. Theunion of a finite collection of closed sets is closed. The empty set ∅ and the entiremetric space X are closed.

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Let X be any metric space, and let S be a subset of X .

Definition A3.5. The interior of S (inX), denoted S, is the union of all the opensubsets of X that are included in S.

In symbols, we have

S =⋃

U : U ⊆ S and U open in X.

The interior of S is an open set (because it is the union of a family of opensets) and (from the definition) any open subset of X that is included in S is alsoincluded in S. Thus, the interior of S is just the biggest open subset of X that isincluded in X . In particular, S is open if and only if it is equal to its own interior.Dually, we have

Definition A3.6. The closure of S (in X), denoted S, is the intersection of all theclosed subsets of X that include S.

The closure of S is a closed set, and it is the smallest closed set that includesS; in particular, S itself is closed iff it is equal to its own closure. Finally,

Definition A3.7. The boundary of S (in X), denoted ∂X , is the set-theoreticdifference ∂S = S \ S.

The boundary of S is the intersection of two closed sets (S and X \ S), andit is therefore closed.

Example A3.8. Let X = R and let S =((0, 1) ∩Q

)∪ (2, 3]. Then S = (2, 3),

S = [0, 1] ∪ [2, 3], and ∂S = [0, 1] ∪ 2, 3.

Exercise A3.9. Show that if A,B are subsets of X , and A ⊆ B, then A ⊆ B

and A ⊆ B. Now show that for any subset S of X ,((S))

=(S).

One more remark about open sets. Let X be a metric space and let Y be asubset of X . As we said earlier, Y can be considered as a metric space in its ownright (with the metric that it inherits from X). What is then the relation betweenthe open subsets of the new space Y and the open subsets of the original spaceX?

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Proposition A3.10. Let X be a metric space, Y a metric subspace (as above).Then a subset V ⊆ Y is open in Y iff it can be written V = U ∩ Y for someU ⊆ X open in X .

Proof. Notice the following relationship between balls in Y and in X:if y ∈ Ythen

BY (y; r) = BX(y; r) ∩ Y.Let V be open in Y . Then for each y ∈ V there is ry > 0 such thatBY (y; ry) ⊆ V .Let U =

⋃y∈Y BX(y; ry). This is a union of open subsets of X , so open in X ,

and

U ∩ Y =

(⋃y∈Y

BX(y; ry)

)∩ Y =

⋃y∈Y

BY (y; ry) = V.

This proves one direction of the “if and only if” statement in the proposition; theother direction (which is easier) is an exercise.

The definition of continuity is translated in the natural way to the metric spacecontext.

Definition A3.11. A function f : X → Y between metric spaces is continuous atx ∈ X if for every ε > 0 there is δ > 0 such that d(f(x), f(x′)) < ε wheneverd(x, x′) < δ. It is continuous if it is continuous at every x ∈ X .

But there is a very important alternative characterization in terms of open sets.

Theorem A3.12. Let X and Y be metric spaces. Then f : X → Y is continuousiff, for every open U ⊆ Y , the inverse image

f−1(U) := x ∈ X : f(x) ∈ U

is open in X .

Proof. Suppose that f is continuous and let U ⊆ Y be open. Let x ∈ f−1(U);then f(x) ∈ U so by definition of ‘open’ there is ε > 0 such that B(f(x); ε) ⊆ U .By definition of ‘continuous’ there is δ > 0 such that if x′ ∈ B(x; δ) then f(x′) ∈B(f(x); ε) ⊆ U . But this means that B(x; δ) ⊆ f−1(U). Thus the set f−1(U) isopen.

Conversely, let x ∈ X , ε > 0 and suppose that f satisfies the condition inthe theorem. In particular, we may consider U = B(f(x); ε), an open set suchthat x ∈ f−1(U). Our hypothesis tells us that f−1(U) is open, which means thatthere is a δ > 0 such that B(x; δ) ⊆ f−1(U). We have shown that wheneverx′ ∈ B(x; δ), f(x′) ∈ B(f(x); ε). This gives us continuity.

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Lecture A4Continuity and sequences

In the previous lecture we saw that a map f : X → Y between metric spacesis continuous iff f−1(U) is open (in X) whenever U is open (in Y ).

Remark A4.1. It is equivalent to say that f−1(F ) is closed in X whenever F isclosed in Y . But in general nothing can be said about the behavior of the directimage f(S) of an open or a closed set S under a continuous map.

Definition A4.2. A map f : X → Y between metric spaces is called a homeo-morphism if it is a bijection and both f and f−1 are continuous. (Equivalently,f is a continuous map with a continuous inverse.) If there is a homeomorphismbetween X and Y , then we say that these spaces are homeomorphic.

A homeomorphism is a “topological equivalence” (it is easy to check that therelation of homeomorphism is an equivalence relation in the sense of algebra). Intopology, we consider homeomorphic spaces to be ”essentially the same“. A fun-damental question in topology is to devise processes for answering the question,Are two concretely given spaces homeomorphic or not?

Example A4.3. The map x 7→ x(1 + x2)−12 is a homeomorphism from R to the

open interval 9− 1, 1). Its inverse is y 7→ y(1− y2)− 12 .

Example A4.4. The map t 7→ (cos 2πt, sin 2πt) is a continuous bijection from thehalf-open interval (0, 1] onto the unit circle in R2, but it is not a homeomorphism.

Example A4.5. Let (X, d) be a metric space and let (X, d′) be the same setequipped with the discrete metric. Then the identity map (X, d′) → (X, d) isa continuous bijection, but it is usually not a homeomorphism.

Example A4.6. Are Rn and Rm homeomorphic if m 6= n? This and relatedquestions were problematical in the early days of topology. (Note that there arecontinuous maps of Rm onto Rn even if m < n — “space-filling curves”.) Butthe answer, as expected, is indeed “no”, as was shown by Brouwer and others atthe beginning of the 20th century.

We’ll now study the properties of sequences. A sequence in a metric space Xis a mapping from the natural numbers N toX: we’ll follow the usual abbreviationof referring to “the sequence (xn)” rather than “the sequence which maps thenatural number n to xn ∈ X .”

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Definition A4.7. Let (xn) be a sequence in the metric space X . We say (xn)converges to x ∈ X if for every ε > 0 there is an integer N such that d(x, xn) < εwhenever n > N . We write then x = limn→∞ xn.

Exercise A4.8. Let T be the metric space 1, 12, 13, . . . , 0 (with the metric it in-

herits as a subspace of R). Show that a sequence (xn) in the metric space X isconvergent if and only if there exists a continuous function f : T → X such thatf(1/n) = xn; and in that case f(0) = limn→∞ xn.

Proposition A4.9. A function f : X → Y between metric spaces is continuousat x if and only if, whenever (xn) is a sequence converging to x, the sequence(f(xn)) converges to f(x).

Proof. Suppose that f is continuous at x and let ε > 0 be given. By definition ofcontinuity there is δ > 0 such that d(x, x′) < δ implies d(f(x), f(x′)) < ε. Bydefinition of convergence there is N such that for all n > N , d(x, xn) < δ. Thusd(f(x), f(xn)) < ε and f(xn) converges to f(x).

Suppose that f is not continuous at x. Then there is ε > 0 such that for allδ > 0 there is x′ with d(x, x′) < δ and d(f(x), f(x′) > ε. Let xn be a value of x′

corresponding to δ = 1/n. Then xn → x, but d(f(x), f(xn)) > ε for all n, hencef(xn) does not converge to f(x).

Remark A4.10. This result illustrates how sequences can be used to probe thetopological properties of metric spaces. At the same time, it shows how the use-fulness of these “probes” depends on the countable local structure of a metricspace — specifically, on the fact that any ball around x ∈ X contains one ofthe countably many balls B(x; 1/n). When we come to consider more generaltopological spaces, this “countable local structure” may not be available, and thensequences will be of less use.

Definition A4.11. Let X be a metric space, A ⊆ X . A point x ∈ X is a limitpoint of A if it is the limit of a sequence of distinct points of A.

Lemma A4.12. x is a limit point of A if and only if every open set containing xalso contains infinitely many points of A.

Proof. Suppose that x is a limit point of A. Then there is a sequence (xn) ofdistinct points of A, converging to x. Let U be an open set containing x. Thenthere is some ε > 0 such that B(x; ε) ⊆ U . By definition of convergence, there

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is N > 0 such that xn ∈ B(x; ε) ⊆ U for all n > N . In particular, U containsinfinitely many of the xn, all of which are members of A.

Conversely suppose that every open set containing x also contains infinitelymany points of A. Define a sequence (xn) inductively as follows: x1 is anypoint in B(x; 1) ∩ A, and, assuming that x1, . . . , xn−1 have been defined, let xnbe any point in B(x; 1/n) ∩ A that is distinct from the (finitely many!) pointsx1, . . . , xn−1. Then (xn) is a sequence of distinct points of A and it converges tox.

Proposition A4.13. The closure of a subset A is the union of A and the set ofall its limit points. In particular, A is closed if and only if it contains all its limitpoints.

Proof. A point x belongs to the closure ofA iff it is not in the interior ofX\A. Bydefinition of “interior”, this is the same as to say that each ball B(x; ε) must meetA. By the previous lemma, then, any limit point of A must belong to A. Con-versely, suppose that x ∈ A \ A. Define a sequence (xn) inductively as follows:x1 is any point in B(x; 1)∩A, and, assuming that x1, . . . , xn−1 have been defined,let xn be any point in B(x; δn) ∩ A where δn = mind(x1, x), . . . , d(xn−1, x).Then (xn) is a sequence of distinct points of A and it converges to x, so x is alimit point of A.

Definition A4.14. An isolated point of A is a point of A that is not a limit pointof A. A closed subset of a metric space is perfect if it has no isolated points.

Exercise A4.15. Show that a ∈ A is isolated if and only if there is ε > 0 such thatB(a; ε) ∩ A = a.

Exercise A4.16. A closed ball in a metric space is a set

B(x; r) := x′ ∈ X : d(x, x′) 6 r.

Show that a closed ball is closed. Must every closed ball be the closure of theopen ball of the same center and radius?

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Lecture A5Compactness

I’ll take for granted the notion of subsequence of a sequence of points in amatric space. (Formally speaking, if a sequence in X is a map N → X , a subse-quence of the given sequence is the result of composing it with a strictly mono-tonic map N→ N.)

Lemma A5.1. Every sequence of real numbers has a monotonic subsequence.

Proof. Let (xn) be a sequence of real numbers. Suppose that it has no monotonicincreasing subsequence. Then there must be some n such that xm < xn for allm > n. (Otherwise, we could build a monotonic increasing subsequence byinduction.)

But now take this n and call it n1. Apply the previous argument to the “tail” ofthe original sequence beginning at n1, i.e. the subsequence xn1+1, xn1+2, . . .. Thisalso has no monotonic increasing subsequence, so by the same argument there isn2 > n1 such that xm < xn2 for all m > n2. Continuing in this way by inductionwe get a monotonic decreasing subsequence xn1 , xn2 , . . ..

It is a standard consequence of completeness that every bounded, monotonicsequence of real numbers is convergent. Thus we get the Bolzano-Weierstrasstheorem: every bounded sequence of real numbers has a convergent subsequence.

Definition A5.2. A Cauchy sequence in a metric space is a sequence (xn) withthe following property: for every ε > 0 there exists N > 0 such that, for alln,m > N , d(xn, xm) < ε. A metric space is called complete if every Cauchysequence in it converges.

A Cauchy sequence is necessarily bounded. Moreover, if a Cauchy sequencehas a convergent subsequence, then the whole sequence is in fact convergent.These facts combine with the Bolzano-Weierstrass theorem to show that everyCauchy sequence of real numbers converges: i.e., R is (Cauchy) complete.

Definition A5.3. A metric space X is (sequentially) compact iff every sequenceof points of X has a subsequence that converges in X .

Proposition A5.4. Every closed, bounded subset of Rn or Cn is sequentially com-pact. (A subset is bounded if it is contained in some ball.)

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Proof. For R this is just the Bolzano-Weierstrass theorem. For Rn, this followsfrom the fact (easily proved) that a sequence of points in Rn is convergent if andonly if each of its coordinate sequences is convergent.

Proposition A5.5. If A is a subset of any metric space X , and A is compact (inits own right), then A is bounded and closed in X .

Proof. Let x ∈ X be a limit point of A. Then there is a sequence (an) in Aconverging to x. But, by compactness, (an) has a subsequence converging in A.Thus x ∈ A, and A is closed.

Suppose that A is not bounded. Then one can construct by induction a se-quence (an) in A such that d(an, a0) > 1 + d(an−1, a0) for n > 1. The triangleinequality shows that d(an, am) > 1 for n 6= m. Clearly (an) has no convergentsubsequence.

Exercise A5.6. Let f : X → Y be continuous and surjective. Show that if X iscompact then Y is compact also. What has this got to do with the familiar calculusprinciple that ‘a continuous function on a closed bounded interval is itself boundedand attains its bounds’?

Despite the above evidence, it’s not in general true that ‘closed and bounded’equals ‘compact’. (Counterexample?) We shall analyze this in detail.

Definition A5.7. Let X be a metric space. An open cover U for X is a collection(finite or infinite) of open sets whose union is all of X . A Lebesgue number forU is a number δ > 0 such that every open ball of radius δ is a subset of somemember of U .

Theorem A5.8. (Lebesgue) Every open cover of a (sequentially) compact metricspace has a Lebesgue number.

Proof. Suppose that U does not have a Lebesgue number. Then for every nthere is xn ∈ X such that B(xn; 1/n) is contained in no member of U . If X iscompact, the sequence (xn) has a subsequence that converges, say to x. Now xbelongs to some member U of U , since U is a cover. Thus there is ε > 0 suchthat B(x; ε) ⊆ U . There is n > 2/ε such that d(xn, x) < ε/2. But then

B(xn; 1/n) ⊆ B(xn; ε/2) ⊆ B(x; ε) ⊆ U

which is a contradiction.

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Proposition A5.9. The following conditions on a metric space X are equivalent.

(a) Every sequence in X has a Cauchy subsequence.

(b) For every δ > 0 there is a finite cover of X by balls of radius δ.

In this case we say X is totally bounded (some writers say precompact).

Proof. (a) implies (b): Choose x1 ∈ X , and then inductively choose xn ∈ Xsuch that d(xn, xm) > δ when m < n, so long as this is possible. The processmust terminate because if it didn’t it would produce a sequence with no Cauchysubsequence. When it does terminate, with xN say, it does so because the ballsB(xn; δ), n = 1, . . . , N , cover X .

(b) implies (a): Notice the following implication of (b): given any δ > 0, anysequence in X has a subsequence all of whose members are separated by at mostδ (call this a ‘δ-close subsequence’).

Let (xn) be any sequence in X . Let (x1n) be a 2−1-close subsequence of (xn),let (x2n) be a 2−2-close subsequence of (x1n), and so on by induction. Then (xnn) isa Cauchy subsequence of the original sequence.

A metric space X is said to be (covering) compact if every open cover of Xhas a finite subcover.

Proposition A5.10. The following conditions on a metric spaceX are equivalent:

(a) X is sequentially compact;

(b) X is complete and totally bounded;

(c) X is covering compact.

Proof. It is easy to see that (a) and (b) are equivalent.Suppose (a). Let U be an open cover of X . Let δ > 0 be a Lebesgue number

for U (which exists because of Theorem A5.8). Since X is totally bounded it hasa finite cover by δ-balls. But each such ball is a subset of a member of U ; so Uhas a finite subcover.

In the other direction, suppose (c). Let (xn) be a sequence without convergentsubsequence. Then for each x ∈ X there is some εx such that xn /∈ B(x; εx) forall but finitely many n. The B(x; εx) form a cover of X . Picking a finite subcoverwe obtain the contradiction that xn /∈ X for all but finitely many n.

17

Remark A5.11. For general topological spaces, the notions ‘covering compact’and ‘sequentially compact’ are not equivalent, and ‘covering compact’ is usuallythe most appropriate one.

Exercise A5.12. A map f : X → Y between metric spaces is uniformly contin-uous if for each ε > 0 there is δ > 0 such that d(f(x), f(x′)) < ε wheneverd(x, x′) < δ. (The extra information beyond ordinary continuity is that δ doesnot depend on x.) Show that if X is compact, every continuous f is uniformlycontinuous. Hint: use Theorem A5.8.

18

Lecture A6Compactness and Completeness

(The lecture will begin with a review of the basic results about compactness,from the previous section).

Definition A6.1. LetX be a compact metric space. Then C(X) denotes the spaceof all continuous functions X → C. It is a metric space equipped with the metric

d(f, g) = sup|f(x)− g(x)| : x ∈ X.

(Compare Example A2.10.)

Compactness of X ensures that the supremum exists (why)? We can alsoconsider the space CR(X) of continuous real-valued functions.

Proposition A6.2. For a compact X , the metric spaces C(X) and CR(X) arecomplete.

Proof. Most completeness proofs proceed in the same three-stage way: Given aCauchy sequence, identify a candidate for its limit; show that the candidate is inthe space in question; and show that the sequence approaches the candidate limitin the metric of the space in question.

Let (fn) be a Cauchy sequence in C(X). Then, for each x ∈ X , fn(x) is aCauchy sequence in C. Since C is complete this sequence converges. Denote itslimit by f(x).

We show that f is a continuous function on X . Fix x′ ∈ X and let ε > 0be given. Because (fn) is Cauchy there is N such that for n′, n′′ > N we have|fn′(x) − fn′′(x)| < ε/3 for all x. Take n′ = N and let n′′ → ∞ to find that|fN(x)− f(x)| 6 ε/3 for all x. Now, fN is continuous at x′ so there is δ > 0 suchthat |fN(x) − fN(x′)| < ε/3 whenever |x − x′| < δ. It follows that, whenever|x− x′| < δ,

|f(x)− f(x′)| 6|f(x)− fN(x)|+ |fN(x)− fN(x′)|+ |fN(x′)− f(x′)| < ε.

This shows that f is continuous.Finally to show that fn → f in C(X) we must prove that for every ε > 0

there is N such that |fn(x)− f(x)| < ε for all x whenever n > N . But in fact wealready proved this in the previous paragraph.

19

It is an interesting and important question what are the compact subsets ofthe complete space C(X). The answer is given by the Ascoli-Arzela theorem.“Closed and bounded” is not enough.

20

Lecture A7Applications of completeness

Recall that a metric space X is said to be complete if every Cauchy sequencein X converges. The space (0, 1) is not complete, whereas the homeomorphicspace R is complete. This proves that completeness is not a topological property,i.e., is not preserved under homeomorphism.

Exercise A7.1. A uniform homeomorphism between metric spaces is a homeo-morphism f : X → Y such that both f and f−1 are uniformly continuous. Showthat if X is complete and there is a uniform homeomorphism X → Y , then Y iscomplete.

Proposition A7.2. Let X be a complete metric space and let U ⊆ X be an openset. Then U is homeomorphic to a complete metric space (it is topologicallycomplete).

Proof. Let f : U → R be the function defined by

f(x) =1

infd(x, y) : y ∈ X \ U.

This is a well-defined, continuous function with the property that f(xn) → ∞whenever xn is a sequence in U that converges to some x ∈ X \ U .

Now let V be the metric space with the same points as U but with the metric

dV (x, x′) = d(x, x′) + |f(x)− f(x′)|.

It is easy to see that this is indeed a metric. Moreover, it is complete: if (xn) isa Cauchy sequence for the metric dV , then it is also a Cauchy sequence for themetric d (so it converges in X , say to x ∈ X) and, in addition, the values f(xn)are bounded (so the limit x in fact belongs to V ).

Finally I claim that the identity map U → V is a homeomorphism. Supposex ∈ U and let ε > 0 be given. Since f is continuous, there is δ1 > 0 such thatd(x, x′) < δ1 implies |f(x) − f(x′)| < ε/2. Put δ = minδ1, ε/2). Then ifd(x, x′) < δ, we have

dV (x, x′) <ε

2+ |f(x)− f(x′)| < ε.

So the identity map U → V is continuous, and the continuity of the inverse iseasy.

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Remark A7.3. A subset of a metric space is called a Gδ-set (German: Gebeit-Durschnitt) if it is the intersection of countably many open sets. For example, theirrational numbers form a Gδ subset of R. Generalizing the above construction,it can be shown that any Gδ subset of a complete metric space is topologicallycomplete. In particular, the topology on the irrational numbers can be given by acomplete metric. But the same is not true for the rational numbers, as will followfrom the Baire category theorem.

Remark A7.4. A countable union of closed sets is called a Fσ-set (French: ferme,somme). This is the only known example of dual mathematical concepts beingdescribed using dual languages.

Definition A7.5. Let X be a metric space. A mapping f : X → X is a (strict)contraction if there is a constant a < 1 such that d(f(x), f(x′)) 6 ad(x, x′) forall x, x′ ∈ X .

Note that a contraction must be continuous.

Theorem A7.6. (Banach) A contraction on a complete metric space has a uniquefixed point (a point x such that f(x) = x).

Proof. A fixed point is unique because if x, x′ are two such then d(x, x′) =d(f(x), f(x′)) 6 ad(x, x′), which implies d(x, x′) = 0.

To prove existence, start with any x0 ∈ X and define x1 = f(x0), x2 = f(x1)and so on. If d(x0, x1) = r then d(xn, xn+1) 6 anr and so

d(xn, xn+k) 6 (an + · · ·+ an+k−1)r 6anr

1− awhich tends to 0 as n→∞. Thus (xn) is a Cauchy sequence, which converges toa point x. We have

f(x) = lim f(xn) = lim xn+1 = x

so x is a fixed point.

Exercise A7.7. Banach’s fixed point theorem can be extended as follows: iff : X → X is a map and some power fN = f · · · f is a strict contraction, thenf has a unique fixed point. Prove this.

Banach’s fixed point theorem is one of the most important sources of existencetheorems in analysis. Another important source of existence theorems, which alsouses completeness, is the Baire category theorem.

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Definition A7.8. A subset of a metric space is dense if its closure is the wholespace

Example A7.9. The rational numbers Q are dense in R. (It is often important thatR has a countable dense subset. A space with this property is called separable.)

Theorem A7.10. (Baire) In a complete metric space, the intersection of countablymany dense open sets is dense.

Proof. Let x ∈ X and let ε > 0. Let Un, n = 1, 2, . . ., be dense and open.Choose x1 ∈ U1 ∩ B(x; ε/2). There is r1 such that B(x1; r1) ⊆ U1; without lossof generality take r1 < ε/4. Since U2 is dense there is x2 ∈ U2 ∩ B(x1; r1).Proceed inductively in this way choosing xn and rn such that

xn ∈ Un ∩B(xn−1; rn−1), B(xn; rn) ⊆ Un, rn < rn−1/2.

Then (xn) is a Cauchy sequence, say converging to a point x′. Since xn ∈B(xj; rj) for all j 6 n, the point x′ belongs to B(xj; rj) for every j, and hence to⋂Un. Moreover, d(x, x′) < ε. Since x and ε were arbitrary,

⋂Un is dense.

Remark A7.11. The name ‘category theorem’ comes from the following (tradi-tional) terminology. A set A is nowhere dense if the complement of its closure isdense. A set is of first category if it is the countable union of nowhere dense sets.Baire’s theorem then says that in a complete metric space the complement of a setof first category (sometimes called a residual set) is dense.

Remark A7.12. Notice that this gives us another proof that R is uncountable (com-pare Exercise A1.5). For the complement of a point in R is a dense open set. Moregenerally, any complete, perfect (Definition A4.14) metric space is uncountable.For in a perfect metric space, the complement of any point is dense.

Exercise A7.13. Consider the complete metric space CR[0, 1] of continuous real-valued functions on [0, 1]. For a, b ∈ [0, 1] show that the set of functions fwhich are nondecreasing on the interval [a, b] is nowhere dense in CR[0, 1]. UsingBaire’s theorem, deduce that there exist functions f ∈ CR[0, 1] that are nowheremonotonic, that is, monotonic on no subinterval of [0, 1].

23

Lecture A8Normed Spaces

Definition A8.1. Let V be a vector space (real or complex). A norm on V is afunction V → R, denoted v 7→ ‖v‖, such that

(a) ‖v‖ > 0 for all v, and ‖v‖ = 0 iff v = 0.

(b) ‖λv‖ = |λ|‖v‖, where λ is a scalar (real or complex).

(c) ‖v + v′‖ 6 ‖v‖+ ‖v′‖.

Clearly, d(v, v′) = ‖v − v′‖ is then a metric. Because of (b), this metric hasan ‘affine’ structure not present in a general metric.

Example A8.2. Let x = (x1, . . . , xn) be a vector in Rn or Cn. The expressions

‖x‖1 = |x1|+ · · ·+ |xn|

‖x‖2 =(|x1|2 + · · ·+ |xn|2

)1/2‖x‖∞ = max|x1|, . . . , |xn|

all define norms.

Exercise A8.3. Let a and b be positive real numbers and let p > 1. Show that

inft1−pap + (1− t)1−pbp : t ∈ (0, 1) = (a+ b)p.

Hence (or otherwise) show that the expression

‖x‖p = (|x1|p + · · ·+ |xn|p)1/p

is also a norm on Rn or Cn.

Example A8.4. If V = C(X), we may define

‖f‖ = sup|f(x)| : x ∈ X.

This is a norm, and it gives rise to the usual metric on C(X). Since it is complete(theorem A6.2) it is a Banach space as defined below.

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Definition A8.5. If a normed vector space is complete in its metric, it is called aBanach space.

Proposition A8.6. A linear mapping T : V → W between normed vector spacesis continuous if and only if there is a constant k such that

‖Tv‖ 6 k‖v‖

(one then says that T is bounded).

Proof. If T is bounded then ‖Tu− Tv‖ 6 k‖u− v‖ so T is continuous.If T is continuous then there is some δ > 0 such that ‖u‖ < δ implies ‖Tu‖ <

1. Take k = 1/δ.

Definition A8.7. The best constant k in the above proposition, that is the quantity

sup‖Tv‖ : ‖v‖ 6 1

is called the norm of T and denoted ‖T‖.

Exercise A8.8. Show that, with the above norm, the collectionL(V,W ) of boundedlinear maps from V to W is a normed vector space, and that it is a Banach spaceif W is a Banach space.

Exercise A8.9. Show that the norm of linear maps is submultiplicative under com-position, i.e. ‖S T‖ 6 ‖S‖‖T‖.

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Lecture A9Differentiation in Normed Spaces

The basic idea of differentiation is that of ‘best linear approximation’. Normedspaces provide a systematic way to express this.

Notation A9.1. Let f be a function defined on some ball B(0; ε) in a normedspace V and having values in another normed space W . We shall write “f(h) =o(‖h‖)” to mean that the limit limh→0

(‖h‖−1‖f(h)‖

)exists and equals zero. Sim-

ilarly for expressions like “f(h) = g(h) + o(h)”.

Exercise A9.2. Suppose that T : V → W is a linear map. Show that T (h) =o(‖h‖) if and only if T = 0.

Now let f : V → W be a continuous (need not be linear) map between normedvector spaces. In fact, we need only suppose f is defined on some open subset Uof V .

Definition A9.3. With above notation, let x ∈ V . We say that f is differentiableat x if there is a bounded linear map T : V → W such that

f(x+ h) = f(x) + T · h+ o(‖h‖).

By the exercise, T is unique if it exists. It is called the derivative of f at x andwritten Df(x).

Example A9.4. If V = W = R then every linear map V → W is multiplicationby a scalar, i.e. we identify L(V,W ) = R. Under this identification our definitionof the derivative corresponds to the usual one from Calculus I.

Example A9.5. If V = Rm and W = Rn then L(V ;W ) is the space of n × mmatrices. The matrix entries of the derivative of f are the partial derivatives of thecomponents of f as defined in Calculus III. Remark: The existence of the partialderivatives does not by itself imply differentiability in the sense of our definitionabove. This just shows that our definition is right and partials are wrong!

Example A9.6. If V = R and W is arbitrary, the space of (bounded) linear mapsL(V ;W ) can be identified with W itself. Under this identification the derivativeof f is given by the usual formula

Df(x) = limh→0

f(x+ h)− f(x)

h.

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Exercise A9.7. Show that the derivative of a linear map is always equal to themap itself.

Proposition A9.8. (Mean value theorem) Suppose that the function f is differ-entiable throughout a ball B(x0; r) ⊆ V , and that ‖Df(x)‖ 6 k for all x ∈B(x0; r). Then

‖f(x)− f(x0)‖ 6 k‖x− x0‖

for all x ∈ B(x0; r).

Proof. Let ε > 0. By definition of the derivative each y ∈ B(x0; r) has a neigh-borhood Uy ⊆ B(x0; r) such that

‖f(y′)− f(y)‖ 6 (k + ε)‖y′ − y‖ ∀y′ ∈ Uy.

Now we will “creep along” the line segment [x0, x] ⊆ B(x0; r). Let xt = (1 −t)x0 + tx, t ∈ [0, 1], and let T be the set of all those t ∈ [0, 1] such that ‖f(xs)−f(x0)‖ 6 (k+ε)‖xs−x0‖ for all s 6 t. Clearly, T is an interval and the displayedequation above shows that it contains [0, θ) for some θ > 0. Let τ = supT ∈(0, 1] and let y = xτ . If τ < 1 then there exists δ > 0 such that

‖f(xt)− f(y)‖ 6 (k + ε)‖xt − y‖ ∀t ∈ (τ − 2δ, τ + 2δ).

For any c ∈ [0, 1] we have τ − cδ ∈ T and so

‖f(xτ+cδ)− f(x0)‖ 6‖f(xτ+cδ)− f(xτ )‖+ ‖f(xτ )− f(xτ−cδ)‖+ ‖f(xτ−cδ)− f(x0)‖ 6

(k + ε) (‖xτ+cδ − xτ‖+ ‖xτ − xτ−cδ‖+ ‖xτ−cδ − x0‖) =

= (k + ε) (‖xτ+cδ − x0‖) .

The final equality is because the three vectors appearing here are positive multiplesof the same vector. If follows that τ+δ ∈ T , contradicting the definition of τ as thesupremum. Therefore it is impossible that τ < 1 and the theorem is proved.

Remark A9.9. From the proof one sees that the theorem is true if we replace theball by any region that is star-shaped about x0.

It is easy to see that the derivative of f ± g is Df ± Dg. Another familiarcalculus result that generalizes easily is the chain rule.

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Proposition A9.10. (Chain rule) Let V,W,X be normed vector spaces, and letx ∈ V . Let f : V → W be differentiable at x and let g : W → X be differentiableat y = f(x). Then g f is differentiable at x and

D(g f)(x) = Dg(y) Df(x).

Note that f need only be defined near x, and g need only be defined near y.

Proof. Let k(h) = f(x + h) − f(x) = Df(x) · h + o(‖h‖). Note that there is aconstant A such that ‖k(h)‖ 6 A‖h‖ for small ‖h‖. Now write

(g f)(x+ h)− (g f)(x) = g(y + k(h))− g(y)

= Dg(y) · k(h) + o(‖k(h)‖)= Dg(y) ·Df(x) · h+ o(‖h‖)

giving the result.

Remark A9.11. Note that the derivative of f ,Df , is itself a function having valuesin a normed space (namely the space L(V,W )). Thus we can differentiate it anddefine second and higher derivatives if we require.

Exercise A9.12. (Open-ended) Formulate precisely the advanced calculus resultthat “the mixed derivatives are symmetric”, that is ∂2f/∂x∂y = ∂2f/∂y∂x, andprove it (under suitable hypotheses) for maps between normed vector spaces.

Let F be a function R × V → V , where V is a normed vector space. Let[a, b] ⊆ R. A function f : [a, b] → V is a solution to the differential equationdefined by F if it is differentiable and f ′(t) = F (t, f(t)) for all t ∈ [a, b]. Thevalue f(a) ∈ V is called the initial condition for the solution.

Theorem A9.13. Suppose that V is a Banach space and suppose also that thefunction F is continuous and satisfies a Lipschitz condition, that is, there exists aconstant C > 0 such that

‖F (t, v)− F (t′, v′)‖ 6 C‖v − v′‖.

Then for any a ∈ R and any v0 ∈ V there exists a nontrivial interval [a, b] onwhich there is a solution to the differential equation f ′(t) = F (t, f(t)) with initialcondition f(a) = v0; moreover, this solution is unique.

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Proof. We need to use some facts about integration for V -valued functions. Itisn’t the job of this course to teach you about integration theory — for that seeMath 501 — so I will just state the various facts that are needed. Moreover, in afirst course on integration these facts are usually proved only for real or complexvalued functions; you will need to take it on faith that the same facts are true forfunctions having values in a normed vector space V . Finally, we shall use thefact that if V is a Banach space (that is, complete) then the space of continuousfunctions [a, b]→ V , equipped with the sup norm, is also complete. This is a factthat we have already proved for real or complex valued functions, and the proofin the general case is exactly the same.

We will use the fundamental theorem of calculus: a function f ∈ C[a, b] is asolution of the differential equation if and only if it is a solution of the equivalentintegral equation

f(s) = v0 +

∫ s

a

F (t, f(t)) dt, t ∈ [a, b].

We will show that a solution to this equation exists (for b sufficiently close to a)by applying Banach’s contraction mapping theorem (A7.6) to the space C[a, b] ofcontinuous functions on [a, b] (with values in V , if you like). Let I : C[a, b] →C[a, b] be defined by the right hand side of the display above, that is,

(I f)(s) = v0 +

∫ s

a

F (t, f(t)) dt, t ∈ [a, b].

If f, g ∈ C[a, b] then the integrands in I f and I g differ by C‖f − g‖ at most,where C is the Lipschitz constant, and therefore

‖I f −I g‖ 6 C(b− a)‖f − g‖

by standard estimates on integrals. Therefore, if (b − a) < C−1, the map I is astrict contraction and Banach’s fixed point theorem provides a unique solution toI f = f , which is a solution to the differential equation with the specified initialcondition.

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Lecture A10The inverse function theorem

Let V, V ′ be normed vector spaces. The space L(V, V ′) is then a normedvector space also. Say T ∈ L(V, V ′) is invertible if it has an inverse which is abounded linear map from V ′ to V .

Proposition A10.1. Let V, V ′ be Banach spaces. Then the invertibles form anopen subset of L(V, V ′), and the inverse operation T 7→ T−1 is continuous (onthis subset) from L(V, V ′) to L(V ′, V ).

Proof. Without loss of generality V ′ = V . Look first at a nbhd of the invertibleoperator I . Suppose ‖S‖ < 1. For y ∈ V consider the map

φy : V → V, v 7→ y − Sv.

Since ‖S‖ < 1 this map is a strict contraction, so it has a unique fixed point x(Theorem A7.6). This fixed point is an x such that (I + S)x = y. By the triangleinequality,

‖y‖ > (1− ‖S‖)‖x‖, so ‖x‖ 6 (1− ‖S‖)−1‖y‖

and the map y 7→ x is bounded. We have shown that if ‖S‖ < 1, I+S is invertible.Moreover

‖y − (I + S)−1y‖ = ‖S(1 + S)−1y‖ 6 ‖S‖(1− ‖S‖)−1‖y‖

which shows that the map S 7→ (I + S)−1 is continuous at S = 0.We have shown that the identity is an interior point of the set of invertibles

and that the inverse is continuous there. However, left multiplication by a fixedinvertible is a homeomorphism from the set of invertibles to itself; so the sameresults apply to any point of the set of invertibles.

Exercise A10.2. Let V be a normed space and letW be the normed spaceL(V, V ).Show that the mapping i : T 7→ T−1 is differentiable (where defined) on W , andthat its derivative is

Di(T ) ·H = −T−1HT−1.

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Definition A10.3. A map f from an open subset of a normed space V to a normedspace W is continuously differentiable or C1 if it is differentiable (everywhere)and the map x 7→ Df(x) is continuous.

The inverse function theorem says that under suitable conditions, if Df(x) isinvertible then f is ‘locally invertible’ near x.

Theorem A10.4. Let f be as above, defined near x ∈ V , and suppose that fis continuously differentiable and that Df(x) ∈ L(V,W ) is invertible. Supposemoreover that V,W are Banach spaces. Then there is an open set U containing xsuch that f is a bijection of U onto f(U), which is an open set containing f(x),and such that its inverse g : f(U)→ U is also continuously differentiable.

Proof. It is an application of Banach’s fixed point theorem A7.6. We will con-struct the inverse function by looking at the fixed points of a suitable map.

Let A = Df(x)−1. For y ∈ W define a map φy : V → V by

φy(z) = z + A · (y − f(z)).

A fixed point of φy is a solution to f(z) = y. We have

Dφy(z) = I − A Df(z) = A (Df(x)−Df(z)

).

Since Df is continuous there is r > 0 such that if z ∈ U = B(x; r) then‖Dφy(z)‖ < 1

2. From the mean value theorem we conclude that if F is a closed

subset of U , and if we know for some reason that φy maps F to F , then φy isactually a contraction of this complete metric space and so has a unique fixedpoint.

Fix z0 ∈ U and let y0 = f(z0). Thus y0 ∈ f(U) and we want to provethat there is some ε > 0 such that B(y0; ε) ⊆ f(U); this will show that f(U) isopen. Choose δ > 0 such that the closed ball B(z0; δ) is contained in U . Chooseε = 1

2‖A‖−1δ. If z ∈ B(z0; δ) and y ∈ B(y0; ε) then we may compute

‖φy(z)− z0‖ = ‖φy(z)− φy0(z0)‖6 ‖φy(z)− φy(z0)‖+ ‖φy(z0)− φy0(z0)‖

6 12‖z − z0‖+ ‖A(y − y0)‖ 6 1

2δ + 1

2δ = δ

using the mean value theorem A9.8. We conclude that, for these y, the map φy isa contraction of the complete metric space B(z0; δ) and thus has a (unique) fixedpoint there. We have shown that each y0 ∈ f(U) has an ε-neighborhood contained

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in f(U); thus, f(U) is open. The contraction property of the φy ensures that eachy ∈ f(U) has only one inverse image in U ; thus f is a bijection of U onto f(U).

Now to show that the inverse function g = f−1 : f(U) → U is differentiableat y = f(x) write g(y + h) − g(y) = u(h), say. The calculations of the previousparagraph show that ‖u(h)‖ 6 2‖A‖‖h‖ for ‖h‖ small. Then

h = f(g(y + h))− y = f(g(y) + u(h))− y = Df(x) · u(h) + o(‖h‖).

Apply A = Df(x)−1 to get

u(h) = A · h+ o(‖h‖).

This gives differentiability of g with Dg(f(x)) = Df(x)−1. Continuous differen-tiability now follows from the continuity of the inverse operation on the space ofbounded linear maps (Proposition A10.1).

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Lecture A11Topological Spaces

We have been emphasizing that “up to homeomorphism” properties of metricspaces depend only on their open sets. This motivates a further abstraction whichremoves the numerical notion of metric entirely.

Definition A11.1. A topology on a set X is a family T of subsets of X with thefollowing three properties:

(i) If Uα is any family of members of T , then the union⋃α Uα is also a member

of T ;

(ii) IfU1, . . . , Un is a finite family of members of T then the intersection⋂ni=1 Ui

is also a member of T ;

(iii) The sets ∅ and X are members of T .

A set equipped with a topology is called a topological space. The members of Tare called open subsets of the topological space X .

Notice that the open subsets of a metric space automatically satisfy (i)–(iii);thus, every metric space carries a topology (called its metric topology). Differentmetrics can however give rise to the same topology (as we have already seen);moreover, there are topologies that do not arise from any metric.

Example A11.2. The discrete topology on X has T = P(X): every subset isopen. This topology arises from the discrete metric.

Example A11.3. The indiscrete topology on X has just two open sets, ∅ and X .This topology does not arise from a metric as soon as X has more than one point.(Why not?)

Example A11.4. The cofinite topology on X consists of ∅ together with all thecofinite subsets of X (a subset is cofinite if its complement is finite).

It is natural to say that a closed set is one whose complement is open, so thatthe closed sets in the cofinite topology are just the finite sets together with thewhole space. Notice that this topology has many “fewer” open (or closed) setsthan the topologies with which we are familiar. For example, in the cofinite topol-ogy on an infinite set, any two nonempty open sets have a nonempty intersection.A similar but more sophisticated example is the next one.

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Example A11.5. LetK be an algebraically closed field (for example, the complexnumbers) and let A = Kn. A subset V of A is called an affine algebraic variety ifthere is a set S ⊆ K[X1, . . . , Xn] of polynomials in n variables such that

(x1, . . . , xn) ∈ S ⇔ p(x1, . . . , xn) = 0∀p ∈ S;

in other words, a variety is the set of common zeroes of a collection of polynomi-als. (The Hilbert basis theorem says that S can always be taken to be finite.) It isa simple exercise (for an algebraist!) to show that any intersection of varieties is avariety, and a finite union of varieties is a variety. In other words, the varieties canbe considered as the closed sets of a topology on A, called the Zariski topology.

Example A11.6. Let T be a family of topologies on the set X . Then⋂

T isalso a topology on X . In particular, let F be any family of subsets of X at all.Let T be the family of all topologies T such that T ⊇ F (there always is onesuch topology, the discrete topology). The intersection

⋂T is then the smallest

or “weakest” topology containing F . It is called the topology generated by F(you’ll also see F called a subbasis for T ).

Exercise A11.7. Show that the topology on X generated by F can be concretelyexpressed as follows: a set U is open in the generated topology if and only if itcan be written as a union

⋃Uα of some family of sets Uα, where each Uα is the

intersection of some finite family (depending on α) of members of F . (The emptyset and the whole space X are included among these sets by convention: we takethe intersection of an empty family to be X and the union of an empty family tobe ∅.)

Remark A11.8. The following language is conventional in this context: the fam-ily B of subsets of X is called a basis if, for every point x ∈ X , there is somemember of B containing x and every finite intersection of members of B con-taining x includes some member of B containing x. Thus the topology generatedby a basis is just the collection of all unions of families of members of the basis.(For example, the metric balls form a basis for the topology of a metric space.)Moreover, the topology generated by a subbasis F is the topology generated bythe associated basis consisting of finite intersections of members of F .

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Lecture A12Compactness and Hausdorffness

Definition A12.1. LetX be a topological space, with topology T , and let Y ⊆ X .Then the family of subsets of Y given by Y ∩ U : U ∈ T defines a topologyon Y . It is called the relative topology or subspace topology induced by the giventopology on X .

(Compare Proposition A3.10.)Any metric space notion which is defined simply in terms of open sets can

be taken over without change to the more general context of topological spaces.Thus, for example, we can define the interior, closure, and boundary of a subsetjust as in the metric space context.

Exercise A12.2. Let X be a topological space and Y a subspace of X , equippedwith the subspace topology. Let S be a subset of X , with closure S. Show thatthe closure of S ∩ Y in the subspace topology of Y is a subset of S ∩ Y , and thatthese two sets are equal if S itself is a subset of Y ; but that they need not be equalin general.

Just as in metric spaces we can also define a continuous function f : X → Y(X and Y being topological spaces) as one which has the property that, for everyopen subset U of Y , the inverse image f−1(U) is open in X . The proof of thefollowing lemma is then obvious:

Lemma A12.3. The composite of continuous functions is continuous.

Exercise A12.4. Let f : X → Y be a map of spaces (from now on, “space” willmean a topological space, unless otherwise specified). Show that f is continuousas a map from X to Y if and only if it is continuous as a map from X onto f(X),where f(X) ⊆ Y is equipped with its subspace topology.

Definition A12.5. A homeomorphism is a continuous bijection whose inverse isalso continuous.

Definition A12.6. A space X is compact if every open cover of X has a finitesubcover.

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This is the same as the definition of “covering compact” that was given inLecture 5. We won’t make use of the notion “sequentially compact” outside therealm of metric spaces.

Example A12.7. Any finite topological space is compact. A discrete topologicalspace is compact if and only if it is finite.

Proposition A12.8. Let f : X → Y be a continuous map. If X is compact, thenf(X) ⊆ Y is also compact.

Proof. By Exercise A12.4 it suffices to consider the case where f(X) = Y . LetU be an open cover of Y . The sets f−1(U), U ∈ U , are then open in X andthey form a cover since for each x ∈ X , the image f(x) belongs to some memberof U . Since X is compact, there is a finite subcover: there is a finite subsetU1, . . . , Un of U such that the sets f−1(Ui) cover X . But for every y ∈ Y , thereis x ∈ X such that f(x) = y; so x ∈ f−1(Ui) for some i and therefore y ∈ Ui.That is, the U1, . . . , Un cover Y and we have shown that U has a finite subcoveras required.

Proposition A12.9. Let X be a compact space and let Y be a closed subset of X .Then Y is also compact (in its relative topology).

Proof. We must show that every open cover of Y (in the relative topology) has afinite subcover. Let U be such a cover. Each open set U ∈ U is (by definitionof the relative topology) of the form VU ∩ Y , where VU is some open subset ofX . The union of all the sets VU includes Y ; therefore, the open sets VU , togetherwith the open set X \ Y , form an open cover of X . By hypothesis this cover hasa finite subcover, comprising VU1 , . . . , VUn , together possibly with X \ Y . ThenU1, . . . , Un cover Y .

In metric spaces, compact subsets are necessarily closed (Proposition A5.5).This need not be true in general topological spaces. Indeed, let X be any topo-logical space. From Example A12.7, any finite subset of X — in particular, anyone-point subset — must be compact in its subspace topology. But we saw severalexamples in the last lecture of topological spaces in which one-point subsets arenot closed.

What is missing is the Hausdorff property.

Definition A12.10. A topological space X is Hausdorff if, for any two distinctpoints x, y ∈ X , there exist open sets U, V with x ∈ U , y ∈ V , and U ∩ V = ∅.

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Any metric space is Hausdorff (take U = B(x, ε) and V = B(y, ε) for 0 <ε < d(x, y)/2). The Hausdorff property ensures a large supply of open sets.

Proposition A12.11. Let X be a Hausdorff space and let Y ⊆ X be compact (inthe subspace topology). Then Y is a closed subset of X .

Proof. We will show that X \ Y is open. Let x ∈ X \ Y . For each y ∈ Y thereexist disjoint open sets Uy containing x and Vy containing y, by the Hausdorffproperty. The sets Vy ∩ Y form an open cover of Y (in the subspace topology)which therefore has a finite subcover, say Vy1 ∩ Y, . . . , Vyn ∩ Y . Let Ux = Uy1 ∩· · ·∩Uyn . Then Ux is an open set, contains x, and does not meet Y . The set X \Yitself is the union of all these open sets Ux, so it is open.

In particular, a one-point subset of a Hausdorff space is closed.

Corollary A12.12. Let f : X → Y be a continuous bijection, where X is a com-pact space and Y is a Hausdorff space. Then f−1 is continuous (i.e., f is ahomeomorphism).

Proof. To show that f−1 is continuous it is enough to show that for any closedsubset K of X , the image f(K) is closed in Y . But K is compact (Proposi-tion A12.9); therefore f(K) is compact (Proposition A12.8); therefore f(K) isclosed (Proposition A12.11).

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Lecture A13Regular and Normal Spaces

Let A and B be disjoint subsets of a topological space X . One says that A andB are separated by open sets if there exist disjoint open sets U and V such thatA ⊆ U , B ⊆ V . Thus the Hausdorff property says that any two distinct pointscan be separated by open sets.

This is in fact just one (the most important one) of a series of separation prop-erties that a topological space may have.

Definition A13.1. A Hausdorff topological space X is regular if disjoint subsetsA and B of X can be separated by open sets whenever A is a point and B is aclosed set. It is normal if any two disjoint closed subsets can be separated by opensets.

Remark A13.2. A regular Hausdorff space is also called a T3 space and a normalHausdorff space is a T4 space; a plain vanilla Hausdorff space is a T2 space. TheT stands for the German Trennungsaxiom (separation axiom). There are also T0,T1, T2 1

2, T3 1

2, T5, and T6 spaces, and maybe some others that I forgot.

Example A13.3. Every metric space is regular, and indeed normal. To see this weneed to make use of the distance function from a closed set. Let (X, d) be a metricspace and let K be a closed subset. Define a real valued function dK : X → R+

bydK(x) = infd(x, k) : k ∈ K.

It is easy to see that dK is a continuous function and that dK(x) = 0 iff x ∈ K.Now letA andB be disjoint closed sets inX and consider the continuous functionf(x) = dB(x)− dA(x) on X . The sets U = f−1((0,∞)) and V = f−1((−∞, 0))are open (by continuity), disjoint, and A ⊆ U , B ⊆ V . This shows that X isnormal.

Proposition A13.4. Every compact Hausdorff space is regular, and indeed nor-mal.

Proof. It’s best to do this proof in two installments: first regularity, then normality.Let X be a compact Hausdorff space, p a point of X , and B a closed subset notcontaining p. Then B is compact (Proposition A12.9). For each x ∈ B there

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are disjoint open sets Ux containing p and Vx containing x. The Vx form an opencover of B, so there is a finite subcover, say Vx1 , . . . , Vxn . Then

Up = Ux1 ∩ · · · ∩ Uxn , V p = Vx1 ∪ · · · ∪ Vxnare disjoint open sets, the first containing p, the second including V . Thus X isregular.

To prove normality, we iterate the argument. Let A and B be disjoint closed(hence compact) sets. By regularity, for each p ∈ A there are disjoint open Up

containing p and V p including B. The Up form an open cover of A and have afinite subcover say Up1 , . . . , Upn . Then

U = Up1 ∪ · · · ∪ Upn , V = V p1 ∩ · · · ∩ V pn

are disjoint open sets including A and B respectively.

In Example A13.3 we used suitably chosen continuous functions to establishregularity, and indeed normality, for metric spaces. It’s also possible to proceedin the other direction: starting with normality, construct a supply of continuousreal-valued functions.

Lemma A13.5. Let X be a normal Hausdorff space, F a closed subset of X , andW an open subset of X including F . Then there is an open subset V of X suchthat F ⊆ V ⊆ V ⊆ W .

Proof. Let E = X \W . The closed sets E and F are disjoint, so by normalitythere are disjoint open sets U and V with E ⊆ U and F ⊆ V . Since V ⊆ X \ Uwhich is closed, V ⊆ X \U also. In particular, V doesn’t meetE, so it is includedin W .

Theorem A13.6. (Urysohn’s lemma) Let E,F be disjoint closed subsets of a nor-mal Hausdorff space X; then there exists a continuous function f : X → [0, 1]such that f(x) = 0 for x ∈ E, f(x) = 1 for x ∈ F .

Warning: It is not asserted that the sets E,F are exactly equal to f−10,f−11 respectively; in general this stronger statement is not true. We will use thefollowing exercise.

Exercise A13.7. In order that a function f : X → R be continuous, it is necessaryand sufficient that for every x ∈ X and every ε > 0 there exists an open set Wx

containing x such that

f(Wx) ⊆ (f(x)− ε, f(x) + ε).

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Proof. It’s an induction based on lemma A13.5. Set U1 = X \ F ; then U1 isan open set containing E. We are going to construct by induction a collectionof open sets Ut parameterized by dyadic rational numbers t ∈ (0, 1], having theproperties that Ur ⊆ Us whenever r < s and E ⊆ Ut for all t. At the nth stageof the induction we assume that the sets Ut have been constructed for t = m2−n,m = 1, . . . , 2n. Let us make the convention that U0 refers to the closed set E(even though we have not defined a set U0). By lemma A13.5, for each m there isan open set V such that

Um2−n ⊆ V ⊆ V ⊆ U(m+1)2−n ;

call the set V so defined U(m+ 12)2−n . We have now defined the sets Ut for all

t = m′2−(n+1) and the induction continues.Now define f : X → [0, 1] by f(x) = inft : x ∈ Ut ∪ 1. Clearly f

maps X → [0, 1] with f(E) = 0 and f(F ) = 1, so we must show that f iscontinuous, and we will use Exercise A13.7. Let x ∈ X with f(x) = t; supposethat 0 < t < 1 (the cases t = 0 and t = 1 are handled by similar arguments). Letε > 0. There exist dyadic rationals r and s such that

t− ε < r < t < s < t+ ε.

Put Wx = Us \ Ur. Then Wx is an open set and f(Wx) ⊆ [r, s] ⊆ (t− ε, t+ ε) asrequired.

Of course, [0, 1] can be replaced in the statement of Urysohn’s lemma by anyclosed interval [a, b].

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Lecture A14Spaces of Continuous Functions

Let X be a set, Y ⊆ X , and let f : Y → Z be a function. A function g : X →Z is an extension of f if g(y) = f(y) for all y ∈ Y ; in other words, if f = g|Y .WhenX, Y, Z are topological spaces we will naturally be interested in continuousf and g.

Theorem A14.1. (Tietze extension theorem) Let X be a normal Hausdorff space,and let Y be a closed subspace. Let f : Y → R be a bounded continuous func-tion. Then f can be extended to a bounded continuous function g : X → R, withsup |g| = sup |f |.

Proof. Let us temporarily call a bounded continuous function h : Y → R ex-tendible if it extends to a bounded continuous function X → R with the samesup norm. What we have to prove is that all bounded continuous functions areextendible.

If h : Y → [−c, c] is a continuous function, let E and F be the disjoint subsetsh−1([−c,−c/3]) and h−1([(c/3, c]) of Y . They are closed in Y (by continuity),and Y is closed inX , so they are closed inX (Exercise A12.2). Thus by Urysohn’slemma there is a continuous function k : X → [−c/3, c/3] with k(E) = −c/3,k(F ) = c/3. Notice that the supremum norm ‖h−k‖ is at most 2c/3. We haveproved the following statement: for each bounded continuous h : Y → R thereexists an extendible bounded continuous k : Y → R with ‖k‖, ‖h−k‖ 6 2‖h‖/3.

Now we proceed inductively. Let f : Y → R be given and suppose inductivelythat extendible functions g1, . . . , gn have been constructed with ‖gi‖ 6 (2/3)i‖f‖and

‖f − g1 − . . .− gn‖ 6 (2/3)n‖f‖

. Apply the italicized claim to h = f − g1 − . . .− gn and let gn+1 be the k that isconstructed. The function gn+1 then has the desired properties so the induction iscompleted.

By the Weierstrass M-test the series∑∞

i=1 gi converges uniformly on X to afunction g. But the display above shows that, when restricted to Y , f = g. Thusf is extendible as required.

Recall that a subset of a topological space is dense if its closure is the wholespace. A classical method in analysis is to prove a result first for a dense subset ofsome space, and then to extend it to the whole space ‘by continuity’.

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How can we find dense subsets of C(X), the continuous real-valued functionson a compact Hausdorff space X?

Definition A14.2. A collectionL of continuous real-valued functions on a setX isa lattice if, whenever it contains functions f and g, it also contains their pointwisemaximum and minimum — usually written f ∨ g and f ∧ g in this context.

Proposition A14.3. (Stone) Let L be a lattice of continuous functions on a com-pact space X . If, for all x, x′ ∈ X and any a, a′ ∈ R, there is a function f ∈ Lhaving f(x) = a and f(x′) = a′, then L is dense in C(X).

The property appearing in the statement is called the two point interpolationproperty.

Proof. Let h ∈ C(X) be given and let ε > 0. We are going to approximate hwithin ε by elements of L. By hypothesis, for each x, x′ ∈ X there exists fxx′ ∈ Lsuch that fxx′(x) = h(x) and fxx′(x′) = h(x′). Fixing x for a moment, let

Vx′ = y ∈ X : h(y)− ε < fxx′(y).

These sets are open (why?) and x′ ∈ Vx′ so they cover X . Take a finite subcoverand let gx be the (pointwise) maximum of the corresponding functions fxx′ ∈ L.Because L is a lattice, gx ∈ L and by construction,

h(y)− ε < gx(y) ∀y, h(x) = gx(x).

So we have approximated h from one side by members of L.Now we play the same trick again from the other direction: let

Wx = y : gx(y) < h(y) + ε.

Again these form an open cover of X; take a finite subcover and let g be the(pointwise) minimum of the corresponding gx. Then g ∈ L and by constructionh− ε < g < h+ ε, as required.

Exercise A14.4. Use Stone’s theorem to give another proof of the Tietze extensiontheorem in the case of compact Hausdorff spaces. (Show that the collection ofextendible functions is a lattice. . . )

There is a more classical formulation which makes use of algebraic rather thanorder-theoretic operations.

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Lemma A14.5. There is a sequence of polynomials pn(x) on [−1, 1] converginguniformly to |x|.

Proof. Writing |x| =√

1 + (x2 − 1), this follows from the fact that the binomialseries for (1 + t)1/2 converges uniformly for t ∈ [−1, 1].

One says that a subset of C(X) is a subalgebra if it is closed under pointwiseaddition, subtraction, multiplication of functions, and multiplication by scalars.

Lemma A14.6. A closed subalgebra of CR(X) that contains the constant func-tions is a lattice.

Proof. Let A be the given subalgebra and f, g ∈ A; let h = f − g. There isno loss of generality in assuming (by rescaling) that |h| 6 1 everywhere. Anypolynomial in h belongs to A and hence so does |h| = lim pn(h), by closure andlemma A14.5. But now

f ∧ g =f + g − |h|

2, f ∨ g =

f + g + |h|2

belong to A as well.

A subalgebra of C(X) is said to separate points if for every x, x′ ∈ X there isa function in the subalgebra taking different values at x and at x′.

Theorem A14.7. (Stone-Weierstrass) A subalgebra of CR(X) which contains theconstants and separates points is dense.

Proof. The closure of the subalgebra is a closed subalgebra, contains the con-stants, and separates points. Using the algebraic operations, it has the two pointinterpolation property. The previous lemma shows that it is a lattice. Hence byStone’s theorem A14.3 it is all of C(X).

The original result of Weierstrass was

Corollary A14.8. (Weierstrass) Every continuous function on a closed boundedinterval is the uniform limit of polynomials.

Proof. The polynomials form an algebra which contains the constants and sepa-rates points.

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Exercise A14.9. In the complex case the Stone-Weierstrass theorem takes thefollowing form: a ∗-subalgebra of C(X) which contains the constants and sepa-rates points is dense, where the ∗-condition means that the algebra is closed under(pointwise) complex conjugation. Prove this. Try to give an example to show thatthe complex theorem is not valid without the extra condition (we shall discuss thisin detail later).

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Lecture A15Connectedness

What does it mean that a topological space is “all in one piece”?

Proposition A15.1. Let X be a topological space. The following properties of Xare equivalent:

(a) The only subsets of X that are both open and closed are the empty set and Xitself.

(b) The only continuous functions from X to a discrete topological space areconstant.

(c) The only equivalence relation on X with open equivalence classes is the triv-ial one (every point is equivalent to every other point).

Proof. It is clear that (b) and (c) are equivalent, since if f : X → D is a continuousfunction to a discrete space, then the relation “x ∼ y iff f(x) = f(y)” is anequivalence relation with open equivalence classes, and conversely.

Assuming (c), ifA is an open-and-closed subset ofX , then the relation “x ∼ yiff either none, or both, of x, y belong to A” is an equivalence relation whoseequivalence classes A and X \ A are open. Conversely, assuming (a), let ∼ bean equivalence relation with open equivalence classes, and let A be a nonemptyequivalence class. A is open; but, since the complement of A is a union of equiva-lence classes and therefore open, it is also true that A is closed. Thus A = X andthe equivalence relation is trivial.

Definition A15.2. A topological space is connected if it satisfies the equivalentproperties of Proposition A15.1. A subset of a topological space is called a con-nected subset if it is a connected space in the relative topology. A space (or subset)that is not connected is called disconnected.

Proposition A15.3. Let f : X → Y be continuous and surjective. If X is con-nected, then Y is connected.

Proof. Let g : Y → D be a continuous map to a discrete space D. Then f g : X → D is continuous, hence constant because X is connected. But now sincef is surjective, g must be constant itself.

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Proposition A15.4. Let X be a space and let Xαα∈A be a family of subsetseach of which is connected. If

⋂αXα 6= ∅ then

⋃αXα is connected.

Proof. Let Y =⋃αXα and let p ∈

⋂αXα. Let f : Y → D be a continuous map

to a discrete space. Since each Xα is connected, the restriction of f to Xα takesa constant value (say dα ∈ D); since p ∈ Xα for every α, dα = f(p) for each α.Thus f takes the constant value f(p) on Y .

Proposition A15.5. Every interval in R is connected, and every connected subsetof R is an interval.

Proof. We begin by proving that every compact interval [a, b] is connected. Sup-pose that ∼ is an equivalence relation with open equivalence classes; then theequivalence classes form an open cover U for the compact interval [a, b]. By theLebesgue covering theorem (Theorem A5.8)there is δ > 0 such that any ball ofradius less than δ is included in some member of the cover. In particular, any twopoints that are less than δ apart belong to the same equivalence class. But anypair of points in the interval [a, b] can be joined by a chain of points whose suc-cessive members are less than δ apart; so, any two points in [a, b] are equivalent.This proves the connectedness of closed intervals; other kinds of intervals can bewritten as the increasing union of suitable families of closed intervals, so we mayapply Proposition A15.4.

Conversely, let S be a subset of R that is not an interval. Then there is somea /∈ S such that S has members which are less than a and also members that aregreater than a. So, S ∩ (−∞, a) and S ∩ (a,∞) are disjoint open subsets of Swhose union is S; so S is not connected.

Corollary A15.6. (Intermediate value theorem) Let f : [a, b]→ R be continuous.Then the range of f includes all values between f(a) and f(b).

Proof. The range of f , being the image of a connected space under a continuousfunction, is connected. Hence it is an interval. It must contain f(a) and f(b) so,being an interval, it must also contain everything in between.

If X is a space, the relation “x ∼ y iff there is a connected subset of Xcontaining both x and y” is an equivalence relation (by Proposition A15.4). Theequivalence classes are called the connected components of X . The connectedcomponent of x is the largest connected subset of X that contains x.

Exercise A15.7. Show that the closure of a connected subset of X is connected,and deduce that the connected components of X are closed. Give an examplewhere they are not open.

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Exercise A15.8. The quasicomponent of x ∈ X is the intersection of all theopen-and-closed subsets that contain x. Show that the connected component of xis included in the quasicomponent. Can you give an example where they are notequal?

Definition A15.9. A path in a topological space X is a continuous map [0, 1] →X . The start of the path is the point γ(0) and the end of the path is the point γ(1).We say that γ is a path from its start to its end.

Lemma A15.10. Let X be a topological space, γ1 a path in X from a to b, γ2 apath in X from b to c. Then the map γ : [0, 1]→ X defined by

γ(t) =

γ1(2t) (t 6 1

2)

γ2(2t− 1) (t > 12)

is a path in X from a to c.

Proof. All that needs to be checked is the continuity of γ at t = 12. Let U be

an open set in X containing b. By continuity of γ1, there is δ1 > 0 such that if|s−1| < δ1, s ∈ [0, 1] then γ1(s) ∈ U ; by continuity of γ2 there is δ2 > 0 such thatif |s| < δ2, s ∈ [0, 1] then γ2(s) ∈ U . Now let δ = 1

2minδ1, δ2; if t − 1

2| < δ,

t ∈ [0, 1], then γ(t) ∈ U . This gives the required continuity.

Corollary A15.11. The relation “There is a path in X from a to b” is an equiva-lence relation.

Proof. The lemma gives transitivity. Reflexivity and symmetry are easy.

The equivalence classes under this equivalence relation are called path com-ponents of X .

Proposition A15.12. Every path component is connected. Consequently, the pathcomponent of x ∈ X is included in the component of x.

Proof. By Proposition A15.3, the image of a path from a to b is a connected setcontaining a and b. Consequently, the path component of X is the union of afamily of connected sets all of which contain x; so it is connected by Proposi-tion A15.4.

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A space which has only one path component (i.e., any two points can be joinedby a path) is called path connected. By the preceding proposition, a path con-nected space is connected. The converse is false, however, as is shown by thefollowing classical example.

Exercise A15.13. Let Z be the set of points (x, y) ∈ R2 such that either x = 0and y ∈ [−1, 1] or 0 < x 6 1 and y = sin(1/x). Show that Z is a compactconnected metric space with 2 path components.

The same example shows that, in contrast to components, path componentsneed not be closed in general.

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Lecture A16Local Properties of Spaces

Let X be a topological space.

Definition A16.1. An open neighborhood of x ∈ X is an open subset of X thatcontains x. A neighborhood of x ∈ X is a subset that includes an open neighbor-hood of x, i.e. contains x in its interior.

Suppose that P is some kind of topological property (e.g. connectedness).

Definition A16.2. We say that a space X has the property P locally if, for everyx ∈ X , every open neighborhood of x includes another open neighborhood of xthat has property P .

For instance, a space X is locally connected if every open neighborhood ofeach x ∈ X includes an open connected neighborhood. Informally we may saythat x has “arbitrarily small connected neighborhoods”.

Proposition A16.3. In a locally connected space, the connected components areopen.

(It follows that the connected components are equal to the quasicomponents,see Exercise A15.8.)

Proof. Let C be a connected component and let x ∈ C. There is a connectedneighborhood N of x. Then N ⊆ C and so N ⊆ C. Since x ∈ N, x is aninterior point of C. Since x was arbitrary, C is equal to its own interior and so isopen.

Proposition A16.4. In a locally path-connected space, the connected componentsare equal to the path components (and are open).

Proof. The same argument as in the previous proposition shows that the path com-ponents in a locally path-connected space are open. Since the complement ofeach path component is a union of other path components, they are also closed.Now the connected component of x must be included in any open-and-closed setthat contains x, so in particular it must be included in the path component of x.The path component is always included in the connected component, so they areequal.

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The most important example of a locally connected and locally path-connectedspace is furnished by an open subset of Rn (or any normed vector space).

Exercise A16.5. There is a weaker local path connectedness notion, sometimescalled semilocal path connectedness: X is semilocally path connected if for eachx ∈ X and each open neighborhood V of x, there is an open neighborhood U ⊆ Vof x such that for each y ∈ U there is a path γ : [0, 1] → V starting at x andending at y. Show that a space which is locally connected and semilocally pathconnected must be locally path connected. Can you give an example of a spacewhich is semilocally path connected but not locally path connected?

We now turn to consider local compactness. Here it is necessary to modifythe definition somewhat: we say that a space is locally compact if every openneighborhood of each point contains a compact neighborhood (which need not beopen - there may be very few compact, open sets). But in fact there is a simpler,equivalent definition.

Proposition A16.6. A Hausdorff space is locally compact iff each point has acompact neighborhood.

Proof. Local compactness says that each point has “arbitrarily small” compactneighborhoods, a stronger condition than that of the proposition. Suppose that x ∈X has a compact neighborhood K, and let N be any neighborhood of x. We needto find a compact neighborhood that is included in N . Let U = K ∩ N. ThenU is a neighborhood of x, and U and ∂U are closed subsets of a compact space(namely K), hence are compact. By regularity (or, more exactly, by the proof ofregularity) there exists an open subset V of U , containing x, whose closure doesnot meet the compact set ∂U . Then V ⊆ U is the desirec compact neighborhood.

Let X be a compact Hausdorff space and x0 ∈ X . Then X \ x0 is a locallycompact Hausdorff space (since each point of X distinct from x0 has a closed,hence compact, neighborhood that does not meet x0). In fact, every locally com-pact Hausdorff space arises in this way.

Proposition A16.7. Let X be a locally compact Hausdorff space. Let X+ be theunion of X and a disjoint point, denoted∞. The following sets form a topologyon X+:

(a) The open subsets of X in its original topology.

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(b) The subsets of the form ∞ ∪ (X \K) where K ⊆ X is compact.

The topology so defined on X+ is compact and Hausdorff and the identity mapX → X+ is a homeomorphism onto its image.

The space X+ is called the one-point compactification of X . We will need thefollowing easy exercise.

Exercise A16.8. Let Y be a Hausdorff space. If K,L are compact subspaces ofY then so is K ∪ L.

Proof. Let T be the family of subsets of X+ described in the proposition. Weshow it is a topology.

(i) Let Uα be a family of members of T . If they are all of type (a), then theirunion is an open subset of X and is of type (a) also. If one of them is of type(b), say equal to ∞∪(X\K), then their union is of the form ∞∪(X\L)where L is a closed subset of K, hence compact; so the union is of type (b).

(ii) Let U1, . . . , Un be a finite family of members of T . If one of them is oftype (a), then their intersection if of type (a). If they are all of type (b), theirintersection is the complement of a finite union of compact sets, which iscompact by Exercise A16.8 above. Thus the intersection id of type (b).

(iii) Clearly ∅ and X+ belong to T .

Thus T is a topology, and the relative topology that it induces on X is the sameas the original topology of X . It remains to show that this topology is compactand Hausdorff. To show that T is compact, let W be a family of members ofT covering X+. One of them (say W0) must contain ∞, and hence must be ofthe form ∞ ∪ (X \ K), K ⊆ X compact. Now the remaining Wα intersectX in open sets which cover K, so some finite subcollection covers K also; thissubcollection, together with W0, forms a finite subcover of W .

Finally to show that T is Hausdorff: Any two points not at infinity can beseparated by open sets of type (a) (by the Hausdorff property of X). Thus itsuffices to show that ∞ can be separated from a point x ∈ X . But by localcompactness, x ∈ X has a compact neighborhood, K say; then K and ∞ ∪(X \K) are disjoint members of T containing x and∞ respectively.

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Lecture A17Some set theory

Further topics in our discussion will require some more advanced backgroundfrom set theory. We will take a rather “naive” approach to set theory. To learnabout the paradoxes this gives rise to, and how to avoid them, you need a coursein Logic.

Definition A17.1. A partial order on a set S is a relation 6 which is reflexive,antisymmetric, and transitive: that is, for all x, y, z ∈ S, x 6 x, x 6 y and y 6 ximply x = y, and x 6 y and y 6 z together imply x 6 z.

A set with a partial order is called a (partially) ordered set. If S is such a setthen

(a) An upper bound for a subset T ⊆ S is an element x such that y 6 x for ally ∈ T .

(b) S is directed if every two-element subset has an upper bound (equivalently,every finite subset has an upper bound).

(c) S is totally ordered if for all x, y, either x 6 y or y 6 x.

(d) A subset of S that is totally ordered (in the induced ordering) is called a chain.

(e) S is inductively ordered if every chain in S has an upper bound; it is strictlyinductively ordered if every chain has a least upper bound.

(f) A maximal element for S is an element m ∈ S such that x > m impliesx = m (i.e., there isn’t anything strictly greater than m.) Notice that this is aweaker notion than that of upper bound.

(g) S is well-ordered if it is totally ordered and every non-empty subset has a leastmember (necessarily unique).

Exercise A17.2. Give simple examples to illustrate all these notions.

Let S be an ordered set. A map f : S → S will be called an inflator (I justmade this word up) if f(x) > x for all x ∈ S.

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Lemma A17.3. (Bourbaki fundamental lemma) Any inflator on a nonempty strictlyinductively ordered set has a fixed point, i.e. a point x such that f(x) = x.

We’ll prove this in a moment. For now, let’s draw some consequences.

Corollary A17.4. (Weak form of Zorn’s lemma) Every nonempty strictly induc-tively ordered set contains a maximal element.

Proof. Let S be strictly inductively ordered and suppose it contains no maximalelement. Then for each x ∈ S there is a yx ∈ S with yx > x. Consider the map1 fsending x to yx. Then f is an inflator with no fixed point, contradicting Bourbaki’slemma.

Corollary A17.5. (Hausdorff maximality principle) Let S be a nonempty partiallyordered set. Each chain in S is included in a maximal chain (i.e., a chain whichis not itself included in any larger chain).

Proof. Let C be the collection of all chains in S, partially ordered by inclusion.Then C is partially ordered, and indeed is strictly inductively ordered (with leastupper bound for a chain in C given by the union of its members). Now ap-ply A17.4.

Corollary A17.6. (Zorn’s lemma) Every nonempty inductively ordered set (notnecessarily strict) contains a maximal element.

Proof. Let S be such a set and let C be a maximal chain in S (provided by A17.5).Let m be an upper bound for C. Then m is maximal, since if x > m then C ∪xis a chain larger than the supposedly maximal chain C.

Now let us see to proving the Bourbaki lemma A17.3. Let S be nonempty andstrictly inductively ordered and f : S → S an inflator. Pick a point a ∈ S (fixedthroughout the discussion). We’ll say that a subset F of S is closed if a ∈ F , iff(x) ∈ F whenever x ∈ F , and if, whenever C ⊆ F is a chain, the least upperbound of C belongs to F .

Obviously, the intersection of any family of closed sets is closed. Moreover,S itself is closed. Thus, there is a minimal closed set M , namely, the intersectionof all the closed sets that there are. The strategy of our proof is to show that M istotally ordered, i.e., a chain. This will complete the proof because if x is the least

1For those in the know, this is where we use the axiom of choice.

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upper bound of M , then x ∈ M and f(x) ∈ M also, hence f(x) 6 x; since f isan inflator, it follows that f(x) = x.

Note that the collection of all x > a is closed. Thus, a is the least element ofthe minimal closed set M .

Let us call u ∈ M unavoidable if, whenever x ∈ M and x < u, we havef(x) 6 u. If u is an unavoidable point, then the set

x ∈M : either x 6 u or f(u) 6 x

is readily seen to be closed. Since it is a subset of M , it must be the whole of M ,by minimality.

We can now prove that every point is unavoidable. Let U be the set of unavoid-able points inM . It contains a. We will show that it is closed; then, by minimality,it must equal M itself. Suppose then that u is unavoidable and consider f(u). Ifx < f(u) then, by the paragraph above, either x < u, or x = u, or f(u) 6 x, andthe last possibility cannot occur. The other two possibilities yield f(x) 6 f(u) bythe unavoidability of u. So, f(U) ⊆ U . A similar argument shows that the leastupper bound of any chain in U belongs to U , so U is closed.

We now know that each point of M is unavoidable. It is easy to deduce thatM is totally ordered. For, let x, y ∈ M . Since x is unavoidable, either y 6 x orf(x) 6 y, but the latter certainly implies that x 6 y becasue f is an inflator. Asalready remarked, this completes the proof of Lemma A17.3.

Proposition A17.7. (Well ordering principle) Every nonempty set can be well-ordered.

Proof. Let X be a set. Let Y be the set of pairs (S,) where S is a subset ofX and is a well-ordering of S. Y is nonempty and is inductively ordered byinclusion. By Zorn’s lemma, there is a maximal element in Y , say (S,). Butnow we must have S = X; if not, we can choose some x ∈ X \ S and well-orderS ∪ x by insisting that x is greater than every element of S. This contradictsmaximality.

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Lecture A18Countability and Metrization

We now briefly discuss the theory of cardinal numbers. Let X and Y be sets.

Proposition A18.1. There exists an injection from X to Y if and only if thereexists a surjection from Y to X .

Proof. Without loss of generality assume that X 6= ∅. Let i : X → Y be aninjection. Let I = i(X) ⊆ Y . Fix an element x0 ∈ X . Define s : Y → X asfollows: if y ∈ I , then s(y) is the unique x ∈ X such that i(x) = y; otherwise,s(y) = x0. This is a surjection.

Conversely, suppose that s : Y → X is a surjection. For each x ∈ X choosesome yx ∈ Y such that s(yx) = x. Let i be the map that sends x to yx. Then i isan injection.

Exercise A18.2. The last part of this argument uses the Axiom of Choice directly.Reformulate it to use Zorn’s Lemma.

Theorem A18.3. (Schroder-Bernstein) If there exists an injection from X to Yand an injection from Y to X , then there exists a bijection from Y to X .

Proof. (Konig) Assume (wlog) that X and Y are disjoint sets and let Z be theirdisjoint union. let f : X → Y and g : Y → X be injections and let h be the unionof these maps, considered as a map Z → Z. A subset S of Z is stable if z ∈ Ziff h(z) ∈ Z. Let Sz be the minimal stable set containing z. Explicitly, Sz is theset hn(z) : n ∈ Z, where we note that h−1(p) denotes an element q such thath(q) = p; such an element may not exist, but it is unique if it does exist. By theinjectivity of h, stable sets either coincide or are disjoint; so they partition Z.

It suffices now to establish a bijection Sz∩X → Sz∩Y for each Sz separately.The stable sets Sz are of four types:

(a) X-stopping: hn(z) is defined only for n = −k,−k + 1, . . . and h−k(z) ∈ X;

(b) Y -stopping: hn(z) is defined only for n = −k,−k + 1, . . . and h−k(z) ∈ Y .

(c) Two way infinite: the hn(z) are all distinct for different values of n ∈ Z.

(d) Cyclic: hm(z) = z for some m (necessarily even).

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In the first case, f gives a bijection Sz ∩X → Sz ∩ Y . In the second case, g givesa bijection Sz ∩ Y → Sz ∩ X . In the third and fourth cases, either f or g willdo.

To each setX associate a “number”, its cardinality cardX . We define cardX =cardY iff there is a bijection between X and Y (one says that X and Y areequinumerous). We define cardX 6 cardY iff there is an injection X → Y . TheSchroder-Bernstein theorem shows that this defines a partial order on the cardi-nalities.

Definition A18.4. The cardinality of the natural numbers is denoted ℵ0 (read:aleph-zero). A cardinal that is 6 ℵ0 is called countable. (We also call a setcountable if its cardinality is countable.)

Example A18.5. The integers Z and the rationals Q are countable sets.

Proposition A18.6. For every cardinal there is a strictly larger cardinal.

Proof. Let X be a set and let P(X) (the power set) be the set of all subsets of X .We will show that there is no surjection X → P(X). Indeed, suppose that f issuch a surjection. Consider

S = x ∈ X : x /∈ f(x).

This is a subset of X . If S = f(y) for some y ∈ X we obtain (from the definitionof S) the contradiction

y ∈ f(y)⇔ y /∈ f(y).

Thus f is not surjective after all. On the other hand, there is an obvious injectionX → P(X) (as one-element subsets). It follows that card(P(X)) > card(X).

We denote the cardinality of P(X) by 2cardX . More generally, the cardinalityof the set of all maps X → Y is denoted (cardY )cardX . The cardinal 2ℵ0 isdenoted c (the continuum).

Once can do “arithmetic” with cardinals: ifX, Y have cardinalities x, y respec-tively then xy is the cardinality of X×Y , x+ y is the cardinality of X tY , and soon. Arithmetic laws like xy·z = (xy)z then follow from corresponding set-theoreticproperties (a map from Y ×Z toX is the same as a map from Z to the set of mapsfrom Y to X).

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Example A18.7. Let us prove that cℵ0 = c. It’s a one-liner:

cℵ0 =(2ℵ0)ℵ0

= 2ℵ0·ℵ0 = 2ℵ0 = c.

Proposition A18.8. The set of real numbers has cardinality c.

Proof. The set of all sequences of rational numbers has cardinality ℵ0ℵ0 = c. Thesubset of convergent sequences must then have cardinality 6 c, and this subsetmaps onto R, so cardR 6 c. On the other hand, the set of all sequences a = anof zeroes and ones also has cardinality c, and the assignment a 7→

∑∞n=1 3−nan is

an injection from this set of sequences to R, so cardR > c. Schroder-Bernsteinnow completes the proof.

Let X be a topological space.

Definition A18.9. X is separable it it has a countable dense subset. It is sec-ond countable if there is a countable basis for the topology of X: i.e., there is acountable family B of open sets such that every open set is a union of sets fromB.

Lemma A18.10. A second-countable space is separable. A separable metricspace is second-countable.

However, there are in general examples of separable spaces that are not secondcountable. For instance, let X be an uncountable set equipped with the cofinitetopology. Then X is separable (any countable subset of X meets every nonemptyopen set) but it is not second countable (a countable family of cofinite sets canomit only countably many points in total, so there must be some cofinite set thatdoes not contain any member of the family).

Proof. Let X be a second countable space. Let Un be a countable base for thetopology, and select a point xn from each Un. The set S = xn is countable, andfor any x ∈ X and any open set U containing x, there is some Un ⊆ U containingx; so U ∩ S contains xn and is in particular nonempty. Thus S is dense.

Let X be separable and metrizable, with S a countable dense set. Then the setB of balls B(s, t) where s ∈ S and t ∈ Q is countable. Moreover, it forms a basefor the topology: if x ∈ X belongs to an open set U , there is a ball B(x, r) ⊆ U ;there is an s ∈ S such that d(x, s) < r/2; there is a rational t with d(x, s) < t <r/2; the ball B(s, t) belongs to B, contains x, and is included in U .

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A neighborhood base at a point x ∈ X is a family of open sets containing xsuch that each open set containing x must include a member of the family.

Definition A18.11. A topological space X is first countable if each point has acountable neighborhood base.

Every metric space is first countable. Every second countable space is firstcountable but the converse is false: e.g. consider the discrete topology on anuncountable set.

Exercise A18.12. Show that a compact metric space is separable and secondcountable.

Theorem A18.13. (Urysohn metrization theorem) A second countable normalHausdorff space is metrizable, i.e. the topology can be given by a metric.

Proof. Let X be such a space and let B be a countable basis for the topology.Whenever U, V ∈ B with U ⊆ V there exists, by Urysohn’s lemma, a continuousfunction f : X → [0, 1] that is equal to 0 on U and is equal to 1 on X \ V . Thecollection F of all such functions is countable: denote it f1, f2, . . ..

Define d : X ×X → R+ by

d(x, y) =∞∑n=1

2−n|fn(x)− fn(y)|.

We will show that d is a metric defining the topology of X .It is clear that d(x, x) = 0, that d(x, y) = d(y, x), and that d satisfies the

triangle inequality. If d(x, y) = 0, then fn(x) = fn(y) for all n. But, if x 6= y,then there is a V ∈ B containing x but not y (by the Hausdorff property); bynormality there is an open set U containing x with U ⊆ V , and we may takeit that U ∈ B also; so then the function fn corresponding to this pair (U, V ) hasfn(x) = 0 but fn(y) = 1, and this is a contradiction. We conclude that d(x, y) = 0iff x = y, so d is a metric.

Being the sum of a uniformly convergent series of continuous functions, d iscontinuous (relative to the original topology on X). It follows that every d-openball is open in the original topology, so every d-open set is open in the originaltopology.

It remains to show that every set open in the original topology is d-open. LetW be open in the original topology, and x ∈ W . There is a set V ∈ B such that

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x ∈ V and V ⊆ W , and there is a set U ∈ B such that x ∈ U and U ⊆ V . Let fnbe the function constructed above corresponding to the sets U and V . Let ε = 2−n.Then, if y ∈ Bd(x; ε), fn(y) < 1 and so y ∈ V ; in particular y ∈ W . Thus Wcontains a d-ball around each of its points, so it is d-open, as required.

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Lecture A19Convergence and Nets

Let xn be a sequence in a topological space X . (Okay, I know that I toldyou that I was not going to use sequences outside metric spaces, but I lied.) Wesay that xn converges to a point x ∈ X if, for every open neighborhood U of x,there is some N such that xn ∈ U for n > N .

If X is first countable then sequences in x contain all the information thatthere is about the topology of X . One way to see this is the following

Lemma A19.1. In a first countable space X , a subset A is closed if and only ifthe limit of every convergent sequence in A is itself a member of A.

Proof. Suppose that A ⊆ X . A point x belongs to the closure of A iff every openneighborhood of x contains points of A. But, since X is first countable, this willbe true iff every member of a countable neighborhood base Un at x containspoints of A. Picking such a point an ∈ A ∩ Un for each n = 1, 2, 3 . . . gives asequence (an) in A converging to x. The converse statement (if x is the limit of aconvergent sequence in A, then x belongs to the closure of A) is easy.

For another example, try the following exercise.

Exercise A19.2. Let X and Y be first countable. Show that f : X → Y is con-tinuous if and only if the sequence f(xn) → f(x) in Y whenever the sequencexn → x in X . (Hint: The proof is exactly the same as the one we already dis-cussed for metric spaces.) Do you need first countability for both spaces, or onlyfor one of them (and, if so, which)?

But in general sequences are “not enough” to probe the structure of a topolog-ical space.

Example A19.3. Any totally ordered set X can be equipped with the order topol-ogy for which a basis is the “half open intervals”

(a, b] := x ∈ X : a < x 6 b.

Let X be uncountable, and well-order it (using the axiom of choice). Let Ω ∈ Xbe the smallest element that has uncountably many predecessors (a.k.a. the “firstuncountable ordinal”); this exists because of well-ordering. Let C = x ∈ X :x < Ω be the set of predecessors of Ω. Then Ω belongs to the closure of C, butnevertheless there is no sequence in C converging to Ω.

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This motivates the following definition.

Definition A19.4. A net in a set X is a map from a directed set to X . (Recall thata directed set is a partially ordered set in which any two elements have an upperbound.)

Let S be a subset of X .

Definition A19.5. A net xi is eventually in S if there is i0 such that xi ∈ S forall i > i0; it is frequently in S if it is not eventually in X \ S.

Now suppose that X is a topological space.

Definition A19.6. A net xi converges to x ∈ X if it is eventually in everyneighborhood of x.

These definitions match the definitions of “sequence” and “convergent se-quence” when we take the net simply to be the natural numbers N. On the otherhand, in Example A19.3 above, we could take the collection C itself in its naturalordering to be a net, and it is then almost a tautology that C converges to Ω eventhough we have seen there is no sequence that will do this.

Proposition A19.7. In any topological space X , a subset A is closed if and onlyif the limit of every convergent net in A is itself a member of A.

Proof. Emulating the proof of Lemma A19.1, what we have to show is that if xis a point of the closure of A, then there is some net in A that converges to x. Anet has to parameterized by a directed set. The trick is to pick the collection of allneighborhoods of x as the directed set! In fact, suppose that U and V are openneighborhoods of x. We define U 6 V if U ⊇ V (notice that the inclusion isreversed!). This makes the collection N = U open : x ∈ U a partially orderedset, and in fact a directed set: an upper bound forU and V is the intersectionU∩V .Since x belongs to the closure of A, for every U ∈ N there exists aU ∈ A ∩ U .The mapping2 from N to A given by U 7→ aU is a net. Moreover, it convergesto x by definition: if V is any neighborhood of A then there is an element of N(namely V itself) such that if U > V then aU ∈ V .

Similarly we have

Proposition A19.8. A map f : X → Y between topological spaces is continuousif and only if the net f(xi)→ f(x) in Y whenever the net xi → x in X .

2Using the axiom of choice here!

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Proof. The proof that a continuous map has this property is essentially the sameas the corresponding proof for sequences.

Suppose that f has the property in question but is not continuous. Then thereexist a point x ∈ X , y = f(x), and an open neighborhood W of y, such thatf−1(W ) is not a neighborhood of x. It follows that for every open neighborhoodU of x there exists xU ∈ U \W . Organize the xU into a net parameterized by thedirected set N of neighborhoods of x, as in the previous proposition. Then, byconstruction, xU converges to x; but f(xU) does not converge to y, since it neverlies in the neighborhood W of y.

Exercise A19.9. Suppose that all we know about the topology of X is that it isgenerated by some family F of subsets. Show that a net xi converges to x iff,for every U ∈ F which contains x, the net xi is eventually in U . (This exerciseshould help you understand why we insist that a net should be parameterized by adirected set.)

One might ask whether we can similarly generalize the definition of com-pactness, which in the metric space case is expressed as “every sequence has aconvergent subsequence.” To do this, we must arrive at the correct definition ofsubnet.

Definition A19.10. Let D and D′ be directed sets. A function h : D′ → D iscalled final if, for any i0 ∈ D, there is j0 ∈ D′ such that j > j0 implies h(j) > i0.Given a net D → X , the result of composing it with a final function h is called asubnet or refinement of the original one.

Every subnet of a convergent net is convergent (with the same limit); the defi-nition of “final function” is cooked up to make this true.

Definition A19.11. A net xi in X is called universal if, given any subset S ofX whatsoever, either xi is eventually in S or else eventually in X \ S.

Lemma A19.12. Every net has a universal subnet.

Proof. This depends unavoidably on the axiom of choice.LetN : D → X be a net inX , parameterized by a directed setD. A collection

F of subsets of X will be called an N -filter if N is frequently in every memberof F , and if F is closed under finite intersections and the formation of supersets.Such objects exist: for example, X is anN -filter. The collection ofN -filters ispartially ordered by inclusion, and every chain in this partially ordered set has an

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upper bound (the union). Thus Zorn’s Lemma provides a maximal N -filter; callit F0.

Suppose that S ⊆ X has the property that N is frequently in A ∩ S for everyA ∈ F0. Then the union of F0 with the set of all sets A ∩ S, A ∈ F0, is again anN -filter. By maximality we deduce that S itself belongs to F0.

We will use this property to construct the desired universal refinement. Let D′

be the collection of pairs (A, i) withA ∈ F0, i ∈ D, andN (i) ∈ A; it is a directedset under the partial order

(B, j) > (A, i)⇔ B ⊆ A, j > i.

The map (A, i) 7→ i is final so defines a refinement N ′ of the net N . We claimthat this refinement is universal.

Let S ⊆ X have the property that N ′ is frequently in S. Let A ∈ F0 and leti be arbitrary. By definition, there exist B ∈ F0, B ⊆ A, and j > i, such thatN (j) = N ′(B, j) ∈ B ∩ S ⊆ A ∩ S. We conclude that N is frequently in A ∩ Sfor every A ∈ F0 and hence, as observed above, that S itself belongs to F0.

Now let S be arbitrary. It is not possible that N ′ be frequently both in S andin X \ S, for then (by the above) both S and X \ S would belong to F0, and thentheir intersection, the empty set, would do so as well, a contradiction. Thus N ′fails to be frequently in one of these sets, which is to say that it is eventually inthe other one.

Proposition A19.13. The following properties of a topological spaceX are equiv-alent:

(i) X is compact;

(ii) (the finite intersection property) if F is a family of closed subsets of X , andthe intersection of any finite number of members of F is nonempty, then infact the intersection of all the members of F is nonempty;

(iii) every universal net in X converges;

(iv) every net in X has a convergent subnet.

Proof. Suppose (i). Let F be a family as in (ii) and let U = X \ F : F ∈ F.If⋂F is empty then U is an open cover of X , hence it has a finite subcover,

so the intersection of some finite subcollection of F is empty. This proves (ii).Conversely, if we assume (ii) and let U be an open cover that is supposed to have

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no finite subcover, we reach a similar contradiction by considering the family ofclosed sets F = X \ U : U ∈ U. Thus (i) and (ii) are equivalent.

Suppose (ii), and let xα be a universal net inX . If it does not converge, theneach x ∈ X has a neighborhood Ux for which xα is not eventually in Ux, hence(by universality) xα is eventually in Fx = X \Ux. It follows that xα is eventuallyin any finite intersection of the Fx. In particular no finite intersection of the Fx isempty. By (ii) the intersection of all the Fx is nonempty, and this is a contradictionsince x /∈ Fx for each x.

(iii) implies (iv) by Lemma A19.12.Suppose (iv). Let F be a family of closed sets with the finite intersection

property. Let D be the directed set comprised of finite intersections of membersof F (directed by reverse inclusion) and for each D ∈ D let xD be a point of D.Then xD is a net in X; choose a subnet convergent to x ∈ X . By definition, anyneighborhood of x meets every F ∈ F ; since F is closed, we deduce that x ∈ F .Thus the intersection of all the F contains x. This proves (ii).

Now let (Xα) be a family of topological spaces, parameterized by α ∈ Awhere A is some index set (possibly infinite or even uncountable). Let X be theCartesian product

∏αXα: that is, an element of X is a mapping x : α 7→ xα,

where xα ∈ Xα. For each α there is a projection map πα : X → Xα, which sendsthe point x to its “coordinate” xα.

Definition A19.14. The product topology on X is the topology generated by allthe sets π−1α (Uα), where α runs over A and Uα runs over all the open subsets ofXα.

Exercise A19.15. Let xi be a net in the product X defined above. Show that xiconverges to x ∈ X if and only if πα(xi) converges to πα(x) for all α ∈ A .

Theorem A19.16. (Tychonoff) The product of any family of compact spaces iscompact.

Proof. We use the characterization (iii) of compactness in Proposition A19.13above.

Let xi be a universal net in X . Then πα(xi) is a universal net in Xα, henceconvergent, say to xα. Let x ∈ X be the unique point such that πα(x) = xα.Then, by Exercise A19.15, xi → x in X . Thus X is compact.

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Lecture A20Quotient Spaces

Let X be a set and ∼ an equivalence relation on it. Recall that the notationX/ ∼ denotes the set of equivalence classes of the equivalence relation. There isa canonical map π : X → X/ ∼ that sends each x ∈ X to its equivalence class.

Now suppose that X is a topological space.

Definition A20.1. The quotient topology on X/ ∼ is defined by declaring that asubset U ⊆ X/ ∼ is open iff the inverse image π−1(U) is open in X .

Of course, there is something to check here, namely that the sets U identifiedby this definition do satisfy the axioms for a topology. But this is easy. Clearly,the quotient topology makes π continuous, and it is the largest topology (mostopen sets) that does so. Moreover, a function f : (X/ ∼) → Y is continuous ifff π : X → Y is continuous.

Remark A20.2. A subset ofX is called saturated (with respect to∼) if it is a unionof equivalence classes. Thus, the open sets of the quotient topology correspondto the saturated open subsets of X . The saturation of an arbitrary subset S is theunion of all the equivalence classes that meet S, i.e. the smallest saturated set thatincludes S.

Example A20.3. Let X = [0, 1] and let ∼ be the equivalence relation that makes0 and 1 equivalent and whose only other equivalence classes are points. Then[0, 1]/ ∼ is homeomorphic to the unit circle S1. To see this, let g : X → S1 bethe continuous map g(t) = e2πit = (cos 2πt, sin 2πt), and define f : X/ ∼→ S1

by f(π(t)) = g(t); this is well-defined since g(0) = g(1). By the remark above,f is a continuous map from X/ ∼ to S1, which is in fact bijective. We must showthat f is also open, and for this it suffices to show that f takes each neighborhoodof each point π(x) of X/ ∼ to a neighborhood of f(π(x)) = g(x). That is, wemust show that any saturated open set containing x maps to a neighborhood ofg(x). When x 6∈ 0, 1 this is easy, since any saturated open set containing xincludes an interval (x − ε, x + ε). What about saturated open sets containing 0or 1? Any such must include U = [0, ε) ∪ (1 − ε, 1] for some ε > 0, and thenf(U) = (cos 2πt, sin 2πt) : t ∈ (−ε, ε) is a neighborhood of (1, 0) in S1. Sowe are done.

The ideas underlying this example can be summarized in the next two propo-sitions. The first is the topological analog of the “first isomorphism theorem” in

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algebra. Let us call a continuous map f : X → Y ∼-respecting if f(x) = f(x′)whenever x ∼ x′. Then we have

Proposition A20.4. A continuous g : X → Y is ∼-respecting iff g = f π forsome continuous f : X/ ∼→ Y .

Proof. If g is∼-respecting then a function f can be defined by setting f(c), for anequivalence class c, equal to f(x) for any representative x of c. Then g = f π,and f is continuous by definition of the quotient topology.

Proposition A20.5. Let X be a compact space and Y a Hausdorff space andsuppose that g : X → Y is continuous and surjective. Let ∼ be the equivalencerelation on X defined by x ∼ x′ iff g(x) = g(x′). Then the map f : X/ ∼→ Yinduced by g is a homeomorphism.

Proof. Tautologically, g is ∼-respecting, so there is a continuous map f as de-scribed, which is in fact a bijection X/ ∼→ Y . But X/ ∼ is a compact space (asthe image of a compact space under the continuous map π, Proposition A12.8),and Y is a Hausdorff space, so this bijection is in fact a homeomorphism by Corol-lary A12.12.

It might be helpful to think of ∼, in this proposition, as the “kernel” of themap g.

One problem with quotient spaces is that they are frequently not Hausdorff.An obvious necessary condition for Hausdorffness is that the equivalence classesshould be closed: after all, each equivalence class is the inverse image of a pointin X/ ∼ under the continuous map π, and we know that one-point sets are closedin a Hausdorff space. But this is by no means a sufficient condition:

Example A20.6. LetX be the unit square [0, 1]× [0, 1], which is a compact Haus-dorff space, and let∼ be the equivalence relation which makes (x1, y1) equivalentto (x2, y2) iff either x1 = x2 and y1 = y2 or y1 = y2 > 0. Then any two saturatedneighborhoods of (0, 0) and (1, 0) must intersect; so the corresponding quotientspace cannot be Hausdorff.

One can even give an example of this kind where all the equivalence classesexcept one consist of single points.

Exercise A20.7. Let X be the real line, with the topology generated by all theopen sets of the usual topology together with the complement of the set A =1/n : n = 1, 2, 3, . . .. Show that X is a Hausdorff space, but that the quotient

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space obtained by identifying all the points of A to a single point (this is usuallydenoted X/A) is not. (This is also an example of a Hausdorff space that is notregular.)

We will give some useful conditions for Hausdorffness of quotients. Let ∼ bean equivalence relation on X . The graph of ∼ is the subset

G = (x1, x2) : x1 ∼ x2 ⊆ X ×X.

Suppose that X/ ∼ is Hausdorff. Then whenever x1 6∼ x2 there are disjoint opensubsets U1, U2 of X/ ∼ with π(x1) ∈ U1 and π(x2) ∈ U2. Thus π−1(U1) ×π−1(U2) is open in X ×X , contains (x1, x2), and does not meet G. We concludethat if X/ ∼ is Hausdorff, then the graph G is closed. (This statement includesour previous requirement that the equivalence classes should all be closed, but itis stronger.)

The previous exercise shows that the converse is not necessarily the case.However, there are additional conditions under which the closure of the graphis both necessary and sufficient for the Hausdorffness of the quotient.

Proposition A20.8. Let X be a Hausdorff space and let ∼ be an equivalencerelation on X whose graph is closed. If in addition the projection π : X → X/ ∼is an open map then X/ ∼ is Hausdorff also.

Proof. Let G denote the graph of ∼. Suppose that π is open. Let x1, x2 ∈ X withπ(x1) 6= π(x2); then the point (x1, x2) ∈ X × X does not meet the closed setG, so there is an open rectangle of the form U1 × U2 which contains (x1, x2) anddoes not meet G. But then π(U1) and π(U2) are disjoint open subsets of X/ ∼containing x1 and x2 respectively.

Proposition A20.9. Let X be a compact Hausdorff space and let∼ be an equiva-lence relation on X whose graph is closed. Then X/ ∼ is Hausdorff also. More-over, the projection π : X → X/ ∼ is a closed map (it need not be open).

Proof. We first show that the saturation of any closed subset ofX is closed (equiv-alently, π : X → X/ ∼ is a closed map). Let C be a closed subset of S. We maywrite the saturation S of C as

S = π2((C ×X) ∩G) ⊆ X,

where π2 : X ×X → X is the second projection map of the product space. Now,C ×X and G are closed subsets of X ×X , which is compact (by the Tychonoff

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theorem). Thus (C ×X) ∩G is a closed subset of a compact space, so compact.It follows that π2((C ×X)∩G) is a compact subset of a Hausdorff space, so it isclosed. This proves our assertion.

Now suppose that x1, x2 ∈ X with π(x1) 6= π(x2), and let C1 and C2 be theequivalence classes of x1 and x2 respectively. Let us show that there are open setsU1, U2, including C1, C2 respectively, whose saturations do not meet — that is tosay, (U1 × U2) ∩ G = ∅. This is the same argument as the proof that compactHausdorff spaces are normal. First, we prove the analog of regularity. Fix x ∈ C1.For each y ∈ C2 there are open sets Py containing x and Qy containing y suchthat (Py × Qy) ∩ G = ∅. Because C2 is compact there is a finite subcover of C2

say by Qy1 , . . . , Qyn . Then P =⋂nj=1 Pyj and Q =

⋃nj=1Qyj are open sets, one

containing x and one including C2, such that P×Q∩G = ∅. Repeat the argumentusing the compactness of C1 to obtain the desired open sets U1 ⊇ C1, U2 ⊇ C2,with (U1 × U2) ∩G = ∅.

Consider now the closed set X \ Ui, and let Si be its saturation; Si is closed(by the first thing we proved above) and does not meet Ci, so the complementX \ Si = Vi is a saturated open set with Ci ⊆ Vi ⊆ Ui. Now π(Vi) and π(V2) aredisjoint open subsets containing x1 and x2 respectively.

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Lecture A21Basics of Topological Groups

Let G be a group. We can think of it as a set provided with a distinguishedelement (the identity e) and two distinguished maps, the multiplication map

G×G→ G, (x, y) 7→ xy

and the inversion mapG→ G, x 7→ x−1.

Definition A21.1. A topological group is a group provided with a topology forwhich the multiplication and inversion maps are continuous.

There are many examples: the (real or complex) numbers under addition,the nonzero (real or complex) numbers under multiplication, the matrix groupsGL(n,R) and GL(n,C) as well as their various subgroups SL(n,R), SL(n,C),O(n), U(n), etc. Any group can be thought of as a topological group when en-dowed with the discrete topology.

In a topological group the left and right multiplication maps by a fixed elementg, defined by Lg(x) = gx, Rg(x) = xg, are homeomorphisms G → G. Conse-quently, every translate gU or Ug of an open subset U is again open (and everytranslate of a closed set is closed).

Proposition A21.2. A topological group G is Hausdorff if and only if the one-point set e is closed.

Proof. Suppose that e is closed. By translation, every one-point set is closed.If g1 6= g2 there exists an open set V = G \ g2 that contains g1 and does notcontain g2. By continuity of the multiplication map at the point (g1, e), togetherwith the definition of the product topology, there are open sets U1, U2 that containg1 and e respectively such that U1 · U2 ⊆ V . But now U1 and g2U−12 are openneighborhoods of e and g respectively. I claim they don’t intersect: if they did,then u1 = g2u

−12 with u1 ∈ U1, u2 ∈ U2, and then g2 = u1u2 ∈ U1U2 ⊆ V , a

contradiction.

This is one example of the “homogeneity” of topological groups.LetG be a topological group andH a subgroup. We denote byG/H the space

of left cosets xH for x ∈ G, or in other words the space of equivalence classes

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for the equivalence relation x ∼ y ⇔ x−1y ∈ H; similarly we denote by H\Gthe space of right cosets Hx, which are equivalence classes for the equivalencerelation x ∼ y ⇔ xy−1 ∈ H . These are called (left and right) homogeneousspaces of G by H , and we give them the quotient topology.

Exercise A21.3. If H is a normal subgroup of G then the left and right cosetspaces are the same. In this case we know from algebra thatG/H has the structureof a group. We have also just equipped it with a topology. Show that G/H is atopological group, i.e., multiplication and inversion are continuous.

Proposition A21.4. The quotient maps G → G/H and G → H\G are openmaps.

Proof. This is equivalent to the statement that the saturation of an open subset ofG is open. But the saturation of an open set U is⋃

u∈U

uH =⋃h∈H

Uh;

the second representation exhibits it as a union of open sets, hence open.

Theorem A21.5. Let G be a Hausdorff topological group and let H be a closedsubgroup. Then the homogeneous spaces G/H and H\G are Hausdorff.

Proof. By A20.8 and A21.4, it suffices to show that the graph Γ of the equivalencerelation x ∼ y ⇔ x−1y ∈ H is a closed subset of G×G.

Suppose that (x, y) /∈ Γ; then x−1y ∈ G \ H which is an open set. Bycontinuity of multiplication there exist open sets U 3 x−1, V 3 y with U · V ⊆G \H . But then U−1×V is an open neighborhood of (x, y) that does not meet Γ.It follows that the complement of Γ is open, so Γ is closed.

Definition A21.6. Let X, Y be topological spaces. A continuous, surjective mapf : X → Y is called a local homeomorphism if every x ∈ X has a neighborhoodU such that the restriction of f to a map U → f(U) is a homeomorphism.

Local homeomorphisms will play a very important role in the study of cov-ering spaces, a key part of algebraic topology. The classic example of a localhomeomorphism is the exponential map t 7→ e2πit from the real line to the circle.This can be envisaged as a special case of the next result:

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Proposition A21.7. Let G be a topological group and suppose that H is a closedsubgroup which is discrete (in the induced topology). Then the quotient mapπ : G→ G/H is a local homeomorphism.

Proof. There is a neighborhood V of e such that H ∩ V = e. By the samekind of argument (continuity of multiplication) that we have already used severaltimes, there exists a neighborhood U of the identity with U−1U ⊆ V . But nowthe restriction of π : G → G/H to the set U is injective: if π(u1) = π(u2) thenu−11 u2 ∈ V ∩ H = e, and so u1 = u2. Thus the restriction of π to a mapfrom U to π(U) is continuous, bijective, and open (A21.4): in other words, itis a homeomorphism. We have proved that π is a homeomorphism near e; bytranslation, the same holds near any point.

A group G acts on a set X if there is given a map G × X → X , written(g, x) 7→ gx, such that ex = x for all x ∈ X and g(hx) = (gh)x for all g, h ∈ G,x ∈ X . If G is a topological group and X is a topological space we define acontinuous action by requiring that the associated mapG×X → X be continuous.An action is transitive if for all x, y ∈ X there exists g ∈ G with gx = y.

Example A21.8. The rotation group SO(n) acts transitively on the unit sphereSn−1.

Proposition A21.9. Suppose that G acts transitively on the Hausdorff space X .For any x ∈ X let Gx denote the stabilizer

Gx = g ∈ G : gx = x,

which is a closed subgroup of G. Then the formula

[g] 7→ g · x

defines a continuous bijection G/Gx → X . It is a homeomorphism if G is com-pact.

Proof. The continuous map G → X given by g 7→ gx respects the equivalencerelation given by the cosets of Gx. Thus it passes to a continuous map G/Gx →X , which is injective by construction, and is surjective because of the transitivityof the action. The final statement follows because G/Gx is compact and X isHausdorff.

For example, considering the stabilizer of a single point (the North pole) givesus the identification Sn−1 ∼= SO(n)/SO(n − 1) (using the fact that SO(n) iscompact).

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Lecture A22Homotopy

Hereafter “map” means “continuous map”.Consider the unit square X = I2 = (x, y) ∈ R2 : 0 6 x, y 6 1. Imagine

two continuous paths γ1, γ2 : [0, 1] → X , with γ1 beginning at (0, 0) and endingat (1, 1), and γ2 beginning at (0, 1) and ending at (1, 0). It seems obvious thatthese two paths must cross somewhere in the interior of the square. But this ishard (maybe impossible?) to prove just with the tools that we have developed sofar. We need a little homotopy theory.

Definition A22.1. Let X and Y be topological spaces and let f0, f1 : X → Y becontinuous maps. A homotopy between f0 and f1 is a continuous map

h : X × [0, 1]→ Y

with h(x, 0) = f0(x) and h(x, 1) = f1(x) for all x ∈ X . If there exists a homo-topy between f0 and f1 we say that these maps are homotopic.

Proposition A22.2. Homotopy is an equivalence relation on the class of mapsX → Y .

Proof. Clearly any map f is homotopic to itself via the constant homotopy h(x, t) =f(x). If h is a homotopy between f0 and f1, then

k(x, t) = f(x, 1− t)

is a homotopy between f1 and f0. Finally, homotopies can be concatenated likepaths: if h′ is a homotopy from f0 to f1 and h′′ is a homotopy between f1 and f2then

h(x, t) =

h′(x, 2t) (t 6 1

2)

h′′(x, 2t− 1) (t > 12)

is a homotopy between f0 and f2.

Exercise A22.3. Check in detail that the homotopy h defined in the last part ofthis proof is continuous.

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Remark A22.4. This proof is obviously closely related to the proof that path-connectedness is an equivalence relation, Lemma A15.10. In fact, path-connectednessis just homotopy of maps where X is a point. Conversely, one can reduce the no-tion of homotopy to path-connectedness in the space of maps X → Y if onedefines a suitable topology on this space. But we won’t do that here.

Proposition A22.5. If f0, f1 are homotopic mapsX → Y , and g0, g1 : Y → Z arehomotopic maps, then g0 f0, g1 f1 are homotopic maps X → Z. (In particular,homotopic maps remain homotopic after composition on the left or the right witha fixed map.)

There is a related notion, that of relative homotopy. Let X be a space and Aa closed subspace (one says sometimes that (X,A) is a pair). Then a homotopyrelative to A from X to Y is a homotopy h : X × [0, 1] → Y such that h(a, t) isindependent of t for all a ∈ A — a homotopy that stays constant on A. This alsodefines an equivalence relation by the same reasoning as above.

Example A22.6. LetX be the unit disc in Euclidean space. Then the identity mapX → X is homotopic to the map that sends every point to the origin. A homotopybetween these maps is defined by h(x, t) = (1 − t)x. A similar argument can beapplied to any star-shaped subset of Euclidean space.

Example A22.7. Let X, Y be spaces and let f : X → Y be a continuous map.The mapping cylinder of f is the space

Cf = X × [0, 1] tf Y

obtained from the disjoint union X× [0, 1]tY by identifying (x, 0) with f(x) forall x ∈ X (and equipped with the quotient topology). As we saw earlier, there isa continuous map F : Cf → Y defined by F (x, s) = f(x), F (y) = y. Moreover,the identity map Cf → Cf is homotopic, relative to Y , to the map Cf → Y → Cfdefined by composing F with the standard inclusion Y → Cf . A homotopy hbetween these maps can be defined by the formula

h((x, s), t) = (x, (1− t)s), h(y, t) = y.

Notice that the second example actually includes the first: the disc is the map-ping cylinder of the constant map from a sphere to a point.

The geometric situation defined in these examples occurs often enough to havea special name.

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Definition A22.8. Let X be a space and A a closed subspace. A retraction of Xonto A is a continuous map X → A which is the identity on A. A deformationretraction of X onto A is a homotopy, relative to A, from the identity map on Xto a retraction X → A. If these exist one says that A is a retract or deformationretract of X .

The examples above establish that a point is a deformation retract of a disk,and that Y is a deformation retract of the mapping cylinder. Deformation retrac-tion is a special case of the general notion of homotopy equivalence.

Definition A22.9. Let X and Y be spaces and let f : X → Y and g : Y → Xbe maps. One says that f and g are homotopy inverses if the composite g f ishomotopic to the identity map on X and f g is homotopic to the identity mapon Y . A map that has a homotopy inverse is called a homotopy equivalence, andone says that the spaces X and Y are homotopy equivalent if there is indeed ahomotopy equivalence between them.

For example, if A is a deformation retract of X , then the inclusion A → Xand retraction X → A form a pair of homotopy inverse maps, so X and A arehomotopy equivalent.

Proposition A22.10. The relation of homotopy equivalence is, indeed, an equiv-alence relation.

Proof. Suppose that f ′ : X → Y and f ′′ : Y → Z have homotopy inverses g′ andg′′ respectively. Then it follows from Proposition A22.5 that g′ g′′ is a homotopyinverse for f ′′ f ′.

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Lecture A23Homotopy Equivalence

The examples above establish that a point is a deformation retract of a disk,and that Y is a deformation retract of the mapping cylinder. Deformation retrac-tion is a special case of the general notion of homotopy equivalence.

Definition A23.1. Let X and Y be spaces and let f : X → Y and g : Y → Xbe maps. One says that f and g are homotopy inverses if the composite g f ishomotopic to the identity map on X and f g is homotopic to the identity mapon Y . A map that has a homotopy inverse is called a homotopy equivalence, andone says that the spaces X and Y are homotopy equivalent if there is indeed ahomotopy equivalence between them.

For example, if A is a deformation retract of X , then the inclusion A → Xand retraction X → A form a pair of homotopy inverse maps, so X and A arehomotopy equivalent.

Proposition A23.2. The relation of homotopy equivalence is, indeed, an equiva-lence relation.

Proof. Suppose that f ′ : X → Y and f ′′ : Y → Z have homotopy inverses g′ andg′′ respectively. Then it follows from Proposition A22.5 that g′ g′′ is a homotopyinverse for f ′′ f ′.

Definition A23.3. A space is called contractible if it is homotopy equivalent to apoint.

The cone CX on a space X is the quotient X × [0, 1]/X × 0, or in otherwords the mapping cylinder of the constant map from X to a point. Any mappingcylinder deformation retracts onto the range of the map from which it is defined(as we have seen), so the cone on any space is contractible.

Homotopy equivalent spaces can be rather different topologically. For exam-ple, the 1-dimensional spaces given by the “theta curve”, the “figure 8”, and the“dumb-bell” are all homotopy equivalent, although no two of them are homeomor-phic. However, there are topological properties that are preserved by homotopyequivalence.

Example A23.4. Suppose X and Y are homotopy equivalent. If X is path-connected, then so is Y .

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Proof. Let f : X → Y and g : Y → X be mutually inverse homotopy equiva-lences. Note that for each y ∈ Y , the points y and f(tg(y)) belong to the samepath component: indeed, if h is a homotopy between f g and the identity, thent 7→ h(y, , t) defines a path joining y to f(g(y)). Now let y, y′ ∈ Y and supposethat X is path connected. Then there is a path γ in X joining g(y) to g(y′). Nowf γ is a path joining f(g(y)) to f(g(y′)). Thus f(g(y)) and f(g(y′)) are in thesame path component. It follows from the earlier discussion that y and y′ are inthe same path component; that is, Y is path connected.

Suppose that X is a space and A a closed subspace. If A is contractible, wemight say that “up to homotopy”A is the same as a point. IfA really were a point,the quotient space X/A would be just the same as X . So we might be led to hopethat ifA is contractible,X/A is “up to homotopy” the same asX , i.e., the quotientmap X → X/A is a homotopy equivalence. Unfortunately this is not true.

Exercise A23.5. Let Z be the “topologist’s sine curve” of Exercise A15.13, andlet A ⊆ Z be the interval (0, y) : −1 6 y 6 1 on the y-axis. Clearly A iscontractible. Show that Z/A is path-connected. Since Exercise A15.13 shows thatZ itself is not path-connected, this proves that Z and Z/A cannot be homotopyequivalent.

What is missing in this example is the homotopy extension property (HEP).

Definition A23.6. Let X be a space and A a closed subspace. One says that(X,A) has the homotopy extension property if any map from the subspace

X × 0 ∪ A× [0, 1] ⊆ X × [0, 1]

to another space Y can be extended to a map from the whole of X × [0, 1] to Y .

The way to think of this is that the initial data consist of a map X → Ytogether with a homotopy of the restriction of that map to the subspace A; theHEP says that this homotopy can be extended to a homotopy defined on the wholespace X . Most “nice” pairs of spaces have the HEP: proofs are often inductivebased on the following example.

Example A23.7. The pair (Dn, Sn−1) has the homotopy extension property. Tosee this, it is enough to construct geometrically a retraction

r : Dn × [0, 1]→ Dn × 0 ∪ Sn−1 × [0, 1]

(which can be done for instance by projection from the point (0, 3/2) in Rn+1);then the desired extension can be defined by composing with r.

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Proposition A23.8. Suppose that (X,A) has the HEP and that A is contractible.Then the quotient map π : X → X/A is a homotopy equivalence.

Proof. The first order of business is to find a potential homotopy inverse. Letp ∈ A be a point and let h : A × [0, 1] → A be a homotopy between the identitymap and the constant map that sends A to p. By the HEP, the homotopy h extendsto a homotopy H between the identity on X and some map G : X → X , such thatG(a) = p for all a ∈ A. Note that G respects the equivalence relation defining thequotient space, so there is a map g : X/A → X such that g π = G. This map gis, I claim, a homotopy inverse for π.

To see this we must show that g π and π g are homotopic to the identity onX and X/A respectively. The first is easy: g π = G is homotopic to the identityby construction. What about the other? The homotopy H has H(a, t) ∈ A forall a ∈ A and all t. Therefore, π H(a, t) ∈ X/A is constant for all a ∈ A andt ∈ [0, 1]; that is, π H respects the equivalence relation. It follows that there is amap

k : (X/A)× [0, 1]→ (X/A)

such that k(π(x), t) = π(H(x, t)) for all x ∈ X , t ∈ [0, 1]. In particular, k givesthe desired homotopy between π g and the identity on X/A.

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Lecture A24The fundamental group

For the next few lectures we are going to consider pointed topological spaces.Such an object is just a topological space X together with a choice of a base pointx0 ∈ X . A map of pointed spaces (X, x0) → (Y, y0) is just a map from X to Ythat takes x0 to y0. Sometimes, to save space, we’ll omit explicit mention of thebase point.

Example A24.1. We regard the unit circle S1 as the quotient space [0, 1]/0, 1.It is a pointed space whose base point ∗ is the equivalence class 0, 1.

Definition A24.2. A loop in a pointed space (X, x0) is a map of pointed spaces(S1, ∗) → (X, x0). In other words, it is a path γ : [0, 1] → X such that γ(0) =γ(1) = x0.

The natural notion of homotopy for maps of pointed spaces is based homotopy:a based homotopy from (X, x0) to (Y, y0) is a homotopy h : X× I → Y such thath(x0, t) = y0 for all t ∈ I . In particular we can consider the collection of all basedhomotopy classes of loops in (X, x0). This object is denoted π1(X, x0) and it iscalled the fundamental group of (X, x0). As we shall see, there is indeed a naturalway to make it into a group.

Lemma A24.3. Let γ be a based loop in (X, x0). Let g : [0, 1] → [0, 1] be acontinuous map with g(0) = 0 and g(1) = 1. Then γ and γ g represent the sameelement of π1(X, x0). (The composite γ g is called a reparameterization of γ.)

Proof. The maph(s, t) = γ((1− t)s+ tg(s))

defines a homotopy between γ and the reparameterized path γ g.

Let γ′ and γ′′ be paths in X . Recall that the concatenation of these paths isdefined by

γ′ ? γ′′(s) =

γ′(2s) (s 6 1

2)

γ′′(2s− 1) (s > 12)

If γ′ and γ′′ are (based) loops then so is γ′ ? γ′′. Moreover, if γ′ or γ′′ is var-ied by a (based) homotopy, then so is γ′ ? γ′′. Thus the operation ? passes to a“multiplication” operation on π1(X, x0).

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Proposition A24.4. Under this operation, π1(X, x0) becomes a group.

Proof. There are three things that must be verified: associativity, the existence ofan identity, and the existence of inverses. The proofs of all three make extensiveuse of the freedom to deform loops by homotopies and, in particular, to reparam-eterize them.

For associativity, we just need to note that the paths (γ1?γ2)?γ3 and γ1?(γ2?γ3)are related by the reparameterization

s 7→

2s s 6 1

4

s+ 14

146 s 6 1

212

+ 12s 1

26 s 6 1

The constant loop (e(s) = x0 for all s ∈ [0, 1]) defines an identity element.Indeed, it is easy to see that e ? γ and γ ? e are simply reparameterizations of γ,for any loop γ.

Finally suppose that γ is a loop and let γ−1 be the loop defined by γ−1(s) =γ(1 − s). Then γ ? γ−1 is the path s 7→ γ(g(s)) where g(s) = 2s for s 6 1

2and

g(s) = 2− 2s for s > 12. Then

h(s, t) = γ((1− t)g(s))

is a homotopy of based loops from γ ? γ−1 to the identity path.

The effect of the choice of basepoint is summed up in the following.

Lemma A24.5. Let X be a path-connected space and let x0, x1 ∈ X . Then thegroups π1(X, x0) and π1(X, x1) are isomorphic.

Proof. Let q : [0, 1] → X be a path from x0 to x1, and let q−1(s) = q(1 − s), asusual. Then, if γ is a loop based at x0, the path

φ(γ) := q−1 ? γ ? q

is a loop based at x1, and the assignment γ 7→ φ(γ) gives a homomorphism ofgroups from π1(X, x0) to π1(X, x1). The assignment γ′ 7→ q ? γ′ ? q−1 gives theinverse homomorphism π1(X, x1) to π1(X, x0), so these two groups are isomor-phic.

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Beware that the isomorphism in the lemma above depends on the choice of thepath q. One says that the groups π1(X, x0) and π1(X, x1) are “isomorphic but notcanonically isomorphic”.

Let f : (X, x0)→ (Y, y0) be a (basepoint-preserving) map. If γ is a based loopin X , the composite f γ is a based loop in Y . Moreover, the assignment [γ] 7→[f γ] passes to homotopy classes and defines a homomorphism π1(X, x0) →π1(Y, y0).

Definition A24.6. The homomorphism so defined is called the induced homomor-phism of the map f , and it is denoted by f∗ : π1(X, x0)→ π1(Y, y0).

Proposition A24.7. The induced homomorphism has the following properties:

(a) The identity map induces the identity homomorphism.

(b) Suppose that f and g are maps (X, x0) → (Y, y0), which are homotopic bya homotopy h. Let u ∈ π1(Y, y0) be defined by the loop t 7→ h(x0, t). Thenf∗(v) = ug∗(v)u−1 for all v ∈ π1(X, x0). In particular, based homotopicmaps induce the same homomorphism.

(c) If f : X → Y and g : Y → Z are based maps, then (f g)∗ = f∗ g∗.

Item (c) is called functoriality: it is an extremely important property in manymathematical contexts.

Proof. (a) and (c) are immediate consequences of the definitions. Part (b) wouldbe easier if we restricted our attention to based homotopies, but we need the gen-eral case. Consider the map from the unit square to Y defined by

(s, t) 7→ h(γ(s), t).

Going along the bottom and right sides of this square defines the path f ?u, goingalong the left and top sides defines u ? g. But these are homotopic by a homotopythat goes diagonally across the square. Explicitly, we want to define

H(r, s) =

h(2r(1− s), 2rs) (r 6 1

2)

h((1− s)(2r − 1) + s, s(2r − 1) + 1− s) (r > 12)

which gives a homotopy from H(·, 0) = f ? u to H(·, 1) = u ? g.

Corollary A24.8. A homotopy equivalence induces an isomorphism of fundamen-tal groups.

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Proof. Let f : X → Y be a homotopy equivalence, with homotopy inverse g.Then fg and gf are homotopic to the identity, and therefore by (a) and (b) abovethey induce isomorphisms on fundamental groups. Thus by (c) above, f∗ g∗ andg∗ f∗ are isomorphisms, which implies that f∗ and g∗ are isomorphisms too.

In particular

Corollary A24.9. If X is contractible, then its fundamental group is trivial.

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Lecture A25Fibrations and the fundamental group

Recall the homotopy extension property for a pair X,A), which we can alsothink of as a property of the inclusion map i : A → X . The pair has the HEP (or,as one also says, the inclusion i is a cofibration) if, given any homotopy of mapsA → Y , together with an extension of the initial data of the homotopy to a mapX → Y , we can always extend the homotopy to a homotopy of maps X → Y ,respecting the given initial data.

The notion of extension of a map can be expressed in diagrammatic termsusing the inclusion i:

X

A?i

OO

// Y

Here the solid arrows represent the given data and the dotted arrows representthe extension to be constructed. Dual to the notion of extension is the notion oflifting: suppose that p : E → B is a surjective map and f : Y → B is any map,then a lifting of f (over p) is a map g : Y → E that makes the following diagramcommute:

E

p

Y

>>

// B

in other words p g = f .

Definition A25.1. We say that p : E → B has the homotopy lifting property or isa fibration if, given any homotopy of maps Y → B together with a lifting of theinitial data of the homotopy over p, the entire homotopy can be lifted over p.

In other words, p has the HLP if, given the data expressed by the solid arrowsin the diagram below

Y × 0 _

// E

Y × [0, 1] //

<<

B

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the dotted arrow can be filled in in such a way as to make the diagram commuta-tive.

Remark A25.2. The homotopy extension property can be expressed in a similardiagrammatic way, but to do so we have to consider a homotopy of maps A→ Yas a single map from A to the space Y I of maps I → Y . We haven’t discussed thetopology of mapping spaces, so we can’t write down the full details of this, but itshould not be hard to figure out the general idea.

For now we are not going to worry too much about where fibrations mightcome from. Instead we will concentrate on what they might tell us about thefundamental group.

Definition A25.3. Let X be a space. If X is path-connected and π1(X, x0) = 1for some (and hence any) basepoint x0, we will say that X is simply connected.

Contractible spaces, for example, are simply connected (Corollary A24.9).But there are many other examples.

Exercise A25.4. Show that the n-sphere Sn is simply connected for n > 2. Hints:First show that it is enough if one knows that any loop in Sn is homotopic toone that is not surjective. Then prove this statement by any one of a numberof approximation techniques, e.g. by showing that any loop is homotopic to a“piecewise linear” loop made up of great circle arcs.

Definition A25.5. Let p : E → B be a fibration, and let B have a fixed basepointb0. The inverse image F = p−1b0 is called the fiber of the fibration.

Suppose now that p : E → B is a fibration and let b0 ∈ B be a base point. Letγ be a based loop in B and let x ∈ F be a point of the fiber F = p−1(b0). We mayconsider the path γ as a homotopy (of maps of a point into E) and, if we do this,γ and x together provide exactly the data for a homotopy lifting problem:

0 _

x // E

[0, 1]

γ //

γ

>>

B

Filling in the dotted arrow provides a lifting γ of the path γ, but this lifting isnot guaranteed to be a loop! All we know is that p(γ(1)) = γ(1) = b0, that is,

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γ(1) is a point of thee fiber F . It is not uniquely determined by this constructionbut we shall see in a moment that the path component of the fiber in which itlies is uniquely determined. The proof begins to build up the connection betweenfibrations and the fundamental group.

Lemma A25.6. Let p : E → B be a fibration, as above, and let γ be a based loopinB. If γ is nullhomotopic (that is, it represents the identity element in π1(B, b0)),then, for any lift γ : [0, 1]→ E of γ, γ(0) and γ(1) are in the same path componentof the fiber F .

(It’s obvious that γ(0) and γ(1) are in the same path component of E — theyare joined by the path γ, after all! — so the significance of the lemma is that apath joining them can be found that lies wholly in F .)

Proof. We apply the HLP to a homotopy between γ and the constant path. Leth(s, t) be such a homotopy with h(s, 0) = γ(s) and h(s, 1) = b0. Then h togetherwith γ provide the data for a homotopy lifting problem:

[0, 1]× 0 _

γ // E

[0, 1]× [0, 1] h //

H

<<

B

The bottom, right-hand and top sides of the map of the unit square defined by hare constant maps with value x0. Therefore the bottom, right-hand and top sidesof the lifted homotopyH are paths in the fiber F . The concatenation of these threeis a path in F from γ(0) to γ(1).

Corollary A25.7. If γ′, γ′′ are two liftings of the same loop, starting at the samepoint γ′(0) = γ′′(0), then their end points are in the same path component of F .

Proof. The concatenation (γ′)−1 ? γ′′ is a lifting of a nullhomotopic loop.

To help keep track of this business of path components let’s introduce somenotation.

Definition A25.8. For any topological space F , let π0(F ) denote the set of pathcomponents of F .

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Suppose that p : E → B is a fibration with fiber F , as discussed above. Forg ∈ π1(B, b0) and c ∈ π0(F ), let us define cg (read: “c acted on by g”) as follows:choose a based loop γ representing g, and a point x ∈ F representing c, let γ beany lift of γ starting at x, and define cg to be the path component containing γ(1).The lemma and corollary above show that this is well defined.

Proposition A25.9. The construction above defines an action of the fundamentalgroup π1(B, b0) on the set π0(F ). Moreover, if E is path connected, then thisaction is transitive; and if in addition E is simply connected, the action is free.

Let us explain the terminology. A (right) action of a group G on a set C isjust a map C × G → C, written (c, g) 7→ cg, such that ce = c and (cg)h = cgh.An action is free if cg = c implies g = e, and it is transitive if for all pairs c, c′ ofelements of C, there exists g ∈ G such that cg = c′. When a group action is bothfree and transitive, the mapping g 7→ cg, for fixed c, is a bijection from the groupG to the set C.

Proof. The action law (cg)h = cgh follows from the fact that a lifting of a concate-nation of loops is a concatenation of liftings of those loops. And ce = c followsfrom the fact that a constant loop can be lifted to a constant.

Suppose now that E is path connected. Then given any two points x, x′ ofthe fiber there is a path in E joining them. Composing with p gives a based loopin B which (tautologically) lifts to a path in E from x to x′. Thus the action istransitive.

Suppose additionally thatE is simply connected. Suppose that cg = c for somec ∈ π0(F ) and g ∈ π1(B, b0); this means that g is represented by a based loopγ in B which lifts to γ such that γ(0) and γ(1) are in the same path componentof F . Construct a loop β in E by concatenating γ with a path in F from γ(0) toγ(1). The composite p β is the concatenation of γ with a constant path, so it stillrepresents g ∈ π1(B, b0). But, on the other hand, β is nullhomotopic in E (sayby a homotopy H) because E is simply connected. Then p H is a nullhomotopyof the path p β, which therefore represents the identity element of π1(B, b0) asclaimed.

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Lecture A26Covering Spaces

Let p : E → B be a surjective map of topological spaces.

Definition A26.1. The map p above is called a covering map (and E is called acovering space of B) if there is a discrete topological space F (the fiber) such thateach point x ∈ B has an open neighborhood U for which there is a homeomor-phism p−1(U) ∼= U × F making the diagram

p−1(U)∼= //

p

##HHHHHHHHHHHHHHHHHU × F

pr1

U

commute.

In other words, a covering map is “locally the projection of a product witha discrete space”. Some authors (e.g. Hatcher) give a definition which appearsmore general than this, but it turns out that the definitions are equivalent for path-connected base spaces B. A neighborhood U with the property described in thedefinition will be called a trivializing neighborhood for the fibration.

Example A26.2. A homeomorphism is a covering map with 1-point fiber.

Example A26.3. The map R → S1 defined by x 7→ x mod 1 (if we think ofthe circle as [0, 1] with endpoints identified) or x 7→ e2πix (if we think of the unitcircle in C) is a covering map with fiber Z.

Example A26.4. The map S1 → S1 defined by x 7→ nx mod 1 (or z 7→ zn inthe complex pictures) is a covering map with fiber an n-point space.

Example A26.5. There are many different possible covering spaces of the “figure8” space; I’ll draw some pictures in class.

Example A26.6. Let B be a compact oriented surface of genus −2 (i.e. a thesurface of a 2-holed donut). Then B can be obtained by a suitable identificationof the edges of a regular octagon. One can tessellate the hyperbolic plane bysuch octagons with vertex angle of 45 degrees. From this tessellation we obtain acovering map from the plane to B.

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Proposition A26.7. A covering map is a fibration, i.e., it satisfies the homotopylifting property.

Proof. Remember what we have to show: for each space Y we can fill in the“homotopy lifting diagram”

Y × 0 _

f // E

Y × [0, 1] h //

H

<<

B

First of all let’s consider the case when Y is a single point (the “path liftingproperty”). By definition, B has an open cover consisting of trivializing neigh-borhoods. The inverse image of this open cover by the map h is then an opencover of the compact space [0, 1], which has a positive Lebesgue number bythe Lebesgue covering theorem. Partition [0, 1] into finitely many subintervals0 6 t1 6 t2 6 · · · 6 1 of length less than the Lebesgue number. We willconstruct the desired lifting by induction over these subintervals.

Suppose then than a liftingH has been constructed on [0, tk]. We have h([tk, tk+1]) ⊆U for some trivializing neighborhood U , and in particular we may identify p−1(U)with U × F where F is the (discrete) fiber. By connectedness, pr2(H(t)) must beconstant for any lift H . Therefore the unique continuous lift H of h on [tk, tk+1]that agrees with the lift already assumed to have been constructed on [0, tk] is

H(t) = (h(t), pr2(H(tk))) ∈ U × F ∼= p−1(U) ⊆ E.

By induction, we can construct H on [0, 1]. Indeed, we have done more thanclaimed: we have shown that there is a unique possible choice for H .

Now we return to the case of general Y . Because of the uniqueness of pathlifting, which we just proved, there is no question how to define H for general Y :namely, for each fixed y ∈ Y , H(y, t) should be the unique lifting of the path t 7→h(y, t) that begins at f(y). The only question is whether the function H definedby this process is continuous as a function of y and t. To see this, fix y0 ∈ Y andconsider the following fact: for each t ∈ [0, 1] there is a product neighborhoodVt ×Wt of (y0, t) such that h(Vt ×Wt) lies in a trivializing neighborhood in B.By compactness, finitely many of these product neighborhoods Vt × Wt cover

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y0 × [0, 1]. Let V be the intersection of the finitely many Vt’s that appear andlet ` be a Lebesgue number for the covering of [0, 1] by the finitely many Wt’s.Partition [0, 1] into finitely many subintervals 0 6 t1 6 t2 6 · · · 6 1 of lengthless than `. Notice that for each k, h(V × [tk, tk+1]) is contained in a trivializingneighborhood.

We’re going to prove by induction over subintervals that H is continuous onV × [0, 1]. Suppose then that we already know that H is continuous on V × [0, tk];in particular, H(y, tk) is a continuous function of y for y ∈ V . But by the formulaabove, for t ∈ [tk, tk+1],

H(y, t) = (h(y, t), π2(H(y, tk))) ∈ U × F ∼= p−1(U) ⊆ E,

and this is a continuous function of (y, t). So the induction is complete and wehave shown (in particular) that H is continuous at (y0, t) for all t ∈ [0, 1]. Sincey0 was arbitrary, this suffices to complete the proof.

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Lecture A27Applications

So, after all this machinery, we can finally compute π1(S1).

Theorem A27.1. The group π1(S1) is isomorphic to Z.

Proof. We make use of the covering space R → S1 described above (Exam-ple A26.3). The fiber F = Z is discrete and the total space E = R is contractible,so according to Proposition A25.9 the action of π1(S1) on π0(F ) = F is free andeffective. We can therefore define a bijection π1(S1) → Z by sending g ∈ G to0g ∈ Z. All that is necessary is to show that this is a group isomorphism.

For this, notice that S1 itself is a topological group. In any topological groupG, the group operation in π1(G) can be defined by pointwise multiplying paths(using the group operation in G), instead of by concatenating them. For the ex-pression

H(s, t) = γ′(max((2− t)s, 1)) · γ′′(min(0, 1− (2− t)(1− s)))

defines a homotopy from the concatenation of γ′ and γ′′ to their pointwise product.However, the lift of a pointwise product of paths (in S1) is clearly their pointwisesum (in R); in particular, the action of π1(S1) on the fiber Z satisfies 0gh = 0g+0h.This proves that the bijection π1(S1) → π0(Z) = Z is indeed an isomorphism ofgroups.

Exercise A27.2. Using a similar argument to the above, show that if G is anytopological group (not necessarily abelian), then π1(G) is an abelian group.

Remark A27.3. The identity map S1 → S1 is a generator of the cyclic groupπ1(S

1) = Z, as the above proof shows. The integer n ∈ π1(S1) is represented bythe map z 7→ zn, which “wraps” the unit circle n times around itself.

Here are some (standard) applications of this calculation.

Proposition A27.4. (No-retraction theorem) There is no retraction of the diskD2 = z ∈ C : |z| 6 1 onto its boundary circle S1 = z ∈ C : |z| 6 1.

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Proof. A retraction is a map r : D2 → S1 that is a left inverse to the inclusioni : S1 → D2, that is. the diagram

S1 p

AAA

AAAA

AAAA

AAAA

= // S1

D2

r

OO

should commute. But if this is so then the induced diagram

π1(S1)

= //

##GGGGGGGGGGGGGGGGπ1(S

1)

π1(D2)

r∗

OO

should commute also, and this is impossible since π1(S1) ∼= Z while π1(D2) istrivial.

Proposition A27.5. Any continuous map f : D2 → D2 has a fixed point.

Proof. Suppose for a contradiction that f : D2 → D2 has no fixed point. Thenfor each x ∈ D2, let u = u(x) be the unit vector (x − f(x))/‖x − f(x)‖. Letλ = λ(x) be the non-negative root of the quadratic equation

‖x+ λu‖2 = λ2 + 2λu · x+ ‖x‖2 = 1.

Then g(x) = x+λu depends continuously on x and it is the point on S1 obtainedby continuing the ray from f(x) through x until it hits the unit circle. Clearly, ifx ∈ S1, then g(x) = x. Thus g is a retraction of D1 onto S1, contradicting theprevious result.

Let a ∈ C and let γ : S1 → C \ a be a loop in the complex plane that doesnot pass through a. The map νa(z) = (z − a)/|z − a| is a homotopy equivalencebetween C \ a and the unit circle. The element of π1(S1) = Z represented bythe composite

νa f : S1 → S1

is called the winding number of f about a, and is denoted n(f ; a). It is a homotopyinvariant. (It does not depend on the choice of base point because π1(S1) = Z isabelian.)

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Lemma A27.6. (Rouche’s theorem) If f : S1 → C\0, g : S1 → C, and |g(z)| <|f(z)| for all z ∈ S1, then n(f ; 0) = n(f + g; 0).

Proof. The map h(z, t) = f(z) + tg(z) is a homotopy between f and f + g inC \ 0.

Theorem A27.7. (Fundamental theorem of algebra) Every non-constant polyno-mial over C has a root.

Proof. Suppose not and let p(z) = zn + an−1zn−1 + · · · + a0 be a polynomial

without root. Let pr : S1 → C \ 0 be the loop defined by pr(eiθ) = p(reiθ).These are homotopic loops in C \ 0. But for r = 0 the winding number n(pr, 0)is zero, whereas for large r we may write

p(reiθ) = f(reiθ) + g(reiθ), f(z) = zn

and simple estimates show |f(reiθ)| > |g(reiθ)|, so by Rouche’s theorem n(pr, 0) =n(zn, 0) = n. Contradiction.

Lemma A27.8. Let f : S1 → S1 be a loop with is antipodal, that is, f(−z) =−f(z). Then f represents an odd multiple of the generator of π1(S1).

Proof. Let g(z) = zf(z); then g(z) = g(−z) so g respects the equivalence rela-tion z ∼ −z. The quotient of S1 by this equivalence relation is again a copy of S1,and the quotient map is identified with the squaring map s(z) = z2. Consequently,g factors through s and thus g∗ factors through s∗:

π1(S1) = Z s∗ // π1(S

1) = Z // π1(S1) = Z .

But s∗ is multiplication by 2, so the winding number of g is even. The windingnumber of f is 1 less than that of g, so it is odd.

Theorem A27.9. (Borsuk-Ulam) For every continuous map f : S2 → R2 thereexists x ∈ S2 such that f(x) = f(−x).

Proof. Suppose not and let f be a counterexample. Then the map

g(x) = (f(x)− f(−x)|/‖f(x)− f(−x)‖

sends S2 to S1 and satisfies g(−x) = −g(x). Consider the restriction of g to theequator on S1. This is an antipodal map S1 → S1, so it represents an odd multipleof the generator; in particular it is not nullhomotopic. This is a contradiction sinceit factors through the simply connected space S2.

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Corollary A27.10. (Ham sandwich theorem) Any three bounded measurable sub-sets of R3 (e.g. ham, bread, cheese) can be simultaneously volume-bisected by acommon plane cut.

Proof. For each unit vector u ∈ S2 let Pu be the plane having u as normal vectorthat volume-bisects one of the ingredients, say the bread. (The existence of sucha plane follows from the intermediate value theorem; in the general case one hasto maneuver a little to deal with the possible non-uniqueness of such a plane; I’msweeping this under the rug.) Define a map S2 → R2 by sending u ∈ S2 to thevector (h(u), c(u)), where h(u) and c(u) are the fractions of the ham and cheese,respectively, that lie on the u side of the plane Pu. Then h(−u) = 1 − h(u)and c(−u) = 1 − c(u). The Borsuk-Ulam theorem gives us a point such thath(u) = h(−u) and c(u) = c(−u). This implies that h(u) = c(u) = 1

2; thus Pu

perfectly divides the ham and the cheese (as well as the bread).

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