1 The topology of metric spaces - Florida Atlantic...

Introductory Analysis IFall 2014

Notes on Metric Spaces

These notes are an alternative to the textbook, from and including”Closed Sets and Open Sets” (page 58) to and excluding Cantor Sets(page 95)

1 The topology of metric spaces

Assume M is a metric space with distance function d. I will depart a bit fromthe book’s notation; instead of using Mr(p), I will denote by B(p; r) the openball of center p ∈ M and radius r > 0; i.e,

B(p; r) = {q ∈ M : d(q, p) < r}.

The reason is that once we get to concrete metric spaces like Rn, it may be sillyto write (Rn)p(r). More importantly, I like my notation better. Incidentallya notation such as ”r > 0” means: ”r is a positive real number.” The basicdefinition is

Definition 1 A subset U of M is said to be open iff for every p ∈ U there isr > 0 such that B(p; r) ⊂ U . In other words, the set U is open if and only iffor every p ∈ U there is r > 0 such that if q ∈ U and d(q, p) < r, then q ∈ U .

Examples:

1. Let M be an arbitrary metric space. Then the empty set ∅ and M areopen.

2. Let M be an arbitrary metric space. Let p ∈ M and r > 0. Then B(p; r)is open.

3. Let M be a discrete metric space; i.e., d(p, q) = 1 whenever p = q. Thenevery subset of M is open.

4. Let M = Rm. The following sets are open:

U1 = {x = (x1, . . . , xm) ∈ Rm : x1 > 0},U2 = {x = (x1, . . . , xm) ∈ Rm : x1 > x2},U3 = {x = (x1, . . . , xm) ∈ Rm : |x1|+ · · ·+ |xm| < 1},

5. Let M = R. A subset U of R is open if and only if for each x ∈ R thereexist a, b ∈ R such that x ∈ (a, b) ⊂ U . In particular, all open intervalsare open.

You should be able to prove that the sets declared open in these examples areindeed open.

1 THE TOPOLOGY OF METRIC SPACES 2

Theorem 1 Let M be a metric space and let T be the family of all open subsetsof M . This family has the following properties:

1. ∅,M ∈ T .

2. If Uα ∈ T for all α ∈ A; A some index set, then∪

α∈A Uα ∈ T .

In words, T is closed under arbitrary unions. An equivalent way thisproperty is sometimes formulated is: Assume U is a collection of opensubsets of M (finite, infinite, countable or non-countable). Then

∪{U :

U ∈ U} is open.

3. If U, V ∈ T , then U ∩ V ∈ T . The intersection of two open sets is open.By induction, it follows that the intersection of any finite number of opensets is open.

Proof.

1. That ∅,M are open was already mentioned.

2. Let Uα ∈ T for all α ∈ A; A some index set, let U =∪

α∈A Uα ∈ T . Letp ∈ U . Then there exists α ∈ A such that p ∈ Uα. Since Uα is open, thereis r > 0 such that B(p; r) ⊂ Uα. Since Uα ⊂ U , B(p; r) ⊂ U .

3. Let U, V be open. If p ∈ U ∩ V , then p ∈ U and p ∈ V . Since U is open,p ∈ U implies that there exists r1 > 0 such that B(p; r1) ⊂ U . Similarly,because V is open, there exists r2 > 0 such that B(p; r2) ⊂ V . Let r =min(r1, r2). Then r > 0 and clearly B(p; r) ⊂ B(p; r1)∩B(p; r2) ⊂ U ∩V .

Exercise 1 Let M = R.

1. Prove that in R singleton sets are NOT open.

2. For n ∈ N let Un = {x ∈ R : −1/n < x < 1/n}. Prove that Un is openfor all n ∈ N, but

∩∞n=1 Un is not open. This shows that, in general, the

intersection of an infinite family of open sets may not be open.

Note: If M is any set, a family of subsets of M satisfying properties 1, 2, and3 of Theorem 1 is called a topology for M .

Definition 2 Let M be a metric space. A subset F of M is closed iff itscomplement M\F is open.

Examples:


1. Let M be an arbitrary metric space. Then the empty set ∅ and M areclosed.

2. Let M be an arbitrary metric space. Let p ∈ M and r ≥ 0. Then theclosed ball of center p, radius r; that is, the set {q ∈ M : d(q, p) ≤ r} isclosed.

3. A particular case of the previous result, the case r = 0, is that in everymetric space singleton sets are closed.

4. Let M be a discrete metric space; i.e., d(p, q) = 1 whenever p = q. Thenevery subset of M is closed.

5. Let M = Rm. The following sets are closed:

U1 = {x = (x1, . . . , xm) ∈ Rm : x1 ≥ 0},U2 = {x = (x1, . . . , xm) ∈ Rm : x1 ≥ x2},U3 = {x = (x1, . . . , xm) ∈ Rm : |x1|+ · · ·+ |xm| ≤ 11},

6. Let M = R. All closed intervals are closed.

As mentioned in the examples, in R open intervals are open and closed intervalsare closed. Intervals of the form (a, b], [a, b) with a, b ∈ R, a < b are neitheropen nor closed.

By taking complements, the following theorem is an immediate consequenceof Theorem 1

Theorem 2 Let M be a metric space.

1. ∅,M ∈ T are closed.

2. If Fα is a closed subset of M for all α ∈ A; A some index set, then∩α∈A Uα is closed.

In words, arbitrary intersections of closed subsets are closed. Equivalently:Assume F is a collection of closed subsets of M (finite, infinite, countableor non-countable). Then

∩{F : F ∈ F} is closed.

3. If F,G, are closed then F ∪G is closed. By induction, it follows that theUnion of any finite number of closed sets is closed.

Our next task will be to show that the basic metric properties can be de-scribed exclusively in terms of open sets. Another way of stating this is tosay that one never has to mention the distance function again, except if onewants to. It will be convenient to first have another topological concept, thatof neighborhood.

Definition 3 Let M be a metric space. Let p ∈ M . A subset V of M is saidto be a neighborhood of p in M (or just a neighborhood of p) iff there existsan open set U such that p ∈ U ⊂ V .


Exercise 2 V is a neighborhood of p if and only if there exists r > 0 such thatB(p; r) ⊂ V .

We also have the following simple lemma

Lemma 3 A subset U of a metric space is open if and only if it is a neighbor-hood of each of its points.

Proof. Assume first U is open and let p ∈ U . Then p ∈ U ⊂ U , showing U is aneighborhood of p. Conversely, assume U is a neighborhood of each of its points.Then for each p ∈ U , there is Wp open, p ∈ Wp ⊂ U . Let W =

∪p∈U Wp. By

property 2 of Theorem 1, W is open. Since Wp ⊂ U for all p ∈ U , we haveW ⊂ U . On the other hand, if p ∈ U , then p ∈ Wp ⊂ W , thus U ⊂ W . Itfollows that U = W is open.

Let us begin with sequences. We’ll state it in the form of a theorem, but itis fairly immediate.

Theorem 4 Let (pn) be a sequence in the metric space M ; let p ∈ M . Thefollowing statements are equivalent:

1. For each ϵ > 0 there exists N ∈ N such that n ≥ N implies d(pn, p) < ϵ.

2. For each ϵ > 0 there exists N ∈ R such that n > N implies d(pn, p) < ϵ.

3. For each open set U in M such that p ∈ U , there exists N ∈ R such thatn > N implies pn ∈ U .

4. For each neighborhood V of p, there exists N ∈ R such that n > N impliespn ∈ V .

5. If V is a neighborhood of p, then {n ∈ N : pn /∈ V } is finite.

As mentioned before, most of the equivalences are obvious. In lieu of aproof I’ll just make some comments about what could be the less obvious parts.Concerning the equivalence of 1 and 2, notice first that if the following statement”For each there exists N ∈ N such that n ≥ N implies ” (fill the blanksany way you wish), is equivalent to ”For each there exists N ∈ N suchthat n > N implies ” In fact, if something holds for n ≥ N , it also holdsfor n > N . But the converse is also true; if we can find N ∈ N so that aproperty holds for n > N , we can also find N ′ so it holds for n ≥ N ′; namelyN ′ = N +1. Maybe in words: if there is an element of N so the property holdswhen n is strictly larger than that element of N, then there is an element of N(namely the previous element plus 1) such that the property holds for n greaterthan or equal this element. Concerning now the equivalence of 1 and 3, it is aconsequence of N ⊂ R and the fact that if r ∈ R, there is n ∈ N, n > r. That3 implies 1 is a consequence of the fact that B(p; r) is open; the converse is aneasy consequence of the definition of open sets. A similar argument (sort of)


shows the equivalence of 3 and 4. To see 5 is equivalent to any and all of theprevious statements notice that if (pn) is a sequence and B is a set, to say thereexists N ∈ N such that pn ∈ B for all n ≥ N is equivalent to saying that theset of indices for which pn /∈ B is at most N − 1, so finite. Conversely, if theset of indices n for which pn /∈ B is finite, let M be the largest index for whichpn /∈ B. Then pn ∈ B for all n ≥ M + 1.

Convergence can be defined exclusively using the concept of open sets. Be-fore we continue, it may be convenient to define a few more topological concepts.

Definition 4 Let S be a subset of the metric space M . A point p ∈ M is alimit point of S iff there is a sequence (pn) of points of S converging to p. It isan accumulation or cluster point of S iff there is a sequence (pn) of pointsof S such that pn = p for all n, converging to p

Note: I use limit point as in our textbook. A lot of other textbooks (maybeall other textbooks?), if they define limit point at all they mean accumulationpoint.

To see the difference between the two concepts notice first the obvious: Acluster point is a limit point. The second obvious thing is that every point of aset is a limit point of the set. If p ∈ S, let pn = p for all n ∈ N to get a sequenceof points of S converging to p. Also fairly obvious is that if p is a limit point ofS and p /∈ S, then p has to be a cluster point of S. Can this happen? Yes, itcan.

Here are a few examples. A convenient notation is the following: Let S bea set. We denote by S′ the set of all cluster points of S; p ∈ S′ if and only ifthere exists (pn), pn ∈ S, pn = p for all n ∈ N such that limn→∞ pn = p.Examples.

1. Let M be a discrete metric space. Then S′ = ∅ for all S ⊂ M .

2. In R2, consider the set S = {(x, y) ∈ R2} such that x < y. ClaimS′ = {(x, y) ∈ R2 : x ≤ y}. There are more elegant ways to prove thisthen how I am about to do it, but at this level brute force may be necessary.Let us see first that S′ ⊂ {(x, y) ∈ R2 : x ≤ y}. Let (x0, y0) ∈ S′. there isthen a sequence ((xn, yn)) in S converging to (x0, y0). I’ll use a calculusresult; if xn → x0, yn → y0 and xn < yn for all n, then x0 ≤ y0, to conclude(x0, y0) ∈ S′. Notice that we did not have to assume (xn, yn) = (x0, y0)for all n ∈ N. Conversely, assume x0 ≤ y0. Then set xn = x0− 1

n , yn = y0for n ∈ N. Clearly (xn, yn) ∈ S, (xn, yn) = (x0, y0) for all n ∈ N, and(xn, yn) → (x0, y0). It follows that (x0, y0) ∈ S′.

3. In R consider the set S = (0, 1]∪{2}. It is then easy to see that S′ = [0, 1].The point 2 is a limit point of S, but not an accumulation point of S.

4. In R consider the set { 1n : n ∈ N}. Then S′ = {0}.


5. In R consider the subset Z of all integers. Then Z ′ = ∅.

The fact is that the useful concept is that of cluster point. Limit points areless interesting, except when they are cluster points. But . . .

Here is a another quite simple result; let’s call it a proposition for variety’ssake.

Proposition 5 Let M be a metric space, let p ∈ M and let S ⊂ M .

1. The following statements are equivalent.

(a) p is a limit point of S.

(b) If V is a neighborhood of p, then V ∩ S = ∅. (Each neighborhood ofp has non-empty intersection with S.)

(c) If r > 0, then B(p; r) ∩ S = ∅

2. The following statements are equivalent.

(a) p is a cluster point of S.

(b) If V is a neighborhood of p, then V ∩(S\{p}) = ∅. (Each neighborhoodof p contains a point of S different from p.)

(c) If r > 0, then B(p; r) ∩ (S\{p}) = ∅.(d) If V is a neighborhood of p, then V ∩ S is an infinite set.

Proof. I’ll only prove the equivalence of the statements about being a clusterpoint; the proof of the equivalence of the statements about being a limit pointis similar, a bit easier because one doesn’t have to worry about not being equalto p. And, of course, there is no fourth statement.a) ⇒ b). Assume p is a cluster point of S and let V be a neighborhood of p.By definition of cluster point, there is (pn), pn ∈ S, pn = p for all n ∈ N, suchthat pn → p as n → ∞. Since V is a neighborhood of p, there is N ∈ N suchthat pn ∈ V for n ≥ N . Then (for example) pN ∈ V ∩ (S\{p}).b) ⇒ c). Since B(p; r) is a neighborhood of p for each r > 0, this is immediate.c) ⇒ d). Assume that for each r > 0, B(p; r) ∩ (S\{p}) is not empty. Let Vbe a neighborhood of p. There is then r0 > 0 such that B(p; r0) ⊂ V . Letr1 = min(r0, 1); then r1 > 0, B(p; r1) ⊂ B(p′r0) ⊂ V , and by assumption c),there is p1 ∈ B(p; r1)∩S such that p1 = p. Then d(p1, p) > 0 (notice that this isthe first time in quite a while that we actually mention the distance function),let r2 = min(d(p1, p), 1/2). Then r2 > 0 and there exists p2 ∈ B(p; r2) ∩ S suchthat p2 = p. In addition, p2 = p1 since d(p2, p) < d(p1, p). Assume now that forsome n ≥ 1 we have found points p1, . . . , pn ∈ V ∩ S such that pi = pj if i = jand such that 0 < d(pi, p) < 1/i for i = 1, 2, . . . , n. We have done this alreadyfor n = 2. Let

rn+1 = min{d(p1, p), . . . , d(pn, p),1

n+ 1}.

Then rn=1 > 0, hence there exists pn+1 ∈ B(p; rn+1) ∩ S such that pn+1 = p.


What we have done here is to show inductively the existence of a sequence(pn) such that the set {pn : n ∈ N} is an infinite set contained in V ∩S. So weare done. As a bonus we also have limn→∞ pn = p and pn = p for all n. Thatis, as a bonus we proved in passing: If p is a cluster point of S, there exists asequence (pn) of points of S such that pn = pm if n = m, pn = p for all n, (pn)converging to p.d) ⇒ a). Let n ∈ N.The set B(p; 1/n) is a neighborhood of p, hence B(p; 1/n)is an infinite set, hence it must contain points other than p. Lots of them,an infinite number of them. Select one of these points, call it pn. Thuspn ∈ B(p; 1/n) ∩ S, pn = p. The sequence (pn) clearly converges to p, pn = pfor all n ∈ N, thus p is a cluster point of S.

We can now give a convenient characterization of closed sets in terms ofcluster or limit points. It is again a TFAE theorem.

Theorem 6 Let S be a subset of the metric space M . The following statementsare equivalent.

1. S is closed.

2. S contains all of its limit points.

3. S′ ⊂ S (S contains all of its cluster points).

Proof. 1) ⇒ 2 Assume S is closed and let p ∈ M\S. Since M\S is open, asequence of points converging to p must have all but a finite number of termsin M\S, thus it cannot be a sequence of points of S. It follows that p is not alimit point of S. Therefore, if p is a limit point of S, p ∈ S.2) ⇒ 3) Since cluster points are limit points, this is obvious.3) ⇒ 1) Assume S′ ⊂ S. Let U = M\S, we need to show U is open. Let p ∈ U .Then p /∈ S, hence p /∈ S′; p is not a cluster point of S. By Theorem 5, thereexists a neighborhood V of p such that V ∩ (S\{p}) = ∅. Since p /∈ S, this isequivalent to V ∩ S = ∅. Thus V ⊂ U . It is (or should be) clear that supersetsof neighborhoods are neighborhoods, thus U is a neighborhood of p. Since p isan arbitrary point of p, this proves U open.

Definition 5 Let S be a subset of the metric space S. The closure of S,denoted by S, is the intersection of all closed subsets of M containing S; insymbols,

S =∩

{F : F is a closed subset of M and S ⊂ F}

Some comments are in order. First of all, because M itself is closed and con-tains (of course!0 all its subsets, the family {F : F is a closed subset of M and S ⊂F} is never empty. The definition of S makes sense. Second, since S is the in-tersection of a family of sets all of which contain S, we have S ⊂ S. Third, sincethe intersection of closed sets is closed, S is a closed set. It is also easy to see


that S = S if and only if S is closed. In fact, if S = S, then S is closed because,as mentioned, S is closed. Conversely, if S is closed, then S is a member of thefamily of closed sets containing S, hence S ⊂ S. The converse inclusion alwaysholds, thus S = S.

One problem with this definition, however, is that it is fairly useless todetermine closures because to find S we need to know the family of closed setscontaining S and, unfortunately, barS is one of these sets. To find S we needto know S. Fortunately, there are intrinsic definitions. We state them as atheorem.

Theorem 7 Let S be a subset of the metric space M . Then

S = S ∪ S′ = {p ∈ M : p is a limit point of S}.

Proof. We see first that S ∪ S′ is closed. This is equivalent to proving thatM\(S ∪ S′) is open. If p ∈ M\(S ∩ S′), then p /∈ S, p /∈ S′ and we are exactlyin the situation of the proof of 3) → 1) of Theorem 6. The same proof showsM\(S ∪ S′) is open. Since S ∪ S′ is closed, we have S ⊂ S ∪ S′. Conversely,let p ∈ S ∪ S′ and let F be a closed set containing S. Since it is obvious thatS′ ⊂ F ′ and F ′ ⊂ F since F is closed, we see that S ∪ S′ ⊂ F for all closedsets F ⊃ S. Thus S ∪ S′ is in the intersection of all these closed sets, henceS ∪ S′ ⊂ S.

Concerning the characterization in terms of limit points, notice that {p ∈M : pis a limit point of S} = S ∪ S′. Since cluster points are limit points, theinclusion ⊃ is obvious. But so is the converse inclusion since a limit point notin S is a cluster point of S.As a simple but useful corollary we get:

Corollary 8 Let S be a subset of the metric space M . Then p ∈ S if and onlyif V ∩ S = ∅ for all neighborhoods V of p. Equivalently, p ∈ S if and only ifV ∩B(p; r) = ∅ for all r > 0.

It might be a moment to do some exercises. It is important to do simple exerciseslike those that follow to get a feeling for all these objects.

Exercise 3 Let M = Rm and let r > 0, x ∈ Rm. Prove that B(x; r) = {y ∈Rm : |y − x| ≤ r}.

Exercise 4 Let M = Rm. Prove that Qm = Rm. A subset of a metric spacewhose closure equals the whole space is said to be dense in the space. Thus Qm

is dense in Rm. In particular Q is dense in R.

Exercise 5 In R2, let S = {(x, sin(1/x)) : x ∈ R, x = 0}. Prove that S =S ∪ {(0, y) : −1 ≤ y ≤ 1}.

Exercise 6 It seems reasonable to assume that the result of Exercise 3 is validin every metric space. But that would be wrong. Assume M is a metric spacewith the discrete metric, and assume that M has at least two points. Let p ∈ M .Determine B(p; 1) and show that B(p; 1) = {q ∈ M : d(q, p) ≤ 1}.


Exercise 7 Assume (pn) is a sequence in a metric space and let S be the rangeof the sequence, i.e., S = {on : n ∈ N}. Prove: If q is a cluster point of S, thenthere is a subsequence of )pn) converging to q. Is the converse true? That is, ifq is the limit of a subsequence of (pn), is p a cluster point of S?

Here are a few other properties of the closure:

Proposition 9 Let M be a metric space.

1. If S ⊂ T ⊂ M , then S ⊂ T .

2. If S is a subset of T , then S = S.

3. If S, T are subsets of M , then S ∪ T = S ∪ T .

Proof. The first statement is obvious, any definition one wants to use. Thesecond statement is also obvious; the closure of a closed set is the set, and S isclosed. For the third statement, notice first that A∪ B is a closed set containingA ∪ B, thus it contains A ∪B. Conversely, A ∪B is a closed set containingA ∪ B, since A ⊂ A ∪ B, A ⊂ A ∪B, hence A ⊂ A ∪B. Similarly, B ⊂ A ∪B,thus A ∪ B ⊂ A ∪B.As a corollary we get: If S1, . . . , Sm ∈ M (a metric space), then

∪mi=1 Si =∪m

i=1 Si.

Exercise 8 Provide an example of a metric space M and a countable family{Sn}n∈N of subsets of M such that∪

n∈N

Un =∪n∈N

Sn.

Here are two final topological concepts (for now).

Definition 6 Let S be a subset of the metric space M . The interior of S isthe set S◦ of all points p of S such that there exists r > 0 with B(p; r) ⊂ S.That is: p ∈ S◦ if and only if there exists r > 0, B(p, r) ⊂ S.

Equivalently: p ∈ S◦ if and only if S is a neighborhood of p. It is also easy tosee that S◦ is the largest open subset of S; that is, if V is open and V ⊂ S,then V ⊂ S◦. Also equivalently

S◦ =∪

{V : V is open and V ⊂ S}.

If we compare this last equivalent way of defining the interior and compare withthe definition of closure, we get at once thanks to our friend DeMorgan,

S◦ = M\(M\S).

We have talked of points very much inside a set, and points outside of it.We need also to talk of points that are halfway between being in and out.

2 SUBSPACES 10

Definition 7 Let M be a metric space, and S ⊂ M . The boundary of S isthe set

∂S = S ∩ (M\M).

Several equivalent definition are: ∂S = S\S0; ∂S is the set of all points psuch that each neighborhood of p has non-empty intersection with S and M\S.Equivalently, p ∈ ∂S if and only if for each r > 0, B(p; r) ∩ S = ∅ = B(p; r) ∩(M\S).

2 Subspaces

As mentioned in class, if M is a metric space, if N ⊂ M , then N becomes ametric space by simply restricting the distance function of M to points of N .We then say N is a (metric) subspace of M . All the topological concepts makesense in N , of course, and we should know how they relate to the similar conceptin M . So suppose S is a subset of N and it is closed in the metric space N . Is itclosed in M? Conversely? Since we now have two spaces, I will write BM (p; r)for the open ball of center p, radius r > 0 in M , and BN (p; r) for the relatedobject in N (which only makes sense if p ∈ N). A first trivial observation isthat if p ∈ N , r > 0, then

BN (p, r) = BM (p, r) ∩N.

This has as consequence that a subset U of N is open in N if and only ifU = W ∩ N for some W open in M . In fact, assume first W is open in M .Let p ∈ W ∩ N . Because p ∈ W , there is r > 0 such that BM (p; r) ⊂ W ,implying that BN (p; r) = BM (p; r) ∩N ⊂ W ∩N . This proves W ∩N is openin N . Conversely, let U be open in N . For each p ∈ U there is rp > 0 such thatBN (p; rp) ⊂ U . The set W =

∪p∈U BM (p; rp) is open in M and W ∩N = U .

A couple of immediate consequences are that if U ⊂ N and U is open in M ,then it is open in N . If N is open in M , then being open in N is equivalent tobeing open in M for subsets of N . One gets similar/exoected results for closedsets.

Suppose F ⊂ N . then F is closed in N (by definition) if and only if N\Fis open in N , which happens if and only if there is W open in M such thatN\F = W ∩N . Observing that N\(W ∩N) = (M\W )∩N , we see that if F isclosed in N , then F = G ∩N , G closed in M . If F = G ∩N , with G closed inM , then N\F = N\(G∩N), showing that N\F is open in N . Thus F is closedin N if and only if there is G closed in M such that F = G ∩N . In particular,if N is closed in M then a subset F of N is closed in N if and only if it is closedin M .

Exercise 9 Let M be a metric space and let N be a subspace of M . Let S ⊂ N .

Prove: The

closureinteriorboundary

of S in N is the intersection of the

closureinteriorboundary

3 EQUIVALENCE OF METRICS 11

of S in M intersection N .

3 Equivalence of Metrics

We saw in a homework that two distance functions for the same set are equiv-alent iff convergence for one is equivalent to convergence for the other one. Anicer definition is:

Definition 8 Assume d1, d2 are distance functions for a set M . They areequivalent, and we write d1 ∼ d2, iff they generate the same tropology.

In other words, a set is open for d1 if and only if it is open for d2. Let us writefor p ∈ M , r > 0,

B1(p; r) = {q ∈ M : d1(p, q) < r}, B2(p; r) = {q ∈ M : d2(p, q) < r}.

Assume d1 ∼ d2. Let r > 0, p ∈ M . Since B1(p; r) is open with respect to d1and p ∈ B1(p; r), the fact that it is also open with respect to d2 implies thatthere exists r′ > 0 such that B2(p, r

′) ⊂ B1(p, r). The same must hold trueif we switch the roles of d1, d2. We see thus that if d1 ∼ d2, for every r > 0,p ∈ M , there exists r′ > 0 such that B2(p, r

′) ⊂ B1(p, r), and for every r′ > 0there is r′′ > 0 such that B1(p, r

′′) ⊂ B2(p, r′).

Conversely, suppose this property holds. If U is open with respect to d1,let p ∈ U . Then there is r > 0 such that B1(p, r) ⊂ U ; by the property thereexists r′ > 0 such that B2(p, r

′) ⊂ B1(p, r), hence B2(p, r′) ⊂ U . Since p was

arbitrary in U , we see that U is open with respect to d2. Since the propertyis symmetric in d1, d2; it also follows that being open with respect to d2 is thesame as being open with respect to d1. Concerning sequences, if two metricsdefine the same open sets, since convergence can be defined exclusively in termsof open sets, convergence will be the same for both metrics. On the other hand,if two metrics have the same notion of convergence then given any subset of thespace, the notion of limit point will be the same for both metrics. It followsthat the concept of being a closed set is the same for both metrics, hence alsothe notion of being the complement of a closed set; i.e., being an open set. Wethus have

Theorem 10 Let M be a set and let d1, d2 be distance functions in M . Thefollowing statements are equivalent.

1. A subset of M is open with respect to d1 if and only if it is open withrespect to d2.

2. For each p ∈ M , r > 0 there exists r′ > 0 such that B2(p; r′) ⊂ B1(p; r),

and for each p ∈ M , r > 0 there exists r′ > 0 such that B1(p; r′) ⊂

B2(p; r).

3. A sequence in M converges with resepct to d1 if and only if it convergeswith resepct to d2.

4 AN EXCURSION INTO SET THEORY 12

4 An Excursion into Set Theory

Let M,N be sets (not necessarily metric spaces) and let f : M → N . If A ⊂ M ,the image of A under f is the set

f(A) = {y ∈ B : ∃ x ∈ M,y = f(x)};

briefly, f(A) = {f(x) : x ∈ A}. The statement y ∈ f(A) is equivalent toy = f(x) for some x ∈ A. In particular, f(M) is the range of f .

If B is a subset of N , one can define the pre-image of B under f . The authorof our textbook denotes it by fpre(B) and gives some excellent reasons why thisnotation is preferable to the usual one, which is f−1(B), except that it isn’treally so much the usual notation as the universal notation, used everywhere Iknow of, except our textbook. So I will stick with the conventional notation.The pre-image of B under f is the set defined by

f−1(B) = {x ∈ M : f(x) ∈ B}.

Notice that x ∈ f−1(B) does NOT imply x = f−1(y) for some y ∈ B; f−1 asa function on B might not even be defined. (Actually, if f−1 is defined, it doesmean that, but that requires a proof.) x ∈ f−1(B) means precisely the same asf(x) ∈ B. Here are some examples.Examples.

1. Let f : R → R be defined by f(x) = 1 for all x ∈ R. Then f(A) = {1} ifA ⊂ R and A = ∅;

f−1(B) =

{∅, if 1 /∈ B,R, if 1 ∈ B.

2. Let f : R → R be given by f(x) = sinx for all x ∈ R. Then f(R) = [−1, 1];f−1{0} = {kπ : k ∈ Z}.

3. If A ⊂ M and A = ∅, then f(A) = ∅. In fact, x ∈ A implies f(x) ∈ f(A).However, if B ⊂ N one can have f−1(B) = ∅ even if B is not empty.f−1(B) = ∅ if and only if B ∩ f(M) = ∅.

Let us prove some basic facts about these operations. A bit of common sensehelps.

Proposition 11 Let f : M → N .

1. If Bλ ⊂ N for λ ∈ Λ (Λ an index set; finite or infinite), then

f−1

(∪λ∈Λ

Bλ

)=

∪λ

∈ Λf−1(Bλ) (1)

f−1

(∩λ∈Λ

Bλ

)=

∩λ

∈ Λf−1(Bλ) (2)

4 AN EXCURSION INTO SET THEORY 13

2. If B,C are subsets of N , then

f−1(C\B) = f−1(C)\f−1(B). (3)

3. If B ⊂ N , thenf−1(N\B) = M\f−1(B). (4)

Proof. These properties prove themselves; if one simply does the obvious (anduses the definitions as provided). To prove (1) notice that x ∈ f−1

(∪λ∈Λ Bλ

)if and only if f(x) ∈

∪λ∈Λ Bλ, which happens if and only if there is λ ∈ Λ

such that f(x) ∈ Bλ; this is equivalent to saying that there is λ ∈ Λ such thatx ∈ f−1(Bλ), which occurs if and only if x ∈

∪λ ∈ Λf−1(Bλ). The proof of

(2) is almost identical; replace “there is” or “there exists” by “for all” (or “foreach”):

x ∈ f−1(∩

λ∈Λ Bλ

)if and only if f(x) ∈

∩λ∈Λ Bλ, which happens if and

only if for each λ ∈ Λ it holds that f(x) ∈ Bλ; this is equivalent to saying thatfor each λ ∈ Λ, x ∈ f−1(Bλ), which occurs if and only if x ∈

∩λ ∈ Λf−1(Bλ).

The proof of (3) is just as immediate:

x ∈ f−1(C\B) ⇔ f(x) ∈ C\B ⇔ f(x) ∈ C and f(x) /∈ B

⇔ x ∈ f−1(C) and x /∈ f−1(B) ⇔ x ∈ f−1(C)\f−1(B).

Property (14) is a consequence of (3); take C = N and notice that f−1(N) = M .

Are similar properties true for images? That is, if Aλ ⊂ M for λ ∈ Λ, is it truethat f applied to the union of the Aλ’s is the union of the f(Aλ); in other words,is the image of a union the union of the images? Is the image of an intersectionthe intersection of the images? The answer is yes for the union, no for theintersection. Proofs are easy. Concerning intersections, suppose A,B ⊂ M . Iff is not one-to-one, there could be a ∈ A\B, b ∈ B\A, such that f(a) = f(b).Then y = f(a) is in f(A)∩f(B), but there is no reason to assume that it also isin f(A∩B). In general, f(A∩B) ⊂ f(A)∩ f(B), but f(A∩B) = f(A)∩ f(B)in general. Equality holds if f is one-to-one. Here are a few more properties, asan exercise. If you understood the definitions, you should have no trouble withthis exercise. If you have any trouble, review the definitions carefully.

Exercise 10 Assume M,N are sets and f : M → N .

1. Let A ⊂ M . Prove: A ⊂ f−1 (f(A)). Show, by a counterexample, thatin general equality does not hold; that is, provide an example in whichA = f−1 (f(A)). After doing this, give a condition on f such that equalitywill hold whenever f satisfies that condition.

2. Let B ⊂ N . Prove: f(f−1(B)

)⊂ B. Show, by a counterexample, that in

general equality does not hold, but give a condition on f such that equalitywill hold whenever f satisfies that condition.

5 CONTINUITY 14

Finally, assume f : M → N is one-to-one and onto so that f−1 : N → M exists.Then it would seem we have two interpretations of f−1(B) for B ⊂ N : As thepre-image of B under f or as the direct image of B under f−1. Fortunately bothinterpretations are the same in this case. Suppose x ∈ f−1(B). If we interpretf−1(B) as the pre-image of B,this means f(x) ∈ B. If we interpret f−1(B) asthe image of B under f−1, this means there exists y ∈ B such that x = f−1(y),hence f(x) = y ∈ B. And, of curse, the only possible value of y is f(x) in thiscase.

5 Continuity

In this section we will assume M,N are metric spaces. I will use the symbol dfor the distance function in M as well as for the distance function in N ; thisshould not cause any confusion; one should know from the context which one isbeing referred to. Anyway, after a while, there probably won’t be any explicitmention of distance functions. I will however denote the open ball of radius rcenter q by BM (q, r) if we are in M , by BN (q, r) in N .

Definition 9 Let f : M → N and let p ∈ M . We say f is continuous at piff for every ϵ > 0 there is δ > 0 such that d(f(q), f(p)) < ϵ whenever q ∈ Mand d(q, p) < δ.

Definition 10 Let f : M → N . We say f is continuous (or, continuous onM)iff f is continuous at each p ∈ M .

We already went over a lot of this in class. A characterization of continuity interms of sequences is given by the following result.

Theorem 12 Let f : M → N and let p ∈ M . Then f is continuous at p ifand only if the following property holds: Whenever (pn) is a sequence in Mconverging to p, the sequence (f(pn)) is a sequence in N converging to f(p).

Proof. Assume first f is continuous at p. Let (pn) be a sequence inM convergingto p. To prove (f(pn)) converges to f(p), let ϵ > 0 be given. By the continuityof f at p, there exists δ > 0 such that d(f(q), f(p)) < ϵ whenever q ∈ M andd(q, p) < δ. Because δ > 0 and limn→∞ pn = p, there is N ∈ N such that n ≥ Nimplies d(pn, p) < δ, hence d(f(pn), f(p)) < ϵ.

Conversely, assume the property holds. We proceed by contradiction. As-sume f is not continuous at p. There exists then ϵ > 0 for which no δ > 0 works,thus for every δ > 0 there exists pδ ∈ M , d(pδ, p) < δ and f(pδ), f(p)) ≥ ϵ. Then(p1/n)n∈N is a sequence in M converging to p; since d(f(p1/n, f(p)) ≥ ϵ for alln; f(p1/n) does not converge to f(p), contradicting our hypothesis.Moving a bit away from explicit mention of the distance functions, we have thefollowing simple lemma. I prove it because perhaps we all need to become a bitmore familiar with this way of proving things.

Lemma 13 Let f : M → N , let p ∈ M . The following are equivalent.

5 CONTINUITY 15

1. f is continuous at p.

2. For each neighborhood W of f(p), there exists a neighborhood V of p suchthat f(V ) ⊂ W .

3. If W is a neighborhood of f(p), then f−1(W ) is a neighborhood of p.

Proof. 1) ⇒ 2) Assume f is continuous at p and assume W is a neighborhoodof f(p). There is then ϵ > 0 such that BN (f(p); ϵ) ⊂ W . By continuity,there is δ > 0 such that q ∈ M , d(q, p) < δ implies d(f(q), f(p)) < ϵ, thusf(q) ∈ BN (f(p); ϵ) ⊂ W . Take V = BM (p; δ); then W is a neighborhood of pin M and if z ∈ f(V ), then z = f(q), some q ∈ V = BM (p, δ), thus z = f(q),d(q, p) < δ, hence d(z, f(p)) < ϵ and z ∈ W . That is, f(V ) ⊂ W .2) ⇒ 3) Assume 2) Let W be a neighborhood of f(p) in N . By 2), there exists aneighborhood V of p, with f(V ) ⊂ W . Applying f−1 and using (10), part 1, V ⊂f−1(f(V )) ⊂ f−1(W ). Since supersets of neighborhoods are neighborhoods, weare done.3) ⇒ 1) Assume 3) Let ϵ > 0 be given. Then BN (f(p); ϵ) is a neighborhood off(p), hence f−1(BN (f(p); ϵ) is a neighborhood of p, hence contains an open setcontaining p, which in turn contains an open ball centered at p. That is, thereis δ > 0 such that BM (p; δ) ⊂ f−1(BN (f(p); ϵ). It follows that if q ∈ M andd(q, p) < δ, then d(f(q), f(p)) < ϵ.

An important but simple result is the following.

Theorem 14 Let M,N,P be metric spaces; let f : M → N , g : N → P . Letp ∈ M . If f is continuous at p and g is continuous at f(p), then the compositiong ◦ f is continuous at p

Proof. Let W be a neighborhood of g ◦ f(p) = g(f(p)) in P . Since g is con-tinuous, g−1(W ) is a neighborhood of f(p), thus (because f is continuous),f−1

(g−1(W )

)is a neighborhood of p. But f−1

(g−1(W )

)= (g ◦ f)−1(W ) and,

since W was an arbitrary neighborhood of g◦f(p), we proved that the pre-imageof every neighborhood of g ◦f(p) under g ◦f is a neighborhood of p. Continuityof g ◦ f at p follows.

An obvious corollary is

Corollary 15 Let M,N,P be metric spaces; let f : M → N , g : N → P . Ifboth f and g are contibuous, then g ◦ f is continuous.

Here are a few exercises, to keep practising.

Exercise 11 Let M,N be metric spaces and let p ∈ M be an isolated pointof M ; that is p ∈ M but p /∈ M ′. Equivalently, there is r > 0 such thatBM (p, r) = {p}. Prove: If f : M → N , then f is continuous at p.

6 ALGEBRA OF CONTINUOUS FUNCTIONS 16

Exercise 12 Let M,N be metric spaces, f : M → N , A ⊂ M . The restric-tion of f to A is the function fA : A → N defined by fA(p) = f(p) for p ∈ A.It makes sense to ask about the continuity of fA since A is a metric space withthe metric of M . Prove: If p ∈ A and f is continuous at p,then fA is continu-ous at p. Show (by a counterexample) that the converse is false; one can havep ∈ A ⊂ M , fA continuous at p, but f is not continuous at p.

Exercise 13 Let M,N be metric spaces, f : M → N , A ⊂ M . Let p ∈ A.Assume there exists U open in M such that p ∈ U ⊂ A. Prove: If fA iscontinuous at p, then f is continuous at p.

Exercise 14 Let M,N be metric spaces, f : M → N . Let A,B ⊂ M be suchthat M = A∪B and assume p ∈ A∩B. Prove: f is continuous at p if and onlyif both fA, fB are continuous at p.

We are already characterizing continuous functions in terms of topologicalconcepts. For functions continuous at all points we have some nicer characteri-zations. The following theorem will be used with some frequency.

Theorem 16 Let M,N be metric spaces and let f : M → N . The followingstatements are equivalent.

1. f is continuous.

2. f−1(U) is open in M whenever U is open in N ; in words: The inverseimage of open sets is open.

3. f−1(F ) is closed in M whenever F is closed in N ; in words: The inverseimage of closed sets is closed.

Proof. 1) ⇒ 2) Assume f is continuous and let U be open in N . Let p ∈ f−1(U).Then f(p) ∈ U . Since U is open, U is a neighborhood of f(p) thus, by Lemma13, f−1(U) is a neighborhood of p. Since p ∈ U was arbitrary, itfollows thatf−1(U) is a neighborhood of each on of its points hence, by Lemma 3, it is open.2) ⇒ 3) Assume the inverse image under f of all open sets is open and assumeF is a closed subset of N . Then F = N\U where U is open; thus f−1(F ) =M\f−1(U), proving F is closed since f−1(U) is open.3) ⇒ 2) TO prove that 3) implies 2) all one needs todo is to exchange the words“open” and “closed” in the proof that 2) implies 3).2) ⇒ 1) Assume the inverse image of all open sets is open. Let p ∈ M . Let Wbe a neighborhood of f(p). There is then U open in N , f(p) ∈ U ⊂ W . Thenp ∈ f−1(U) ⊂ f−1(W ). Since f−1(U) is open, f−1(W ) is a neighborhood of p.By Lemma 13, f is continuous at p.

6 Algebra of continuous functions

Assume f : M → Rm. Then f(p) is an m-vector for each p ∈ M and we candefine functions f1, f2, . . . , fm : M → R by fk(p) is the k=th component of p.In other words, f(p) = (f1(p), . . . , fm(p)). We write f = (f1, . . . , fm). We have


Proposition 17 Let M be a metric space and let f = (f1, . . . , fm) : M → Rm.Then f is continuous at p ∈ M if and only if each fk is continuous at p;k = 1, . . . ,m. It follows that f is continuous if and only if each fk is continuous.

Proof. For once, the proof might be easier using the distance functions. Assumefirst f is continuous at p. Let k ∈ {1, . . . ,m}; we want to prove fk is continuousat p. Let ϵ > 0 be given. Because f is continuous, there exists δ > 0 such that if

q ∈ M and d(q, p) < δ, then |f(q)− f(p)| < ϵ. Since |xk| ≤ |x| =√∑n

j=1 |xj |2

for all x = (x1, . . . , xm) ∈ Rm, q ∈ M , d(q, p) < δ implies |fk(q) − fk(p)| ≤|f(q)− f(p)| < ϵ.

Conversely, assume each fk is continuous at p. To prove f is continuous atp, let ϵ > 0 be given. Since then ϵ/

√n > 0, there exist δ1, . . . , δm > 0 such that

q ∈ M ; d(q, p) < δk implies |fk(q) − fk)(p)| < ϵ/√m. Let δ = min1≤k≤m δk.

Then δ > 0 and d(q, p) < δ implies d(q, p) < δk for k = 1, . . . , n, hence

|f(q)− f(p)| =

√√√√ m∑k=1

(fk(q)− fk(p))2 <

√√√√ m∑k=1

ϵ2

m= ϵ.

Generally speaking, a function from a metric space (or any set, actually) toRm is just a convenient way to deal with m functions at once. Once againgenerally speaking (there could be exceptions), if you have f : M → Rm (Mnot necessarily a metric space, just a set) you can write f = (f1, . . . , fm) wherefk : M → R for k = 1, . . . ,m. Chances are that given a property such an f canhave it is either defined by, or a theorem states it, that f : M → Rm has thatproperty if and only if each fk : M → R has the property. We just saw it is sofor continuity.

I will now define a number of functions. The symbols I use will probablybe reused later on with different interpretations. What I mean is that thesesymbols are just one-use symbols. Here are the functions:

• µ : Rm × Rm → Rm, defined by µ(x, y) = x+ y if x, y ∈ Rm.

• ν : R× Rm → Rm, defined by µ(c, x) = cx if c ∈ R and x ∈ Rm.

• λ : Rm × Rm to R, defined by λ(x, y) = x · y =∑m

i=1 xiyi if x =(x1, . . . , xm), y = (y1, . . . , ym) ∈ Rm.

• abs : Rm → R, defined by abs(x) =√x21 + · · ·+ x2

m if x = (x1, . . . , xm) ∈Rm.

All of these functions are continuous. For a proof we need to know a bit aboutproduct of metric spaces. If (M,d1), (N, d2) are metric spaces, then M × Ncan be made into a metric space in several equivalent ways. One can de-fine the metric d for M × N by d((p, q), (p′, q′)) = d(p, p′) + d(q, q′). Or onecould define d((p, q), (p′, q′)) =

√d1(p, p′)2 + d2(q, q′)2. Or d((p, q), (p′, q′)) =


max(d1(p, p′), d2(q, q

′)). We should all have enough experience by now to re-alize a) All three of these are metrics, and b)they are equivalent. So use theone you like most; the main thing is that if we denote the ball of center pradius r > 0 in M by BM (p; r), the one in N by BN (q, r). then U is openin M × N if and only if for every p, q) ∈ U , there exist r1 > 0, r2 > 0 suchthat BM (p, r1) × BN (q; r2) ⊂ U . One can always assume r1 = r2, if neces-sary. Or if one wants to. Notice that BM (p, r) × BN (q; r) is not the same asBM×N ((p, q), r) if we use the first two metrics we defined above for M ×N , butis equal to BM×N ((p, q), r) for the third of these metrics. When M = Rm andN = Rn, then one sees that if we identify Rm × Rn with Rm+n by

((x1, . . . , xm), (y1, . . . , yn)) 7→ (x1, . . . , xm, y1, . . . , yn),

then they also get identified as metric spaces. In fact, if we use as productmetric the second definition, namely d((p, q), (p′, q′)) =

√d1(p, p′)2 + d2(q, q′)2,

and d1, d2 are the Euclidean metrics for Rm,Rn, respectively, then d is theEuclidean metric for Rn+m. One more thing. The standard way to prove bythe definition f : M×N → P is continuous, where (M,d1), (N, d2), and (P, d)are metric spaces is as fllows. One begins with Let (p0, q0) ∈ M ×N . Let ϵ > 0be given. (or similar). Now one somehow has to find δ > 0; so one will have astatement such as Let δ = . The right hand side of this equality will mostlikely be a function of ϵ, p0, q0. If it isn’t 100% obvious that this d is positive,one has to verify that it is; one should be able to say δ > 0. Next one assumesp ∈ M , q ∈ N satisfy d1(p, p0) < δ and d2(q, q0) < δ. One hasto show that thisimplies d(f(p, q), f(p0, q0)) < ϵ.

With all this said, the continuity of the functions mentioned above becomesquite easy to prove. I will use the Euclidean distance(s) throughout. So |x| isthe absolute value of x if x ∈ R; |x| =

√∑mi=1 x

2i if x ∈ Rm, and (|x, y)| =√∑m

i=1 x2i +

∑mi=1 y

2i if (x, y) ∈ R×Rm ∼= R2m.

Continuity of µ Let (x0, y0) ∈ Rm × Rm. We have to prove µ is continuousat (x0, y0). Let ϵ > 0. Let δ = ϵ/2. Then δ > 0 and |x − x0| < δ, |y − y0| < δimplies

|µ(x, y)−µ(x0, y0)| = |(x+y)−(x0+y0)| = |(x−x0)+(y−y0)| ≤ |x−x0|+|y−y0| < 2δ = ϵ.

Continuity follows.

Continuity of ν Let (c0, x0) ∈ R × Rm. We have to prove ν is continuous at(c0, x0). Let ϵ > 0. Let

δ = min

(1,

ϵ

2(|c0|+ 1),

ϵ

2(|x0|+ 1)

).

Then δ > 0. If |c−c0| < δ, |x−x0| < δ then because δ ≤ 1, we have |c−c0| < 1,|x− x0| < 1, hence

|c| = |c−c0+c0| ≤ |c−c0|+|c0| < 1+|c0|, |x| = |x−x0+x0| ≤ |x−x0|+|x0| < 1+|x0|.


Thus

|cx− c0x0| = |c(x− x0) + (c− c0)x0| ≤ |c||x− x0|+ |c− c0||x0|

≤ (1 + |c0|)|x− x0|+ (1 + |x0|)|x− x0| < (1 + |c0|)δ + (|x0|)δ <ϵ

2+

ϵ

2= ϵ.

Continuity of λ Let (x0, y0) ∈ Rm ×Rm. We have to prove λ is continuous at(x0, y0). The proof uses the same idea as the proof of the continuity of ν, plusthe Cauchy-Schwarz inequality

|x · y| ≤ |x||y|.

Let ϵ > 0. Let

δ = min

(1,

ϵ

2(|x0|+ 1),

ϵ

2(|y0|+ 1)

).

Then |x − x0| < δ implies |x| = |x − x0 + x0| < 1 + |x0|. If now |x − x0| <δ, |y − y0| < δ, then

|λ(x, y)− λ(x0, y0)| = |x · y − x0 · y0| = |x · (y − y0) + (x− x0) · y0| ≤ |x||x− x0|+ |x− x0||y0|

< |x|δ + |y0|δ < (|x0|+ 1)δ + |y0|δ ≤ ϵ

2+

ϵ

2= ϵ.

Notice that if m = 1, then λ is just the usual product.

Continuity of abs Absolutely trivial.

As a corollary we obtain the following theorem.

Theorem 18 Let M be a metric space.

1. Let f, g : M → Rm be continuous. Then the functions f + g, f − g fromM to Rm are continuous. (f ± g : M → Rm is defined by f ± g(x) =f(x)± g(x).)

2. Let f, g : M → Rm be continuous. Then f · g : Rm]toR is continuous,where f · g(x) = f(x) · g(x).

3. Let f : M → Rm be continuous. Then |f | : Rm]toR is continuous, where|f |(x) = |f(x)|.

4. Let f : M → Rm be continuous and let ϕ : Rm → R. Then ϕf : Rm → Rm

is continuous, where ϕf(x) = ϕ(x)f(x).

Proof. If f, g : M → Rm are continuous, the map p 7→ (f(p), g(p) : M →Rm × Rm is continuous (Proposition 17). Now f + g is just this map followedby µ. If we follow it by λ, we get that f · g is continuous. If f : M → Rm,ϕ : M → R are continuous, then p 7→ (ϕ(p)mf(p)) : M → R×Rm is continuous;following by ν we get the continuity of ϕf . Specializing to ϕ(p) = −1 for allp ∈ M we get −g is continuous if g is continuous, hence we get f −g = f +(−g)is continuous. Following a continuous f with abs proves |f | continuous.

7 COMPLETENESS AND SOME REALITY CHECKS 20

7 Completeness and Some Reality Checks

An important property that a metric space may or may not have is completeness.To discuss what is meant by completeness, we need to introduce the notion ofa Cauchy sequence

Definition 11 Lat M be a metric space. A sequence (pn) in M is said to be]extbfCauchy sequence iff for each ϵ > 0 there exists N ∈ N such that d(pn, pk) <ϵ whenever n, k ≥ N .

In a way, the following theorem presents the main example of a Cauchy sequence.

Theorem 19 Assume (pn) is a sequence of points in the metric space M . If(pn) converges, it is Cauchy.

Proof. Assume (pn) converges and let p be the limit. Let ϵ > 0 be given. Thenϵ/2 > 0 and there exists N such that n ≥ N implies d)pn, p) < ϵ/2. If nown, k ≥ N , then

d(pn, pk) ≤ d(pn, p) + d(p, pk) <ϵ

2+

ϵ

2= ϵ.

Is the converse true? The answer is: sometimes. Not in general. For example,consider the following sequence (xn) of rational numbers, defined defined by

xn =n∑

k=1

1

k2, n = 1, 2, 3 . . .

If we consider Q as a metric space with the metric inherited from R, (xn) is aCauchy sequence. To prove this without introducing gadgets like the integraltest, I will prove first that

n+p∑k=n

1

k2≤ 1

n− 1(5)

for all n ∈ N, n ≥ 2, p ∈ N∪{0}. The proof is simple, it relies on the inequalityand equality

1

k2=

1

(k − 1)k=

1

k − 1− 1

k.

The proof of (5) can now be done as follows. Notice, we assume n ≥ 2 to besure that k ≥ 2 and 1/k − 1 is defined.

n+p∑k=n

1

k2≤

n+p∑k=n

1

(k − 1)k=

1

n− 1− 1

n+

1

n− 1

n+ 1+ · · ·+ 1

n+ p− 1− 1

n+ p.

The last sum “telescopes,” all terms but the first and last cancel so that wehave

n+p∑k=n

1

k2≤ 1

n− 1− 1

n+ p≤ 1

n− 1


which is (5). To prove now it is a Cauchy sequence, let ϵ > 0 be given. LetN = 1

ϵ +1. If n,m ≥ N , without loss of generality we may assume m > n, writem = n+ 1 + p for some p ≥ 0. Then

|xm − xn| =n+1+p∑k=n+1

1

k2≤ 1

n≤ 1

N< ϵ

by (5) with n replaced by n+ 1.(We can always assume m > n ≥ N instead of m,n ≥ N when dealing with

Cauchy sequences because if m = n then d(pm, pn) = 0 and there is nothing toprove; if m < n, we can switch the roles of m,n.)

We see that (xn) is a Cauchy sequence in Q. In 1644, the Italian mathe-matician Pietro Mengoli wondered about the limit of this sequence. He couldnot figure it out. It was only almost a century later, in 1735, that Euler provedlimn→∞ xn = π2/6. In 1798, Legendre proved π2 irrational. It follows that thesequence (xn) does not converge in the metric space Q.

Definition 12 A metric space is said to be complete iff all of its Cauchysequences converge.

The example we just did shows that the metric space Q is not complete.

Exercise 15 Prove that if M is a discrete metric space, M is complete.

lack of completeness is not a terribly bad sin. One can show that everymetric space can be embedded into a complete space; essentially, all one doesis to add a limit for each Cauchy sequence. Doing for Q is another way ofproducing R. But let us see that R is complete. There are many proofs; hereis one way of doing it. We will use the concepts of lim sup and lim inf (whichalways exist for every sequence of real numbers) and the fact that a sequenceof real numbers converges if and only if it is bounded and its liminf equals itslimsup. First we will see that Cauchy sequences are always bounded; in fact,let’s have it as an exercise.

Exercise 16 Let M be a metric space, let (pn) be a Cauchy sequence in M .Then (pn) is bounded in the following sense: There is q ∈ M and a non-negativereal number M such that d(pn, q) ≤ M for all n ∈ N.

Some comments may be in order. In the first place, if one q ∈ M works, everyother q in M works. In fact, if q, q′ ∈ M and d(q, pn) ≤ M for all n, thend(q′, pn) ≤ M + d(q, q′) for all n ∈ N. A good choice for q,M for the case ofthe exercise is q = pN , where N is such that d(pn, pm) ≤ 1 if n,m ≥ N , andM = max(d(p1, pN ), . . . , d(pN−1, pN ), 1).

It is now easy to prove:

Theorem 20 The metric space R is complete.


Proof. Let (pn) be a Cauchy sequence in R. To prove it converges, we willprove that lim infn→∞ pn equals lim supn→∞ pn. Let L = lim infn→∞ pn, L =lim supn→∞ pn. Since L ≤ L, to prove L = L it suffices to prove L ≥ L. In thefirst place, by Exercise 16, L,L ∈ R; neither is ∞ or −∞.

Let ϵ > 0 be given. Because the sequence is Cauchy, there is N such thatn,m ≥ N imply |xn − xm| < ϵ/2. Recall the definitions of lim sup and lim inf.First lim sup:

lim supn→∞

xn = limn→∞

supn≤k

xk.

It follows that there is N2 such that |L− supn≤k xk| < ϵ/2 for n ≥ N1. All we’ll

use from this is that n ≥ N1 implies that L < supk≥n xk + (ϵ/2) for n ≥ N1.

Let N2 = max)N,N1). Then n ≥ N implies n ≥ N1, thus L − ϵ2 < supk≥n xk,

hence there exists k ≥ n such that L− ϵ2 < xk. But now k ≥ n ≥ N2 ≥ N , thus

|xn − xk| < ϵ/2, hence

xn > xk − ϵ

2> L− ϵ.

That is, we proved that if n ≥ N2, then xn > L − ϵ. Assume n ≥ N2. Thenk ≥ N2 if k ≥ n so that xk ≥ L − ϵ. It follows that infk≥n xk ≥ L − ϵ for alln ≥ N2, thus

L = limn→∞

infk≥n

xk ≥ L− ϵ.

Since ϵ > 0 is arbitrary, this implies L ≤ L.

Thew next set of exercises are important results from real analysis; theyprovide an alternative proof of the completeness of R. Thus, do not use com-pleteness to prove any of them.

Exercise 17 Let (xn) be a sequence of real numbers and assume it is increasing;xn ≤ xn+1 for all n ∈ N. Prove: If the sequence is bounded above, it convergesto a finite value. Otherwise it diverges to infinity. In fact, in either case

limn→∞

xn = sup{xn : n ∈ N}.

Similarly, if (xn) is a decreasing sequence of real numbers, then

limn→∞

xn = inf{xn : n ∈ N};

it follows that such a sequence converges if and only if it is bunded below; oth-erwise it diverges to −∞.

The proof is quite easy; letting σ = sup{xn : n ∈ N}, assume σ < ∞.Given ϵ > 0 there is some N ∈ N with xN > σ − ϵ; since the sequence increasesxn > σ − ϵ if n ≥ N . If σ = ∞ one argues similarly,but differently. This result,together with is one of the best ways of proving convergence of sequences of realnumbers. Here are is a classical example.


Example 1. Let (fn) be the Fibonacci sequence; the sequence defined induc-tively by f1 = 1, f2 = 1, fn = fn−1 + fn−2 if n ≥ 3. We will prove that

limn→∞

fn+1

fn=

1

2(1 +

√5).

Let us set gn = fn+1/fn. Then, of course, gn = (fn + fn−1)/fn = 1 + (1/gn−1)and taking limits here we conclude that IF there is a limit ℓ it must satisfyℓ = 1 + 1/ℓ, hence ℓ2 − ℓ − 1 = 0, thus ℓ = (1 ±

√5)/2 and, since it can’t be

≤ 0, ℓ = (1 +√5)/2. But why does a limit exist?

Here are the first few terms of the sequence (gn) (up to three decimal places).

1, 2, 1.5, 1.667, 1.6, 1.625, 1.615

Writing out terms, one sees that the odd terms increase, the even terms de-creases. Can we prove this? It might be a good idea to have a formula givinggn interms of gn−2 rather than gn−1. That, of course, is quite easy:

gn = 1 +1

gn−1= 1 +

1

1 + 1gn−2

=2gn−2 + 1

gn−2 + 1= 2− 1

gn−2 + 1.

Let us write, as is customary, φ = 12 (1 +

√5). Notice that

2φ+ 1

φ+ 1= φ.

With this we can prove: If gn−2 < φ, then gn < φ, and conversely, gn−2 > φ,implies gn < φ. In fact, assume gn−2 < φ. Then gn−2+1 < φ+1, 1/(gn−2+1) >1/(φ+ 1), and

gn = 2− 1

gn−2 + 1< 2− 1

φ+ 1=

2φ+ 1

φ = 1= φ.

The proof that gn−2 > φ implies gn > φ is identical. Consider now the differencegn − gn−2. We have

gn − gn−2 =2gn−2 + 1

gn−2 + 1− gn−2 = −

g2n−2 − gn−2 − 1

gn−2 + 1.

Suppose gn−2 < φ. Since f(x) = x2 − x − 1 < 0 for (1 −√5)/2 < x <

(1+√5)/2 = φ, in this case gn− gn−2 > 0 and gn < φ. Notice that g1 = 1 < φ,

thus the sequence of odd terms is increasing. In case that’s not clear, we haveby induction: g2k−1 < φ for all k, and because g2k−1 < φ we also have thatg2k+1−g2k−1 > 0. The sequence (g2k−1) is increasing and bounded above by φ.Thus it converges. Set α = limk→∞ g2k−1. Similarly one sees that the sequence(g2k) decreases and is bounded below by φ. Let β = limk→∞ g2k. Finally, weneed to see that α = β = φ. That is easy. From

g2k =2g2(k−1) + 1

g2(k−1) + 1

we get taking limits α = (2α + 1)/(α + 1), hence α2 − α − 1 = 0 and, since αmust be ≥ 0, α = φ. Similarly β = φ. It now follows that limn→∞ gn = φ.


Exercise 18 Let (pn) be a Cauchy sequence in a metric space M . Prove: If(pn) has a convergent subsequence, then (pn) itself converges.

This is also quite easy because once some terms with a large index of aCauchy sequence are near something, all terms with large indices will be nearthat something.

An alternative proof of completeness can now be obtained as a consequenceof the next exercise, which is interesting but not too frequently used.

Exercise 19 Let (xn) be a sequence of real numbers. Prove: It contains anincreasing or a decreasing subsequence.

Hint: To prove a statement of the form “p or q” one usually assumes that oneof p or q is false, and proves the other one has to hold. So assume, for example,that the sequence contains no decreasing subsequences. To start the increasingsubsequence it might be good to find the smallest term of the sequence and startthere. Why will there be such a term? There would be also a smallest termamong the terms following the smallest term of the sequence. And so forth.

Using these exercises, here is how one proves completeness: Assume (xn) isa Cauchy sequence of real numbers. By Exercise 16 it is bounded. By Exercise19, it has an increasing or decreasing subsequence; such a subsequence, beingbounded, converges. Thus (xn) has a convergent subsequence hence, by Exercise18, it converges.

Exercise 20 Prove Rm is complete.

The proof is quite straightforward. Show that a sequence (xn) = ((xn1, . . . , xnm))is Cauchy in Rm if and only if the m sequences (xnk), k = 1, . . . ,m are Cauchyin R. A similar proof shows that if M,N are complete metric spaces, so isM ×N .

The following result is also quite useful.

Theorem 21 Assume M is a complete metric space. A subspace N of M iscomplete if and only if N is a closed subset of M .

Proof. Assume first N is a closed subset of M and let (xn) be a Cauchy sequencein N . Then is is also a Cauchy sequence in M ,hence converges in M . But sinceN is closed, the limit must be in N . Thus a Cauchy sequence in N convergesin N ,hence N is complete.

Conversely, assume N is complete. Let (xn) be a sequence in N convergingto a point x ∈ M . As a convergent sequence it is Cauchy, thus it must convergein N , hence x ∈ N . Thus x contains all of its limit points, hence it is closed.

8 COMPACTNESS 25

8 Compactness

Here I really differ with the author of the textbook. His definition of a compactset is absolutely not standard, and I think it is better if you get the standardone.

Definition 13 Let M be a metric space, let C ⊂ M . An open covering of Cis a non-empty family U of open subsets of M such that their union includes C:C ⊂

∪U∈U U .

Here are some comments. Frequently the open cover of a set C is indexed;an open covering can be described as a family {Uλ}λ∈Λ of open sets such thatC ⊂

∪λ∈Λ Uλ. An open covering of the whole space M is a family {Uλ}λ∈Λ of

open sets such that M =∪

λ∈Λ Uλ. A typical way open coverings are definedis as follows: Suppose C is a subset of a metric space. For each x ∈ C selectrx > 0 (usually, satisfying some specific property). Then {B(x; rx)}x∈C is anopen covering of C.

For example, if M = R, the following families are open coverings of R:

1. U1 = {(−n, n) : n ∈ R}.

2. U2 = {(x, x+ 1) : x ∈ R}.

3. U3 = {R}.

4. U4 = {(−1, 1)} ∪ {(1/n,∞) : n ∈ N} ∪ {(−∞,−1/n) : n ∈ N}.

Definition 14 Let M be a metric space, C ⊂ M and U and open covering ofC. A subcovering (of U) is a family V ⊂ U that is still a covering of C.

For example, consider the covering U1 of the first example above. The familyV = {(−p, p) : p ∈ N, p prime} is a subcovering of U1. Of course, every coveringis a subcovering of itself.

Here is the main definition of this section.

Definition 15 A subset C of a metric space M is compact iff every opencovering of C has a finite subcovering.

Before wondering what compact spaces are good for, let us see some examples.

The proof that a set C is compact, by the definition (not a very desirablething to do, as it turns out), goes somewhat like this.

Let U be an open covering of C. Nothing else can be assumed of U .Next somehow one produces sets U1, . . . , Un ∈ U and proves C ⊂

∪ni=1 Ui. One

can also assume the initial covering is indexed, so one might start by saying let{Uλ}λ∈Λ be anopen covering of C. Then one has to produce λ1, . . . , λn ∈ Λsuch that C ⊂

∪nj=1 Uλj . Here is a very simple example.

If C is a finite subset of a metric space, then C is compact.

8 COMPACTNESS 26

Proof. Assume U is an open covering of C. If C = ∅, select any U ∈ U ;then {U} is an open subcovering of U . If C = ∅, write C = {p1, . . . , pn}. SinceU covers C, for each i = 1, . . . , n there is U1 ∈ U such that pi ∈ Ui. Then{U1, . . . , Un} is a finite subcovering of U .

On the other hand, to prove a set not compact it suffices to exhibit a singlecovering without a finite subcovering. For example, R is not compact. In fact,consider the covering {(−n, n) : n ∈ N} it is not hard to show it has no finitesubcovering. In fact, a finite subcovering would be of the form Ui = (−ni, ni),i = 1, . . . , N , where n1, . . . , nN ∈ N. If x ∈ R, x > max(n1, . . . , nN ), then

x /∈∪N

i=1 Ui.

What are compact sets good for? One important thing for which they aregood for is to make local properties global. To be a bit more precise, they aresets in which local properties are global. For example, a typical compactnessresult is that if M is a compact metric space, f : M → N continuous (whereN is another metric space), then it is uniformly continuous. Continuity impliesthat given ϵ > 0, for every p ∈ M there is a δ > 0, possibly depending onp, such that if q ∈ B(p, δ) then something happens. In the case of continuitywhat happens is d(f(q), f(p)) < ϵ. If we emphasize the dependence of δ on pand call it δp, the family of open balls {B(p, δp)}p∈M is an open covering of M .Compactness allows us to extract from this possibly humongous family of ballsa finite number, and from there we can get a δ > 0 that works for all points.We’ll see a complete proof later on. First some basic stuff about compactness.We have to begin taking care of a certain ambiguity. Suppose C is a subset ofa metric space M . If it is a compact subset of M , is it also compact as a metricspace? If it is compact as a metric space (with the metric inherited from M , isit also compact as a subset of M? The answer to both questions is yes. It reallyis a trivial matter, but it needs to be done. Going through it might providefurther practice in dealing with compactness.

Suppose C is compact as a metric space. To prove it is also compact as asubset of M , we assume U is an open covering of C in M ; that is a family of opensubsets of M whose union includes C: C ⊂

∪U∈U U . Then C =

∪U∈U (U ∩ C)

so that {U ∩ C : U ∈ U} is an open covering of the compact metric space C,hence there is a finite subcovering {U1 ∩ C, . . . , Un ∩ C}. Then {U1, . . . , Un}is a finite subcovering of U . For the converse we assume C is compact as asubset of M and assume that U is an open covering of the metric space C; thusU is a family of subsets of C, open in C, such that C =

∪U∈U U . Each U in

U isopen in C, thus it is of the form U = WU ∩ C, where WU is open in M .This is sort of nasty notation so I’ll start the proof of the converse again, usingan indexed family as covering. To prove C is a compact metric space, assume{Uλ}λ∈Λ is an open covering of the metric space C; that is Uλ is open in C foreach λıΛ and C =

∪λ∈Λ Uλ. For each λ ∈ Λ, because Uλ is open in C, there

exists Wλ open in M such that Uλ = Wλ ∩ C. Then C ⊂∪

λ∈Λ Wλ and, sinceit is a compact subset of M , we see that there exist λ1, . . . , λm ∈ Λ such that

8 COMPACTNESS 27

C ⊂∪m

i=1 Wλi. Intersecting wit C we get C =

∪mi=1 Uλi

, proving {Uλi}ni=1is a

finite subcovering of {Uλ}λ∈Λ. Since {Uλ}λ∈Λ was an arbitrary covering of C,compactness follows.

One consequence is that when dealing with compactness we can always as-sume that the compact set is the whole metric space. Another thing to noticeis that since the definition of compactness is given exclusively in terms of opensets, if a set is compact for one metric, it is compact for all equivalent metrics.

Proposition 22 If C is compact, C is bounded.

Note: A subset S of a metric space M is bounded iff there exists a ball B(p; r)in M such that S ⊂ B(p; r). A simple application of the triangle inequalityshows that if there is p ∈ M , r > 0 with S ⊂ B(p, r), then for every q ∈ Mthere is r′ > 0 such that S ⊂ B(q; r′). Boundedness is not a big deal, as theexercise following this proposition shows. But boundedness with respect to awhole class of equivalent metrics is a big deal.

Proof. Assume C is not bounded. Let p ∈ M . Then for every r > 0 itholds that C ⊂ B(p, r). It follows that the family {B(p;n) : n ∈ N} is an opencovering of C without a finite subcovering. It follows that if C is not bounded,it is not compact.

As mentioned before, being bounded in a metric space is not a big deal.Given any metric, one cam always find an equivalent one that is bounded. Weleave the proof as an exercise.

Exercise 21 Let (M,d) be a metric space. For p, q ∈ M define

d1(p, q) = min(d(p, q), 1).

Prove that d1 is a metric for M , d1 is equivalent to d, and d1 is bounded.

What is a bigger deal is that a compact set is bounded no matter whatequivalent metric one uses.

Proposition 23 Assume C is a compact subset of the metric space M . ThenC is a closed subset of M .

Proof. Again we proceed by the contrapositive, proving that if C is not closedit is not compact. So assume C is not closed. That means that there is atleast one cluster point of C not in C. So assume p ∈ C ′\C. For r > 0 letUr = {q ∈ M : d(q, p) > r}. It is clear that each Ur is open and∪

r>0

Ur = M\{p} ⊃ C.

Thus {Ur}r>0 is an open covering of C. Suppose it admits a finite subcov-ering; say there exist r1, . . . , rn all positive such that C ⊂

∪mi=1 Uri . If r =

8 COMPACTNESS 28

min(r1, . . . , rn), then r > 0 and∪m

i=1 Uri = Ur, so C ⊂ Ur. But then B(p; r/2)∩C = ∅; by part 2(c) of Proposition 5 this contradicts that p is a cluster point ofC. This contradiction proves C is not compact.

Another not too hard result is.

Proposition 24 Assume M is a compact metric space and C is a closed subset.Then C is compact.

Proof. We have to show that every open covering of C admits a finite subcover.So assume U is an open covering of C. We know that M is compact, suggestingextending the covering of C to one of M . That is easily done, just throw inM\C. Since C is closed, M\C is open and U ∪ {M\C} is an open coveringof M . Thus there exists a finite number of sets in that covering which alsocover M ; there is no harm in assuming M\C is one of them; that is, there existU1, . . . , Um ∈ U such that M =

∪mi=1 Ui∪M\C. Thus C ⊂

∪mi=1 Ui∪M\C and,

since C ∩ (M\C) = ∅, C ⊂∪m

i=1 Ui.

The following is an important property of compactness:

Theorem 25 Let M be a compact metric space and let (Fn) be a decreasingsequence of closed non-empty subsets of M . Then ∩∞

n=1Fn = ∅.

Proof. We proceed by contradiction; assume ∩∞n=1Fn = ∅. In this case, let

Un = M\Fn is open for each n ∈ N and

∞∪n=1

Un = M\ ∩∞n=1 Fn = M\set = M ;

i.e., {Un}n∈N is an open covering of M . There exist thus n1, . . . , nN ∈ N such

that∪N

k=1 ∞Unk= M . Without loss of generality we may assume n1 < · · · <

nN ; since the Fn’s decrease the Un’s increase. Thus M =∪N

k=1 ∞Unk= UnN

.But then FnN = ∅, a contradiction that proves the theorem.

Exercise 22 This exercise generalizes considerably the result of Theorem 25.It can be used as another definition of compactness. First a definition. A metricspace M is said to satisfy the finite intersection property (FIP) iff the followingholds. Whenever {Fλ}λ∈Λ is a family of closed sets such that the intersectionof any finite subfamily is not empty, then the intersection of all the sets in thefamily is not empty. In symbols (mostly): If {Fλ}λ∈Λ is a family of closed setsthen ∩

λ∈Γ

Fλ = ∅ ∀Γ ⊂ Λ,Γ finite ⇒∩λ∈Λ

Fλ = ∅.

Prove: A metric space satisfies FIP if and only if it is compact.

8 COMPACTNESS 29

Hint: De Morgan laws and contradiction. Or contradiction and De Morganlaws.

Let’s summarize what we know of compact sets. They are bounded andclosed. Suppose C is a compact subset of a metric space and A is a subset ofC. As we know from the section of subsets, because C is closed in M , the setA is closed in C if and only if it is closed in M . Because C is also compactas a subspace of M , Proposition 24 applies and we can say that A is compact.A closed subset of a compact set is compact. The property that every opencovering has a finite subcovering is very important. Iy is very useful to knowone has it, but frequently very difficult to verify directly. To compensate for thisdifficulty there are several other equivalent (or mostly equivalent) definitions ofcompactness, one of the most common being in terms of sequences. It is thedefinition in our text; this definition is equivalent to compactness in metricspaces, but not in general topological spaces. That is why it may be useful togive it a slightly different name.

Definition 16 Let M be a metric space. We say M is sequentially compactiff each sequence in M has a convergent subsequence.

Sequentially compactmetric spaces are compact, and we’ll prove the equiva-lence. But for the time being, you may amuse yourself proving that Propositions22, 23, and 24 are valid if one replaces “compact” by “sequentially compact.”Obviously you should do this using only the definition of sequential compactness.

Exercise 23 1. If C is sequentially compact, C is bounded.

2. Assume C is a sequentially compact subset of the metric space M . ThenC is a closed subset of M .

3. Assume M is a compact metric space and C is a closed subset. Then Cis sequentially compact.

Sequential compactness is usually easy to verify; compactness is not so easy.We will prove the equivalence of the two concepts for metric spaces. Thatcompactness implies sequential compactness is easy; the converse is hard. Iprove it in several steps; each step will be a property shared by sequentiallycompact (hence also compact) sets; these properties are important in their ownright.

Theorem 26 Let M be a metric space. It is compact if and only if it is se-quentially compact.

Proof. To avoid getting stuck in trivial nonsense, assume M = ∅. The result istrivial if M = ∅.

Let us do the hard part first, so assume M is sequentially compact.Step 1. For every ϵ > 0, there exists a finite number of points p1, . . . , pk ∈ Msuch that

M =

k∪ν=1

B(pν ; ϵ).

8 COMPACTNESS 30

In fact, let ϵ > 0 and assume there is no such finite set. By induction weconstruct a sequence (pn) with the property that d(pn, pm) ≥ ϵ for all n,m ∈ N,n = m. To start, let p1 be any element of M (here is where the assumptionM = ∅ is convenient). By our assumption (no finite set of balls of radius ϵ coversM) M = B(p1; r). Thus there is p2 ∈ M , p2 /∈ B(p1, r); i.e., d(p2, p1) ≥ ϵ.

Assume now found p1, . . . , pn ∈ M for some n ≥ 2 such that d(pk, pℓ) ≥ ϵfor 1 ≤ k = ℓ ≤ n. By our assumption, M =

∪ni=1 B(pi; ϵ), hence there is

pn+1 ∈ M\ (∪n

i=1 B(pi; ϵ)). Then d(pn+1, pi) ≥ ϵ for i = 1, . . . , n.By induction, we proved such a sequence exists. But this sequence can-

not have any convergent subsequence; convergent subsequences are Cauchy se-quences implying that there have to be terms (lots of them!) at distance< ϵ(also at distance < ϵ/2, < ϵ/4; etc), but any two terms of the sequence (henceof any subsequence) are at distance ≥ ϵ from each other. Step 1 is done.Step 2. M has a countable dense subset.

That is, there exists a subset D of M such that D is countable (its elementscan be listed as a sequence) and such that D = M ; every point of M is a limitpoint of D. One says M is separable.

For n ∈ N let Dn = {pn,1, . . . , pn,kn} such that M =∪kn

ν=1 B(pn,ν ; 1/n). LetD =

∪∞n=1 Dn. As a countable union of finite sets, D is countable.

Let p ∈ M ; we need to see p is a limit point of D. For this it suffices to seethat B(p; r)∩D = ∅ for all r > 0; so let r > 0. Let n ∈ N be such that 1/n < r.There is ν ∈ {1, . . . , kn} such that p ∈ B(pn,ν ; 1/n) thus d(pn,ν , p) < 1/n < r.Since pn,ν ∈ D, we are done.Step 3. This is an intermediate step, of little importance once the theorem isproved. Every open covering of U has a countable subcovering. Assume U isan open covering of M . Let D be a countable dense subset of M . Here is atempting argument, but it would be wrong. Let p ∈ D. Them by definitionof covering, p ∈ U for some U ∈ U ; select one such member of U and denoteit by Up. That is, for p ∈ D, p ∈ Up ∈ U . Then {Up : p ∈ D} is a family ofopen sets; it seems almost obvious that it covers M . Think of it in terms ofthe real line, where D could be the rationals. For each rational you take a setcontaining a whole interval around the rational; how much of the reals can theunion of such sets miss? The obvious answer is nothing, and it remains obviousuntil one tries to prove it, and one sees one can’t. Because the correct answer is“a lot.” I will get back to this question after the theorem is proved; I mentionthis here to ex[lain a bit why the arguent is a bit convoluted.

So let’s return to the beginning; we have U and D. We need to get not justany set Up ∈ U but large ones, so for p ∈ D we let

rp = sup{r > 0 : ∃ U ∈ U , B(p, r) ⊂ U}.

Since U is a covering, it is clear that rp > 0 for all p ∈ D. Moreover, for eachp ∈ D there exists Up ∈ U such that B(p; 3rp/4) ⊂ Up. We claim {Up}p∈D is acovering of M . It suffices to prove that {B(p; rp/2)}p∈D covers M . Let q ∈ M .Now (please, try to follow this argument, it is mildly tricky), there is U ∈ Usuch that q ∈ U . SInce U is open, there is r > 0 such that B(q, r) ⊂ U . Because

8 COMPACTNESS 31

D is dense, there is p ∈ D such that p ∈ B(q, r/4). This implies B(p, 3r/4) ⊂ U .In fact, we have for x ∈ B(p, r/s), d(x, q) ≤ d(x, p) + d(p, q) < 3r/4 + r/4 = r,thus x ∈ B(q, r) ⊂ U . There is no reason to assume this U is Up but, becauseU(p, 3r/4) ⊂ U ∈ U , we conclude that 3r/4 ≤ rp. Since p ∈ B(q; r/4), wehave q ∈ B(p; r/4) ⊂ B(p; rp/3) ⊂ B(p; 3rp/4). This part of the proof can berewritten in a nicer way.Step 4. M is compact. Assume U is an open covering of M . By Step 3, we mayassume that U is countable, and write U = {U1, U2, U3, . . .}. We now basicallyrepeat the argument of Step 1. If no finite subcovering exists we can constructa sequence (pn) in M such that p1 /∈ U1, p2 /∈ U1 ∪U2, p3 /∈ U1 ∪U2 ∪U3, dots;

in general pn /∈∪n−1

j=1 Uj . By sequential compactness, there is a convergentsubsequence (pnk

); say limk→∞ pnk= p. Then p ∈ Uℓ for some ℓ ∈ N. It follows

that there is K such that pnk∈ Uℓ for k ≥ K. However, the construction of

the sequence implies that pn /∈ Uℓ for n > ℓ; only a finite number of terms ofthe sequence (at most) can be in Uℓ so the assumption that there is no finitesubcover has led to a contradiction.

The hard part of the theorem is proved.

Conversely assume M is compact and assume (pn) is a sequence in M . As-sume it has no convergent subsequence. It is then easy to see that the setsFn = {pk : k ≥ n} are all closed, for the simple reason that they have nocluster points. Since thery are all not empty, we conclude by Theorem 25 that∩∞n=1Fn = ∅. But clearly ∩∞

n=1Fn = ∅, so our assumption of the existence of asequence with no convergent subsequences has lead us to a contradiction.

Here is a strange set that will probably appear again in the measure theorypart of this course. Write out the rationals as a sequence; Q = {r1, r2, r3, . . .}.Let ϵ > 0. Consider the open intervals In =

(rn − ϵ

2n+1, rn +

ϵ

2n+1

)for n ∈ N.

Let U =∪∞

n=1. Now the set U contains an interval around each rational number;the rationals are dense in R. How can U not be all of R? Think about it. Tryto imagine how U looks.

Here is a simple fact. If you take a number of intervals of a given length,their combined length will be at most the sum of their lengths; it will be thesum of their length only if there are no overlaps. So the total length of the setcovered by say the first N of these intervals satisfies (writing ℓ(A) for the lengthof the set A)

ℓ(

N∪n=1

In ≤n∑

i=1

21

ϵ2n+1= ϵ

N∑n=1

1

2n= (1− 2−N−1)ϵ < ϵ.

So as we keep on adding intervals we never get past ϵ in length. The set U hasa total length of < ϵ. Try again to picture how U looks.

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