Lecture notes of MA3209: Metric space WANG Dong October 27, 2014 1 Definition and examples of metric spaces Definition 1.1. A metric space R =(X, ρ) is the pair of two things: a set X , whose elements are called points, and a distance function ρ(x, y), defined for arbitrary x and y in X , such that 1. ρ(x, y) ≥ 0 (nonnegativity), 2. ρ(x, y) = 0 if and only if x = y (identity of indiscernibles), 3. ρ(x, y)= ρ(y,x) (symmetry), 4. ρ(x, y)+ ρ(y,z ) ≥ (x, z ) (triangle inequality). Example 1.1. X = R and ρ(x, y)= |x - y|. Example 1.2. X = R n and the distance between x =(x 1 ,...,x n ),y =(y 1 ,...,y n ) is ρ ∞ (x, y) = max n k=1 (|x k - y k |). Example 1.3 (‘ ∞ space). X is the set of sequences x =(x 1 ,x 2 ,...,x n ,... ) of real numbers that satisfy the condition sup ∞ k=1 |x k | < ∞. The distance function is ρ ∞ (x, y)= ∞ sup k=1 |x k - y k |. Example 1.4 (C ∞ [a, b] space). X is the set of all continuous functions on interval [a, b]. The distance between two continuous functions x(t) and y(t) is ρ ∞ (x, y) = max t∈[a,b] (|x(t) - y(t)|). The proofs that the metric spaces in Examples 1.1, 1.2, 1.3 and 1.4 are well defined are left for homework. The key step is to show that the distance function ρ(x, y)(ρ ∞ (x, y)) satisfies the triangle inequality (Condition 4), and the proofs are based on the inequality |a - c|≤|a - b| + |b - c|. Example 1.5 (Euclidean n-space). X = R n and for x =(x 1 ,x 2 ,...,x n ), y =(y 1 ,y 2 ,...,y n ), ρ(x, y)= p (x 1 - y 1 ) 2 +(x 2 - y 2 ) 2 + ··· +(x n - y n ) 2 . See Figure 1 for the n = 2 case. 1
Transcript
Lecture notes of MA3209: Metric space
WANG Dong
October 27, 2014
1 Definition and examples of metric spaces
Definition 1.1. A metric space R = (X, ρ) is the pair of two things: a set X, whoseelements are called points, and a distance function ρ(x, y), defined for arbitrary x and yin X, such that
1. ρ(x, y) ≥ 0 (nonnegativity),
2. ρ(x, y) = 0 if and only if x = y (identity of indiscernibles),
Example 1.2. X = Rn and the distance between x = (x1, . . . , xn), y = (y1, . . . , yn) isρ∞(x, y) = maxnk=1(|xk − yk|).
Example 1.3 (`∞ space). X is the set of sequences x = (x1, x2, . . . , xn, . . . ) of realnumbers that satisfy the condition sup∞k=1|xk| <∞. The distance function is
ρ∞(x, y) =∞
supk=1|xk − yk|.
Example 1.4 (C∞[a, b] space). X is the set of all continuous functions on interval [a, b].The distance between two continuous functions x(t) and y(t) is
ρ∞(x, y) = maxt∈[a,b]
(|x(t)− y(t)|).
The proofs that the metric spaces in Examples 1.1, 1.2, 1.3 and 1.4 are well defined areleft for homework. The key step is to show that the distance function ρ(x, y) (ρ∞(x, y))satisfies the triangle inequality (Condition 4), and the proofs are based on the inequality|a− c| ≤ |a− b|+ |b− c|.
Example 1.5 (Euclidean n-space). X = Rn and for x = (x1, x2, . . . , xn), y = (y1, y2, . . . , yn),
ρ(x, y) =√
(x1 − y1)2 + (x2 − y2)2 + · · ·+ (xn − yn)2.
See Figure 1 for the n = 2 case.
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Figure 1: When n = 2, we see from the figure that ρ(x, y) is the usual distance and thetriangle axiom has a simple geometric meaning.
In the Euclidean n-space, it is straightforward to check that Conditions 1, 2 and 3 aresatisfied. To show that the triangle inequality also holds, we denote
In this example, conditions 1, 2 and 3 are obviously satisfied, and the triangle in-equality, based on arguments similar to those in the for Example 1.5, is equivalent to thefollowing theorem.
Theorem 1.2 (Minkowski inequality). Let a1, a2, . . . , an and b1, b2, . . . , bn be real numbersand and p ∈ [1,∞). Then
The proof of the Minkowski inequality when p > 1 depends on Holder’s inequality,while the p = 1 case turns out to be much easier and is left as an exercise.
Theorem 1.3 (Holder’s inequality). Let x1, x2, . . . , xn and y1, y2, . . . , yn be real numbersand p, q ∈ (1,∞) such that 1
p+ 1
q= 1,
n∑k=1
|xkyk| ≤
(n∑k=1
|xk|p) 1
p(
n∑k=1
|yk|q) 1
q
. (2)
Proof of Theorem 1.2 (Minkowski inequality) for p > 1 by Theorem 1.3 (Holder’s inequality).We prove a seemingly stronger, but actually equivalent inequality
Sum up the two identities above, we obtain with the help of the simple identity (3)
n∑k=1
(|ak|+ |bk|)p ≤
( n∑k=1
|ak|p) 1
p
+
(n∑k=1
|bk|p) 1
p
( n∑k=1
(|ak|+ |bk|)p)1− 1
p
.
Dividing (∑n
k=1(|ak|+ |bk|)p)1− 1p on both sides, we prove the Minkowski inequality.
Before the proof of the Holder’s inequality, we remark that both sides of (2), asfunctions in either x1, . . . , xn or y1, . . . , yn, are homogeneous functions of degree 1. Herewe say a function f(z1, . . . , zn) is homegeneous of degree k if
f(λz1, . . . , λzn) = λkf(z1, . . . , zn). (4)
Proof of Theorem 1.3 (Holder’s inequality). First we prove a simple inequality
ab ≤ ap
p+bq
q, (5)
for a, b > 0, p, q > 1 and 1p
+ 1q
= 1. Without loss of generality, we prove (5) for ap−1 ≥ b.
(Otherwise bq−1 > a, and it is equivalent to the case that we consider by the symmetry(a, p)↔ (b, q).) The proof of the inequality relies on the integral formula
ap
p=
∫ a
0
xp−1dx,
bq
q=
∫ b
0
yq−1dy =
∫ b
0
y1p−1dy,
where in the second formula we used the relation q−1 = 1p−1
as a consequence of 1p+ 1q
= 1.
Hence ap
pand bq
qare expressed geometrically as the areas of regions in Figure 2.
Noting that the graph of x = y1p−1 coincides with that of y = xp−1, we put the two
graphs in one figure. Then it is clear that ab, geometrically represented by the area ofthe rectangle, is smaller than the sum of ap
pand bq
qby a “corner”, see Figure 3. Thus (5)
is proved.Since both sides of Holder’s inequality (2) are homogeneous in x1, . . . , xn and y1, . . . , yn
separately of degree 1, we have that it holds for x = (x1, . . . , xn) and y = (y1, . . . , yn),then it also holds for λx = (λx1, . . . , λxn) and µy = (µy1, . . . , µyn), if λ 6= 0 and µ 6= 0.
n∑k=1
|xkyk| ≤
(n∑k=1
|xk|p) 1
p(
n∑k=1
|yk|q) 1
q
⇐⇒n∑k=1
|λxkµyk| ≤
(n∑k=1
|λxk|p) 1
p(
n∑k=1
|µyk|q) 1
q
.
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Figure 2: The areas of the two shaded regions represent ap
pand bq
qrespectively.
Figure 3: ap
p+ bq
qis larger than ab by the area of the area of the solid coner.
Thus without loss of generality, we need only to prove it under the condition∑n
k=1|xk|p =1 and
∑nk=1|yk|q = 1. In that case, inequality (5) implies that for each k, |xkyk| ≤
|xk|pp
+ |yk|qq
. Sum up for k from 1 to n, we have
n∑k=1
|xkyk| ≤n∑k=1
|xk|p
p+|yk|q
q=
1
p
n∑k=1
|xk|p +1
q
n∑k=1
|yk|q =1
p+
1
q
= 1 =
(n∑k=1
|xk|p) 1
p(
n∑k=1
|yk|q) 1
q
,
and prove the special case. The theorem in general setting is then proved by the homo-geneous property.
Remark 1.1. The Euclidean n-space is a special case of the metric space in Example 1.6with p = 2, and the distance ρ∞ in Example 1.2 is, in some sense, the limit of ρp asp→∞:
ρ∞(x, y) = limp→∞
ρp(x, y), for all x, y ∈ Rn.
The proof of the above limit identity is left for exercise.
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Example 1.7 (`p space). X is the set of sequences x = (x1, x2, . . . , xn, . . . ) of realnumbers that satisfy the condition
∑∞k=1|xk|p <∞. The distance function is (p ≥ 1)
ρp(x, y) =
(∞∑k=1
|xk − yk|p) 1
p
.
First we show that ρp(x, y) is well defined. For any n,(n∑k=1
|xk − yk|p) 1
p
≤
(n∑k=1
|xk|p) 1
p
+
(n∑k=1
|yk|p) 1
p
≤
(∞∑k=1
|xk|p) 1
p
+
(∞∑k=1
|yk|p) 1
p
= const <∞,
where in the first step we use the Minkowski inequality. Since (∑n
k=1|xk − yk|p)1p increases
monotonically as n → ∞, the limit exists by the monotone convergence theorem, and itis ρp(x, y).
Then we show the condition 4 holds, that is,(∞∑k=1
|xk − zk|p) 1
p
−
( ∞∑k=1
|xk − yk|p) 1
p
+
(∞∑k=1
|yk − zk|p) 1
p
≤ 0.
This is the n→∞ limit of the inequality(n∑k=1
|xk − zk|p) 1
p
−
( n∑k=1
|xk − yk|p) 1
p
+
(n∑k=1
|yk − zk|p) 1
p
≤ 0.
Example 1.8 (Cp([a, b]) space). X is the set of all continuous functions on interval [a, b].The distance is (p ≥ 1)
ρp(x, y) =
(∫ b
a
|x(t)− y(t)|p) 1
p
.
The distance function ρp(x, y) is well defined since x(t) and y(t) are bounded, which isa consequence of the extreme value theorem. To show that the triangle inequality holds,we need the integral form of the Minkowski inequality. The proof is a homework problem.
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Recapitulation
• Three important inequalities: Schwarz, Holder, Minkowski.
• Definition of metric space: nonnegativity, identity of indiscernibles, symmetry, tri-angle inequality.
• Examples of metric spaces that will show up throughout this module:
Definition 2.1. An open ball B(x0, r) in the metric space R is the set of all points x inR which satisfy the condition ρ(x, x0) < r. x0 is called the centre and r the radius. Theopen ball B(x0, r) is often denoted as Nr(x0), where N stands for “neighbourhood”. Aneighbourhood of x0 is a set that contains an open ball that is centered at x0.
Example 2.1. The open balls in R2 with origin as their centres and radii 1, in metricspaces with distance function ρ1, ρ2 (Euclidean 2-space), ρ4 and ρ∞ are shown in Figure4.
Figure 4: The open balls with center (0, 0) and radius 1 in (R2, ρ1), (R2, ρ2), (R2, ρ4) and(R2, ρ∞) respectively.
Definition 2.2. A closed ball B[x0, r] is the set of all points x in R which satisfy thecondition ρ(x, x0) ≤ r.
Example 2.2. The closed balls in R2 with origin as their centres and radii 1, in metricspaces with distance function ρ1, ρ2 (Euclidean 2-space), ρ4 and ρ∞ are shown in Figure5.
Figure 5: The closed balls with center (0, 0) and radius 1 in (R2, ρ1), (R2, ρ2), (R2, ρ4)and (R2, ρ∞) respectively.
Definition 2.3. A point x is called a contact point of the set M if every neighbourhoodof x contains at least one point of M . The set of all contact points of the set M is theclosure of M , denoted by M .
In last example, the closure of B((0, 0), 1), the open ball, is the closed ball B[(0, 0), 1].See Figure 6 for an explanation in the Euclidean 2-space, and other cases are left forexercise.
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Figure 6: The point x1, in the open ball, is in the closure of the open ball since all itsneighbourhoods at least contain itself; the point x2, outside of the closed ball, is not in theclosure of the open ball since it has a small neighbourhood disjoint to the open ball; thepoint x3, lying on the boundary of the ball, is in the closure, since all its neibourhoods,however small, contains a small open ball centered at x3 that intersects with the openball.
However, is the closure of the open ball B(x0, r) the closed ball B[x0, r] in all matricspaces? The answer is NO, and we have a counterexample as follows.
Example 2.3 (discrete metric space). Let X be any set (of more than one element), and
ρ(x, y) =
{1 if x 6= y,
0 if x = y.
Then for any x in X, B(x, 1) = {x}, and B[x, 1] = X. Since for any y 6= x, x 6∈ N1/2(y),
we have B(x, 1) = {x}, and then B(x, 1) 6= B[x, 1].
Theorem 2.1. The closure of the closure of M is equal to the closure of M : M = M .
Proof. Every point of M is a contact point of M , so M ⊇ M , and also M ⊇ M . Then
we need to show that if x ∈ M , it is also in M . If x is in M but not in M , there existsa neighbourhood U of x such that it does not intersect with M . We have an ε > 0 such
that x ∈ Nε(x) ⊆ U and then Nε(x) does not intersect with M . Since x ∈ M , Nε/3(x)contains a point x1 in M . Then the neighbourhood Nε/3(x1) of x1 contains a point x2 inM since x1 ∈M . By the triangle inequality,
ρ(x, x2) ≤ ρ(x, x1) + ρ(x1, x2) <ε
3+ε
3=
2
3ε < ε,
which is contradictory to the assumption that Nε(x) does not intersect with M .
Theorem 2.2. If M1 ⊆M2, then M1 ⊆M2.
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Proof. Exercise.
Theorem 2.3. The closure of a sum is equal to the sum of clusures: M1 ∪M2 = M1∪M2.
Proof. Be Theorem 2.3, M1 ∪M2 ⊇ M1 since M1 ∪M2 ⊇ M1, and similarly M1 ∪M2 ⊇M2. Thus M1 ∪M2 ⊇M1 ∪M2.
Now we show that M1 ∪M2 ⊆ M1 ∪M2. If there exists x in M1 ∪M2 but not inM1 ∪M2, then there exist neighbourhoods U and V of x, such that U is disjoint to M1,and V that is disjoint to M2. Suppose x ∈ Nε1(x) ⊆ U and x ∈ Nε2(x) ⊆ V where ε1, ε2are positive. We have Nmin(ε1,ε2)(x), the intersection between the two neighbourhood,is disjoint to both M1 and M2, which means Nmin(ε1,ε2)(x) ∩ (M1 ∪ M2) = ∅. This iscontradictory to the assumption that x ∈M1 ∪M2.
Definition 2.4. The point x is called a limit point of the set M if an arbitrary neigh-bourhood of x contains an infinite number of points of M , and is called an isolated pointof M if x is in M and has a neighbourhood that does not contain any point of M that isdifferent from x.
Theorem 2.4. Every contact point of the set M is either a limit point or an isolatedpoint of M .
Proof. Let x be a contact point ofM , but not a limit point. Then there is a neighbourhoodU of x such that it contains only finitely many points in M . Suppose x ∈ Nε(x) ⊆ Uand Nε(x) ∩M = {x1, x2, . . . , xk}. Let ρ(x, xi) = εi for i = 1, 2, . . . , k, and take ε be thesmallest one among nonzero values of εi. (One of εi may be 0, which means that x itself isamong the k points.) Then the neighbourhood Nε(x) contains no points among x1, . . . , xkthat are different from x, in other words, it contains no points in M that are differentfrom x. We conclude that x is an isolated point of M , and it is one of xi, (1 ≤ i ≤ k).
Thus the closure of a set can be divided into
M = {isolated points} ∪ {limit points that belong to M}∪ {limit points that does not belong to M}.
Definition 2.5. Let x1, x2, . . . be a sequence of points in the metric space R, we saythis sequence converges to the point x if every neighbourhood of x contains all points xnstarting with some one of them. The point x is called the limit of the sequence {xn}.
Two simple properties:
• No sequence can have two distinct limits.
• If {xn} converges to x, so does any of its subsequence.
Theorem 2.5. • The point x is a contact point of the set M if and only if thereexists a sequence {xn} of points of the set M which converges to x.
• The point x is a limit point of M if and only if there exists a sequence {xn} ofdistinct points of the set M which converges to x.
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Proof. If {xn} ⊂ M converges to x, then in any neighbourhood U of x there are points{xn, xn+1, xn+2, . . . } ⊂ M lying in U . (Here n depends on U). So x is a contact point.Furthermore, if {xn, xn+1, . . . } are distinct points, then U contains infinitely many pointsof M , and it is a limit point.
Conversely, if x is a contact point of M , then we construct x1, x2, . . . as follows:
Step 1: Choose x1 as a point in M that lies in N1/1(x).
Step 2: Choose x2 as a point in M that lies in N1/2(x).
. . . . . .
Step n: Choose xn as a point in M that lies in N1/n(x).
. . . . . .
and we construct the desired sequence in M that converges to x. If x is further a limitpoint, in Step n (n > 1) we “choose xn as a point in M that lies in N1/n(x) that is distinctfrom x1, x2, . . . , xn−1”.
Definition 2.6. Let A and B be two sets in the metric space R = (X, ρ). The set A isdense in B if A ⊇ B, In particular, A is everywhere dense if A = X.
Definition 2.7. The metric space R is separable if it has an everywhere dense subsetthat is countable.
Example 2.4. The metric space R = (R, ρ(x, y) = |x−y|) is separable, and the countableset of rational numbers is everywhere dense.
Proof. For any real number x, let xn be the largest rational number in the form m10n
(m isan integer) that is less than or equal to x. Then {x1, x2, . . . } converges to x. By Theorem2.5, x is a contact point of Q = {rational numbers}, and Q is everywhere dense.
One example of the construction: if x = π, then x1 = 31101
= 3.1, x2 = 314102
= 3.14,x3 = 3141
103= 3.141, . . . .
Example 2.5. The metric space R = (Rm, ρ∞(x, y) = maxnk=1(|xk − yk|)) is separable,and the countable set A = {x = (x1, x2, . . . , xm) : xk ∈ Q} is everywhere dense.
Proof. For any point x = (x1, . . . , xm), let the sequence {x(n)k } be the sequence con-
structed in Example 2.4. Then as n→∞, x(n) = (x(n)1 , x
(n)2 , . . . , x
(n)m ) converges to x.
Example 2.6. `∞ space is not separable.
To show that all everywhere dense subset of `∞ space are uncountable, we need thefact that
Proposition 2.6. The subsets of the set of all positive integers (Z+ = {1, 2, 3, . . . }) isan uncountable set.
The proof of this proposition will be given later.
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Proof of that `∞ is not separable. Let S be any subset of Z+. Define x(S) ∈ `∞ as
x(S) = (x(S)1 , . . . , x(S)
n , . . . ), where x(S)k =
{1 if k ∈ S,
0 if k 6∈ S.
Then we see that
ρ∞(x(S), x(S′)) =
{1 if S 6= S ′,
0 if S = S ′.(6)
Suppose A is an everywhere dense subset of R. For each x(S), there is an a(S) ∈ A lyingin N1/3(x(S)). The points a(S) are distinct. Otherwise if a(S) = a(S′) with S 6= S ′, then
which is contradictory to (6).This the set A has no fewer elements than the subsets of Z+, and it is an uncountable
set.
Proof of Proposition 2.6. Here we give the proof of Proposition 2.6 that the set of subsetsof positive integers is uncountable. This is a special form of the so called Cantor’s theorem.
Suppose S = {all subsets of Z+ = {1, 2, 3, . . . }} is countable. Then there exists an 1-1correspondence between S and Z+. A possible form of the correspondence is as follows:
1
l{1, 2}
2
l{1, 2, 4}
3
lall even numbers = {2, 4, 6, . . . }
4
lall prime numbers
. . .
. . .
. . .
(7)
We consider the subset s of Z+, which is an element of S, defined as
s = {all positive integers that are not in the corresponding subsets of Z+}.
(For example, if the correspondence is constructed as in (7), then 1 6∈ s, 2 6∈ s, 3 ∈ s,4 ∈ s, . . . .)
Then s, as an element of S, has a corresponding integer in Z+, denoted as ns. Belowwe discuss if ns belongs to s.
If ns ∈ s, then by the definition of s, ns is not in s, whici is a contradiction; if ns 6∈ s,then by the definition of s, ns ∈ s, which is also a contradiction.
The two contradictions imply that the initial assumption that S is countable is false.
Definition 2.8. Let x be a point in the metric space R and A be a subset of R, thenthe distance from x to A is defined as
ρ(A, x) = inf{ρ(a, x) : a ∈ A}.
If A and B are subsets of R, then the distance between A and B is defined as
ρ(A,B) = inf(ρ(a, b) : a ∈ A, b ∈ B}.
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We can show that ρ(A, x) = 0 if and only if x is a contact point of A.
Definition 2.9. If A is a set in the metric space R, then all limit points of A constitutethe derived set of A, denoted by A′.
Unlike the closure of a set, the derived set of a derived set is not the derived set, thatis, (M ′)′ 6= M ′ generally.
Example 2.7. In the metric space R = (R, ρ(x, y) = |x− y|), let M = {1/n | n ∈ Z+}.Then M ′ = {0}, and (M ′)′ = ∅.
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Recapitulation
• Definitions: closure, limit point, limit, (everywhere) dense, separable space.
• Theorems:
– M = M .
– M1 ⊆M2 ⇒M1 ⊆M2.
– M1 ∪M2 = M1 ∪M2.
– M = {isolated points} ∪ {limit points in M} ∪ {limit points not in M}.– x ∈M ⇔ there is {xn} ⊆M that approaches x;
x is a limit point of M ⇔ there is {xn} ⊆ M with distinct points that ap-proaches x;
• Clarification: Many, but not all, metric spaces are separable.
Counterexample: `∞ space.
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3 Open and closed sets
Definition 3.1. A set M in a metric space X is said to be closed if it coincides with itsclosure: M = M .
Example 3.1. The closure of any set M is closed, since M = M .
Example 3.2. The closed ball B[x0, r] is a closed set.
Figure 7: The closed ball is closed.
Proof. It is clear that B[x0, r] ⊆ B[x0, r]. We need to show B[x0, r] ⊇ B[x0, r]. Supposex is a point not in B[x0, r], then ρ(x, x0) > r, say, ρ(x, x0) = r + ε with ε > 0. Thenthe neighbourhood Nε/2(x) contains no points in B[x0, r]. This can be shown by thetriangle inequality: if x′ is a point in Nε/2(x), then ρ(x′, x0) ≤ r, and ρ(x, x0) ≤ ρ(x, x′)+ρ(x′, x0) ≤ r + ε/2, which is contradictory to the assumption that ρ(x, x0) = r + ε.
Thus any point x outside of B[x0, r] has a neighbourhood that is disjoint to B[x0, r]and is not a contact point. We prove that B[x0, r] ⊇ B[x0, r].
Theorem 3.1. The intersection of an arbitrary number of closed sets is a closed set; thesum of a finite number of closed sets is a closed set.
Proof. First, let {Mα} be a family of arbitrary number of closed sets. Need to show thatif x is not in the intersection
⋂αMα, then x is not a contact point of
⋂αMα, that is, x
has a neighbourhood disjoint to it.If x is not in
⋂αMα, then x is not in at least one closed set Mα. Then there exists a
neighbourhood U of x that is disjoint to that Mα. This implies that U is disjoint to theintersection
⋂αMα.
Next we assume M1,M2, . . . ,Mn are closed sets, and need to show that if x is notin the sum of these sets, it has a neighbourhood that is disjoint to the sum. For anyi = 1, 2, . . . , n, since x is outside of the closed set Mi, there is a neighbourhood Ui ⊇Nεi(x) 3 x that is disjoint to Mi. Taking the intersection of the neighbourhoods Nεi(x),which is Nε(x) with ε = min(ε1, ε2, . . . , εn), then we can verify that Nε(x) is disjoint tothe sum
⋃ni=1Mi.
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You should think why the second part of the proof does not work for infinite sum ofclosed sets.
Example 3.3. The single point set {xi} is closed (think is as B[xi, 0]). Therefore thefinite set {x1, x2, . . . , xn} is closed, by Theorem 3.1. But an arbitrary infinite set is notnesessarily closed. (A good example of a nonclosed countable set is Q ⊂ R, with theusual distance function.)
Definition 3.2. The point x is said to be an interior point of the set M if there exists aneighbourhood of x that is contained in M . The set M is called an open set if all of itspoints are interior points.
Example 3.4. Any point in an open ball B(x0, r) is an interior point of B(x0, r), so theopen ball B(x0, r) is an open set.
Figure 8: The open ball is open.
Proof. Suppose x is a point in B(x0, r). Then ρ(x, x0) < r, say, ρ(x, x0) = r − ε withε > 0. Then the neighbourhood Nε/2(x) is contained in B(x0, r). This can be shown bythe triangle inequality: Any point x′ in the neighbourhood Nε/2(x) satisfies ρ(x′, x0) ≤ρ(x′, x) + ρ(x, x0) ≤ (r − ε) + ε/2 = r − ε/2 < r, so x′ is in the open ball. Thus we haveshown that any point x in the open ball is an interior point.
Theorem 3.2. The set M in metric space X is an open set if and only if its complementX \M is a closed set.
Proof. The property that M is open is equivalent to that any point x in M has a nei-bourhood U contained in M . A simple rephrase of this statement is that any point xout of X \M has a neighbourhood U disjoint to X \M . The rephrased statement isequivalent to say X \M is a closed set.
Since all steps of the argument are equivalency, the argument above proves both the“if” and “only if” parts.
Example 3.5. The whole space X is both closed and open (think it informally asB[x0,∞] and B(x0,∞)). Thus its complement, the empty set ∅, is both open and closed.
16
Theorem 3.3. The sum of an arbitrary number of open sets is an open set; the inter-section of a finite number of open sets are open sets.
Proof. A direct proof like that of Theorem 3.1 is not very hard and is left as an exercise.Below we show a proof that relies on the result of Theorem 3.1.
We use the identities
X \
(⋃α
Mα
)=⋂α
(X \Mα), X \
(n⋂i=1
Mi
)=
n⋂i=1
(X \Mi).
The statement that {Mα} are open sets is equivalent to that {X \Mα} are closed sets(Theorem 3.3). Then by Theorem 3.1, the intersection
⋂α(X \Mα) = X \ (
⋃αMα) is
closed, which implies by Theorem 3.3 again that⋃αMα is open.
The proof of the latter part of the theorem is similar. The statement that {Mi} areopen sets is equivalent to that {X \Mi} are closed sets. Then by Theorem 3.1, the sum⋃ni=1(X \Mi) = X \ (
⋂ni=1Mi) is closed, which implies that
⋂ni=1Mi is open.
Definition 3.3. A family of open sets {Gα} is called a basis if every open set in X canbe represented as the sum of (possibly infinitely many) sets belonging to this family.
Theorem 3.4. A family of open sets {Gα} is a basis of the metric space X is for any openset G and every point x in G, a set Gα can be found in this family such that x ∈ Gα ⊂ G.
Proof. If {Gα} is a basis and⋃α∈AGα = G, an open set, then for any x ∈ G, it is
contained by at least one Gα. On the other hand, if for any x ∈ G, there is a Gx in thefamily {Gα} such that Gx ⊆ G and x ∈ Gx, then the sum
⋃x∈GGx = G.
Example 3.6. The family of all open balls form a basis, since any x in the open setG is aninterior point, so there is a neighbourhood U of x and ε > 0 such that x ∈ Nε(x) ⊆ U ⊆ G.
Example 3.7. The family of all open balls with rational radii form a basis. To see this,note that in the above mentioned setting, we can take x ∈ Nε′(x) ⊆ G where ε′ is apositive rational number less than ε.
Theorem 3.5. The metric space X has a countable basis if and only if it is separable(that is, X has a countable everywhere dense subset).
In the proof we apply a result as follows:
Proposition 3.6. The union of countably many countable sets is a countable set.
Proof. Suppose the sets A1, A2, A3, . . . which are countable sets individually, and we showthat their union is also countable. Without loss of generality, we assume that there areno redundant elements in An, n = 1, 2, 3, . . . , (otherwise the totall number of elements inthe union can only be fewer). Then label elements in An as
An = {an,1, an,2, an,3, . . . },
17
and arrange the elements of the union⋃∞n=1An as an infinite array
and find that these elements has an 1-1 mapping with positive integers.
Proof of Theorem 3.5. If {x1, x2, . . . } is a countable everywhere dense set, then we showthat the countable family A = {Nr(xm) : m = 1, 2, 3, . . . and r ∈ Q+} is a basis. Toshow that A is countable, we use Proposition 3.6 and that A is the union of the countablymany countable sets Am = {Nr(xm) : r ∈ Q+}.
For any open set G and x ∈ G, there exists a neighbourhood U such that x ∈ Nε(x) ⊆U ⊆ G. Since {xi} is everywhere dense, there is an xn in Nε/3(x). Then take a rationalnumber r such that ε/3 < r < 2ε/3. We have that ρ(x, xn) < ε/3 < r, so x ∈ N1/k(xn).On the other hand, any point x′ in N1/k(xn) satisfies ρ(x′, x) ≤ ρ(x′, xn) + ρ(xn, x) <r + ε/3 < ε. It implies that N1/k(xn) ⊆ Nε(x) ⊆ G. There fore x ∈ N1/k(xn) ⊆ G, andby Theorem 3.4 A is a basis.
Conversely, if {G1, G2, . . . } is a countable basis, we take xi as an arbitrary point inGi, and show that {xi} is everywhere dense. Suppose x is a point in G. For any positiveinteger k, N1/k(x) is an open set, and then it contains one open set in the basis {Gi},say Gnk , that contains x. Then the point xnk in Gk satisfies ρ(xnk , x) < 1
k. Therefore
the sequence {xn1 , xn2 , . . . } converges to x, and then {xn} is an everywhere dense subsetsince any x ∈ X is its contact point.
Definition 3.4. A family of sets {Mα} is a covering of the metric space X if⋃αMα = X.
If all Mα are open (closed resp.), the covering is called an open covering (closed coveringresp.).
Theorem 3.7. If X is a separable metric space, and {Mα} is an open covering, then wecan select a countable subcovering.
Proof. Let {Gα} be a countable basis. For any k = 1, 2, 3, . . . , if Gk is contained bysome Mα, then we pick up one that constains Gk and name it Mk; otherwise we leaveMk undefined. Then we have a countable family of open sets, like {M1,M3,M4,M6, . . . }
18
(assuming that G2, G5, . . . are not contained by any Mα). We claim that this family ofopen sets cover X.
For any x ∈ X, there is at least one Mα that contains x. Since Mα is open, there is aGn such that x ∈ Gn ⊆ Mα, by Theorem 3.4. Then Mn, which is one Mα that containsGn, is well defined, and it contains x. (But Mn may not be the Mα we consideredabove.)
19
Recapitulation
• Definitons: closed set, open set, interior point, basis, (sub)covering.
• Notable results:
– A closed ball is closed.
– An open ball is open.
– The whole space is both closed and open.
– The empty set is both closed and open.
– A finite point set is closed.
• Theorems:
– The arbitrary intersection and finite sum of closed sets are closed; the finiteintersection and arbitrary sum of open sets are open.
– A metric space has a countable basis if and only if it is separable.
– In a separable metric space, each open covering has a countable subcovering.
20
4 Examples of open and closed sets
For simple metric spaces, like the real line R with the usual metric, it is possible to givea “simple” description of all open and closed sets.
Lemma 4.1. In the real line with the usual metric, the open intervals with two rationalend points form a basis.
Proof. We know that the set of rational numbers {xk | xk ∈ Q} is a countable basis.From the proof of Theorem 3.5, we see that the open balls with center xk and rationalradii, namely the open intervals (xk− r, xk + r) where r is a positive rational number, fora countable basis of R. It is clear that such intervals have two rational end points, and allopen intervals with two rational end points can be written in the form of (xk−r, xk+r).
Below we consider generalised open intervals on the real line such that we count(−∞,∞), (a,∞) and (−∞, b) as open intervals.
Theorem 4.2. Every open set on the real line is the disjoint sum of a countable numberof open intervals.
A proof better than that given in class. Let G be an open set, and {x1, x2, . . . } be therational numbers in G. Define the family of open intervals
A′i = {(a, b) | a, b are real numbers or ±∞, and xi ∈ (a, b) ⊆ G}.
Then each A′i is not empty. Let
Ii =⋃
(a,b)∈A′i
(a, b).
Then each Ii is an open interval, and
Ii = ( inf(a,b)∈A′i
a, sup(a,b)∈A′i
b).
To see it, we first check that Ii ⊆ (inf a, sup b) (We omit the subscripts for notational
Figure 9: The definition of Ii.
simplicity). On the other hand, we consider an arbitrary y ∈ (inf a, sup b), and withoutloss of generality assume that inf a < y < x. Since y > inf a, there exists (a′, b′) such thatinf a < a′ < y < x < b′, and then y ∈ (a′, b′) ⊆ Ii. thus we show that (inf a, sup b) ⊆ Ii.See Figure 9 for the formation of Ii. We have
∞⋃i=1
Ii =⋃
a,b∈R, (a,b)⊆G
(a, b) = G. (8)
21
Below we show that if Ii 6= Ij, then they are disjoint. Thus if we count identical Ii as oneinterval, the decomposition (29) gives the desired disjoint open interval decompositionof G. Suppose Ii and Ij are not disjoint, so that Ii ∩ Ij is an open interval, and xk is arational number in their intersection, see Figure 10. Since the intervals Ii, Ij contain xk,we have Ii ∈ A′k and Ij ∈ A′k, and then Ii ⊆ Ik and Ij ⊆ Ik. On the other hand, Ik isan interval such that Ik ⊇ Ii 3 xi, so Ik ∈ A′i and Ik ⊆ Ii. Similarly Ik ⊆ Ij. ThereforeIi = Ik = Ij. We complete the proof.
Figure 10: Intervals Ii, Ij and the rational point xk.
Corollary 4.3. Every closed set on the real line is obtained by removing a countablenumber of disjoint open intervals.
Below we show that the seemingly simple closed sets on R can be complicated. Let
F0 = [0, 1],
F1 = [0,1
3] ∪ [
2
3, 1],
F2 = [0,1
9] ∪ [
2
9,1
3] ∪ [
2
3,7
9] ∪ [
8
9, 1],
...
and generally Fn+1 be defined recursively by removing open intervals of length 3−(n+1)
from the middle of each of the 2n closed intervals with length 3−n in Fn. The Cantorset is defined as F =
⋂∞n=0 Fn. F is closed, since it is the intersection of closed sets. See
Figure 11 for the shapes of Fn. But the Cantor set itself cannot be shown graphically.
Figure 11: The shapes of F0, F1, F2 and F3.
A simple property of F is that F has zero “length”. Although we have not definedrigorously the length for a fractal object like F , we take for granted that length is non-negative, and if A ⊆ B ⊆ R, then the length of A is no more than the length of B. Since
22
the length of each Fn, the disjoint sum of 2n closed intervals whose common length is 3−n,the “length” of F is limn→∞ 2n3−n = 0. (The ad hoc argument will be made regorous inMA4262, Measure and Integration.)
A not so simple property of F is that the number of points in F is uncountable, andit is the same as that of points in [0, 1].
Proof. Recall the binary numerical system. Every number in [0, 1] can be written asx = (0.a1a2a3 . . . )2 if x = a1
121
+ a2122
+ a3123
+ · · · . For example,
0.5 =1
2= (0.100 . . . )2,
0.75 =1
2+
1
4= (0.110 . . . )2,
2
3= 1 · 1
2+ 0 · 1
4+ 1 · 1
8+ 0 · 1
16+ 1 · 1
32+ · · · = (0.10101 . . . )2.
Thus the number of points in [0, 1] is no more than the nuber of 0− 1 sequences. (Notethat one number may have two binary expansion, for example, 0.5 = 1
4+ 1
8+ 1
16+
132
+ · · · = (0.01111 . . . )2.) Similarly, real numbers in [0, 1] have ternary expansionsx = (0.a1a2a3 . . . )3 where ai is 0, 1 or 2, if x = a1
131
+ a2132
+ a3133
+ · · · . For example,
1
3= (0.1000 . . . )3,
2
3= (0.2000 . . . )3,
1
2=
1
3+
1
9+
1
27+ · · · = (0.1111 . . . )3.
From the definition, we have
F0 = {all real numbers x = (0.a1a2a3 . . . )3},F1 = {real numbers x = (0.a1a2a3 . . . )3 with a1 = 0, 2},F2 = {real numbers x = (0.a1a2a3 . . . )3 with a1 = 0, 2, a2 = 0, 2},
...
and then
F = {real numbers x = (0.a1a2a3 . . . )3 with all ai = 0, 2}.
Suppose two numbers a = (0.a1a2a3 . . . )3 and b = (0.b1b2b3 . . . )3 are in F , such thatai = 0, 2 and bi = 0, 2 for all i, and their ternary expansions are not identical, then a 6= b.To see it, suppose ai = bi for i = 1, . . . , n− 1, and an = 0 6= bn = 2, then we have
Hence points in F has an 1− 1 correspondence with the 0− 1 sequences
(0.a1a2a3 . . . )3 ↔a1
2,a2
2,a3
2, . . . .
23
We conclude that the points in F is no fewer than the points in [0, 1].On the other hand, obviously the points in F are no more than the points in [0, 1].
Therefore we prove the property that F contains the same number of points as that ofpoints in [0, 1] by the seemingly-naive reason “if set A has no fewer elements than set Band also no more elements than set B, then A has the same number of elements as Bdoes”. See the Cantor-Bernstein-Schroder theorem. http://en.wikipedia.org/wiki/
Next we consider examples of closed sets in the metric space C∞[a, b] where ρ(x(t), y(t)) =maxt∈[a,b]|x(t)− y(t)|. In this section we denote the space as C[a, b].
Example 4.1. The set {x(t) ∈ C[a, b] | x(t) < K} is open, since it is the open ballB(0, K); the set {x(t) ∈ C[a, b] | x(t) ≤ K} is closed, since it is the closed ball B[0, K].
Example 4.2. Let MK be the set of all functions x(t) ∈ C[a, b] such that
|x(t1)− x(t2)| ≤ K|t1 − t2|, for all t1, t2 ∈ [a, b].
Then MK is closed.
Proof. Let the sequence {xn(t)} ∈MK and xn(t)→ x(t) in C[a, b], then the convergenceis the uniform convergence of functions, and then it is also pointwise. So for any t1, t2,
|x(t1)− x(t2)| = | limn→∞
xn(t1)− limn→∞
xn(t2)| = | limn→∞
(xn(t1)− xn(t2)| ≥ K|t1 − t2|.
Then the limit x(t) ∈MK , and thus MK is closed.
Example 4.3. The set M =⋃K>0MK is that of all Lipschitz continuous function on
[a, b]. M is not the whole set of C[a, b], for example, f(x) =√x is in C[0, 1], but not in
MK for any K, since when both t1 and t2 are close to 0, then∣∣∣∣f(t1)− f(t2)
t1 − t2
∣∣∣∣ =1√
t1 +√t2
can be greater than K. M is not closed.
Proof. We use an indirect way to show that M is not closed. First, all polynomials arein M . Next, by Weierstrass approximation theorem, the set of polynomials is everywheredense in C[a, b]. Thus if M ⊇ {polynomials}, then M ⊇ {polynomials} = C[a, b], and wehave that M = C[a, b] 6= M .
Below we prove the Weierstrass approximation theorem used in last proof.
Theorem 4.4 (Weierstrass approximation theorem). The set of polynomials are densein C[a, b], that is, for any continuous function f(x) on [a, b], there is a sequence ofpolynomials p1(x), p2(x), . . . that uniformly converge to f(x).
Proof. Without loss of generality, we assume that [a, b] ⊂ (0, 1). Suppose f(x) is contin-uous on [a, b], we extend it to be a continuous function on [0, 1] such that f(x) = f(a)for x ∈ [0, a] and f(x) = f(b) for x ∈ [b, 1]. See Figure 12. Consider the function
jn(x) =1
Jn(1− x2)n, where Jn =
∫ 1
−1
(1− x2)ndx,
defined on [−1, 1], see Figure 13. We have that for any δ > 0,
limn→∞
∫ −δ−1
jn(x)dx+
∫ 1
δ
jn(x)dx = 0.
Define
pn(x) =
∫ 1
0
f(u)jn(u− x)du
for all x ∈ [a, b]. Then we see that pn(x) is a polynomial of degree ≤ 2n. Below we showthat pn(x)→ f(x) uniformly on [a, b].
Figure 12: The extension of f(x). Figure 13: The shape of jn(x).
By the change of variable v = u− x, we write∫ 1
0
f(u)jn(u− x)du =
∫ 1−x
−xf(v + x)jn(v)dv,
and
pn(x) =
∫ −δ−x
+
∫ δ
−δ+
∫ 1−x
δ
f(v + x)jn(v)dv = I1 + I2 + I3,
where δ is a small positive number. Furthermore, write
I2 =
∫ δ
−δ[f(x) + (f(v + x)− f(x))]jn(v)dv
= f(x)
∫ δ
−δjn(v)dv +
∫ δ
−δ[f(v + x)− f(x)]jn(v)dv
= I ′2 + I ′′2 .
Since f(x) is uniformly continuous in [0, 1], for any x ∈ [a, b], and any ε > 0, we havethat if δ is small enough, |f(v + x)− f(x)| < ε for v ∈ [−δ, δ]. Then
|I ′′2 | ≤∫ δ
−δεjn(v)dv ≤
∫ 1
−1
εjn(v)dv = εJn.
25
To estimate I1 and I3, we note that f(x) is bounded on [0, 1], say, |f(x)| ≤ M forx ∈ [0, 1]. Then
|I1 + I3| ≤∫ −δ−x
+
∫ 1−x
δ
Mjn(v)dv = M
(∫ −δ−1
+
∫ 1
δ
jn(v)dv
).
Thus
|f(x)− pn(x)| = |f(x)− (I1 + I ′2 + I ′′2 + I3)|≤ |f(x)− I ′2|+ |I ′′2 |+ |I1 + I3|
=
∣∣∣∣f(x)
(1−
∫ δ
−δf(v)dv
)∣∣∣∣+ |I ′′2 |+M
(∫ −δ−1
+
∫ 1
δ
jn(v)dv
)≤M
(∫ −δ−1
+
∫ 1
δ
jn(v)dv
)+ |I ′′2 |+M
(∫ −δ−1
+
∫ 1
δ
jn(v)dv
).
From the results derived above, for any ε > 0, we can find δ depending on f , such that|I ′′2 | < ε. Fix this δ, if n is large enough, we have also
M
(∫ −δ−1
+
∫ 1
δ
jn(v)dv
)< ε.
Then for n large enough, |f(x) − pn(x)| < 3ε, and we prove the uniform convergencepn(x)→ f(x).
As a corollary, we have that
Corollary 4.5. Polynomials with rational coefficients are dense in C[a, b]. ThereforeC[a, b] has a countable everywhere dense subset, and it is separable.
The proof is left as an exercise.
26
Recapitulation
• The open sets on the real line are disjoint union of open intervals.
• Weierstrass approximation theorem.
27
5 Topological spaces, continuous mappings, Homeo-
morphism and isometry
We introduce the concept of topological spaces, which are generalisation of metric spaces.
Definition 5.1. A topological space is a set of points, with a family of subsets calledopen sets that satisfy the following axioms:
1. The whole point set and the empty set are open.
2. The sum of an arbitrary number of open sets is open, and the intersection of a finitenumber of open sets is open.
We denote a topological space by T = (X, τ), where X is the point set and τ is thefamily of open sets.
Definition 5.2. In a topological space, the complement of an open set is called a closedset.
Theorem 5.1. In a topological space, the intersection of an arbitrary number of closedsets is closed, and the sum of a finite number of closed sets is closed.
The proof is left as an exercise.
Definition 5.3. A neighbourhood of the point x ∈ T , a topological space, is a set thatcontains an open set containing x. x ∈ T is a contact point of the set M if everyneighbourhood of the point x contains at least one point of M . x is a limit point of Mif every neighbourhood of x contains infintely many points of M . The set of all contactpoints of M is called the closure of M , denoted as M .
Suppose (X, ρ) is a metric space. Then X together with the family of open setsdefined by the metric is a topological space. On the other hand, a topological space iscalled metrisable if a metric, i.e., well-defined distance function, can be introduced to thepoint set so that the sets which are open with respect of this metric coincide with theopen sets of the topological space, and then we say the topological space is induced bythe metric. It is an exercise that the in a metric space, a set U is a neighbourhood of apoint x with respect to the metric if and only if U is a neighbourhood of x with respectto the induced topology, and a point x is a contact point (resp. limit point) of a set Mwith respect to the metric if and only if x is a contact point (resp. limit point) of the setM with respect to the induced topology.
Example 5.1 (non-metrisable topological space). Let X = {a, b}, with open sets ∅, {a}and {a, b}. This is a well-defined topological space, but it is not induced by any metric.The reason is that any distance ρ on T such that ρ(a, b) = d > 0 defines four open sets:∅, {a} = B(a, d/2), {b} = B(b, d/2) and {a, b}.
Definition 5.4. A topological space is called a Hausdorff space if any two distinct pointsx and y of the space T have disjoint neighbourhoods.
28
Definition 5.5. Let T = (X, τ) and T ′ = (X ′, τ ′) be two topological spaces. Themapping f : T → T ′ is continuous at the point x0 ∈ X if for any neighbourhood V of thepoint f(x0) ∈ X ′, the inverse image f−1(V ) ⊆ X is a neighbourhood of x0 in T . If f iscontinuous at all points of X, f is a continuous mapping from T to T ′.
Theorem 5.2. A mapping f : T = (X, τ)→ T ′ = (X ′, τ ′) between two topological spacesis continuous if and only if the inverse image of any open set in T ′ is open in T .
Proof. First suppose that f : T → T ′ is continuous at all points in T . Let V ⊆ T ′ be anopen set and U = f−1(V ) ⊆ T . Let x be any point in U , and then f(x) ∈ V . Since V isa neighbourhood of f(x), U = f−1(V ) is a neighbourhood of x, that is, there is an openset Ux such that x ∈ Ux ⊆ U . It is clear that
U =⋃x∈U
Ux,
and then U is open.On the other hand, suppose that f : T → T ′ satisfies that f−1(V ) is open in T for
any open set V in T ′. Let x be any point in and N be a neighbourhood of f(x), thenthere is an open set Vf(x) such that f(x) ∈ Vf(x) ⊆ N . Denote Ux = f−1(Vf(x)), we haveby the assumption that Ux is open. We have that
x ∈ Ux ⊆ f−1(N),
and then f−1(N) is a neighbourhood of x, and we prove that f is continuous at thearbitrary point x. Thus we prove that f is continuous.
Corollary 5.3. A mapping f : T → T ′ between two topological spaces is continuous ifand only if the inverse image of any closed set in T ′ is closed in T .
Corollary 5.4. If T , T ′ and T ′′ are topological spaces and f : T → T ′ and g : T ′ → T ′′ arecontinuous mappings, then the composite mapping i : T → T ′′ defined as i(x) = g(f(x))is continuous.
The proofs of Corollaries 5.3 an 5.4 are left as an exercise.The continuity of a mapping between two metric spaces has a more concrete definition.
Definition 5.6. Let R = (X, ρ) and R′ = (X ′, ρ′) be two metric spaces. The mappingf : R → R′ is continuous at the point x0 ∈ R if for arbitrary ε > 0, there exists aδ > 0 such that ρ′(f(x), f(x0)) < ε for all x ∈ R such that ρ(x, x0) < δ, or equivalently,f(Nδ(x0)) ⊆ Nε(f(x0)). A mapping f : R → R′ is continuous if it is continuous at eachpoint of R.
There is an equivalent definition of continuity of mappings between metric spaces thatis easier to work with.
Theorem 5.5. The mapping f is continuous at the point x if and only if for everysequence {xn} that converges to x, the image sequence {yn = f(xn)} converges to y =f(x).
29
Proof. First we prove the “only if” part. Suppose f : R → R′ is continuous at x ∈ X,and {xn} converges to x, we want to show that {yn} converges to y, that is, for any ε > 0,there is an Mε such that yn ∈ Nε(y) for n > Mε. Since f is continuous, for any ε > 0,there is δ > 0 such that f(Nε(x)) ⊆ Nε(y). Since {xn} converges to x, there is an M ′
δ
such that xn ∈ Nδ(x) for all n > M ′δ. Then it suffices to let Mε = M ′
δ.Next we prove the “if” part by proving its contraposition. Suppose f is not continuous
at x0, then there exists ε > 0 such that f(Nδ(x)) 6⊆ Nε(y) for all δ > 0. Thus we define{xn} by choosing xn in N1/n(x) such that y = f(xn) /∈ Nε(y). Then the sequence {xn}conferges to x, but the image sequence {yn} does not converge to y.
The next theorem confirms us that the two concepts of continuity of mappings areequivalent on a metric space with the induced topology.
Theorem 5.6. Let R = (X, ρ) and R′ = (X ′, ρ′) be two metric spaces, and let T and T ′
be the topological spaces induced by the two metric spaces respectively. Then a mappingf : X → X ′ is continuous at a point x0 ∈ X as a mapping between metric spaces R andR′ if and only if it is continuous at the point x0 as a mapping between topological spacesT and T ′.
Proof. Suppose first that f satisfies that for any ε > 0, there is δ > 0 such thatf(Nδ(x0)) ⊆ Nε(f(x0)), which is equivalent to that Nδ(x0) ⊆ f−1(Nε(f(x0))). Let V ⊆ Xbe any neighbourhood of f(x0). Then there is ε > 0 such that f(x0) ∈ Nε(f(x0)) ⊆ V .Suppose U = f−1(V ) ⊆ X ′. Then there exists δ > 0 such that
x0 ∈ Nδ(x0) ⊆ f−1(Nε(f(x0))) ⊆ U,
and we find that U is a neighbourhood of x0. Thus f is continuous as a mapping from Tto T ′.
On the other hand, suppose f is continuous as a mapping from T to T ′. SinceNε(f(x0)) is a neighbourhood of f(x0) for any ε > 0, f−1(Nε(f(x0))) is a neighbourhoodof x0. Hence there is δ > 0 such that
x ∈ Nδ(x0) ⊆ f−1(Nε(f(x0))).
(Note that in the proof we use the old definition of neighbourhood in Definition 2.1 inSecition 2. But it is an exercise that this definition agrees with the new definition ofneighbourhood in an induced topology.) Thus f is continuous at point x0 as a mappingfrom R and R′.
If a continuous mapping is bijective, then it has an inverse mapping, but the inversemapping may not be continuous.
Example 5.2. Let R be the metric space (X = [0, 2π) ⊂ R, ρ(x, y) = |x − y|), andR′ be the metric space (X = {(cos θ, sin θ) | θ ∈ [0, 2π)} ⊆ R2, ρ((x1, y1), (x2, y2)) =((x1− y1)2 + (x2− y2)2)1/2). Then the mapping f : R→ R′ where f(x) = (cos x, sinx) iscontinuous, but the inverse mapping f−1 is discontinuous at (1, 0). See Figure 14.
Proof. Consider the sequence {(cos(2π−1), sin(2π−1)), (cos(2π−12), sin(2π−1
2)), (cos(2π−
13), sin(2π− 1
3)), . . . }. It converges to (1, 0), but the image of the sequence by the mapping
f−1 does not converge to f−1((1, 0)) = 0. (Actually the image sequence has no limit atall.)
30
Figure 14: The inverse function f−1 : R′ → R is well defined, and f−1((1, 0)) = 0. Theinterval [0, ε) is an open set in R, although it is not an open interval. Its inverse imageunder f−1, (f−1)−1([0, ε)) = f([0, ε)) = {(cos θ, sin θ) | θ ∈ [0, ε)}, is not an open set inR′, since the point (1, 0) is nota an interior point of the inverse image.
Remark 5.1. The metric space R is a subspace of the Euclidean 1-space (real line), andR′ is a subspace of the Euclidean 2 space.
Definition 5.7. Let (X, ρ) be a metric space and A be a subset of X. Then the pointset A with the distance function ρ inherited from (X, ρ) form a metric space (A, ρ), andit is a subspace of (X, ρ).
Below we define some relations between metric spaces.
Definition 5.8. The mapping f : R→ R′ is a homeomorphism if it is bijective and bothf and f−1 are continuous. Then we say R and R′ are homeomorphic.
Definition 5.9. The mapping f : R→ R′ is an isometry if ρR′(f(x1), f(x2)) = ρR(x1, x2)for all x1, x2 in R. Note that an isometry is automatically continuous and injective. If fis also surjective, then f is an isomorphism, and R and R′ are isomorphic, or equivalently,isometric.
Two isometric spaces can be considered as the same spaces with different names, andthey are homeomorphic. (The proof is left as an exercise.)
Definition 5.10. The mapping f : R → R′ is a uniform isomorphism if f is bijectiveand there exist a, b > 0 such that
aρR(x, y) ≤ ρR′(f(x), f(y)) ≤ bρR(x, y)
for all x, y in R. If R = (X, ρ) and R′ = (X, ρ′) and f(x) = x is uniformly isomorphic,we say ρ and ρ′ are strongly equivalent metrics.
Theorem 5.7. Two strongly equivalent metrics induce the same topological space, i.e.,a set is open in (X, ρ) if and only if it is open in (X, ρ′) where ρ and ρ′ are stronglyequivalent.
The proof is left as an exercise.
Remark 5.2. The terms defined above are all standard, and they are defined as inWikipedia. But their names are seemingly conflicting. An isometric mapping does notmake two metric spaces isometric, although an isomorphic mapping makes two metricspaces isomorphic. Also, uniform isomorphism is a weaker condition than isomorphism,to the contray of uniform continuity vs continuity or uniform convergence vs conver-gence. You do not need to worry the perplexing terms. Whenever these terms are usedin homework or exam, I will give explicit definition in fear of confusion.
31
Recapitulation
• Definitions: topological space, open set, closed set, neighbourhood, contact point,limit point, closure, (the six preceding definitions are defined in topological space aswell as metric space,) metrisable space, Hausdorff space, continuity at a point, con-tinuous mapping, homeomorphism, isometric mapping, isometrism/isomorphismbetween two metric spaces, uniform isomorphism, strongly equivalent metrics.
• Notable results:
– A topological space may not be metrisable.
– The inverse of a bijective and continuous mapping may not be continuous.
• Theorems:
– Arbitrary sum and finite intersection of open sets are open; finite sum andarbitrary intersection of closed sets are closed.
– A mapping f : R→ R′ is continuous at x0 ∈ R if and only if for any {xn} → x,{f(xn)} → f(x).
– A mapping f : R → R′ is continuous if and only if the inverse image of anyopen/closed set in R′ by f is also open/closed in R.
– The composition of two continuous mappings is continuous.
– A pair of strongly equivalent metrics induces the same topological space.
32
6 Complete metric spaces
Definition 6.1. A sequence {xn} of points in a metric space R is a Cauchy sequence if itsatisfies the Cauchy criteron that for any ε > 0, there exists Nε such that ρ(xm, xn) < εfor all m,n > Nε.
Every convergent sequence is a Cauchy sequence. Suppose {xn} converges to x, thenfor any ε > 0, there is an N ′ε such that ρ(xn, x) < ε/2 for all n > N ′ε. Then we can takeNε = N ′ε, since for all m,n > Nε = N ′ε, ρ(xm, xn) ≤ ρ(xm, x) + ρ(x, xn) < ε/2 + ε/2 = ε.
The converse is not true generally.
Example 6.1. Let R = (Q, ρ(x, y) = |x− y|), a subspace of the real line with the usualmetric. Let the sequence {xn} be defined by xn =
∑nk=0 1/k!. This is a Cauchy sequence,
since for any ε, we can choose Nε to be an integer such that N−1ε < ε. For n > m > Nε,
ρ(xm, xn) =n∑
k=m+1
1
k!=
1
m!
(1
m+ 1+
1
(m+ 1)(m+ 2)+ · · ·+ 1
(m+ 1) . . . n
)≤ 1
m!
(1
2+
1
2 · 2+ · · ·+ 1
2n−m
)<
1
m!< ε.
But this sequence does not converge to a rational number. If it has a rational number asits limit, then it has the same limit as a sequence of real numbers. But it then impliesthat e is rational, contradictory to a well-known fact.
Hence we say that the metric space R considered in Example 6.1 is incomplete.
Definition 6.2. A metric space is complete if every Cauchy sequence has a limit.
Example 6.2. A celebrated result that we learnt in Mathematical Analysis I is that thereal line with the usual metric is complete.
Most other non-trivial results of completeness depend on the completeness of R.
Example 6.3. The Euclidean n-space is complete.
Proof. Let {x(k) = (x(k)1 , . . . , x
(k)n )} be a Cauchy sequence. Then for any ε > 0, there
exists Nε such that for any k, l > Nε,
ρ(x(k), x(l)) =
√(x
(k)1 − x
(l)1 )2 + . . .+ (x
(k)n − x(l)
n )2 < ε.
Then for any i = 1, 2, . . . , n, we have that for all k, l > Nε, |x(k)i − x
(l)i | < ε, and then
the sequence {x(k)i } is a Cauchy sequence in R (with the usual metric) as k runs over
Z+. By the completeness of R, we find that {x(k)i } converges to a real number, denoted
as xi. Then we see that {x(k)} converges to x = (x1, . . . , xn) due to the coordinateconvergence.
Recall that in Mathematical Analysis I we learnt that the completeness of R is equiva-lent to the nested interval theorem. The next theorem is a straightforward generalizationto metric space.
33
Theorem 6.1 (Nested ball theorem). The metrix space R is complete if and only if everysequence of nested closed balls with radii converging to zero has non-empty intersection.
Proof. First we prove the “only if” part, and suppose that the metric space R is complete.Let B[x1, r1] ⊇ B[x2, r2] ⊇ · · · with rn → 0. We show that the sequence of the centers{xn} has a limit, say x, and x is in the intersection of all the closed balls.
For any ε > 0, we choose Nε such that rNε < ε/2. Let m,n be any numbers largerthan Nε. We have xm ∈ B[xm, rm] ⊆ B[xNε , rNε ] and then ρ(xm, xNε) ≤ rNε , and similarlyρ(xn, xNε) ≤ rNε . Hence ρ(xm, xn) ≤ ρ(xm, x) + ρ(xn, x) ≤ 2rNε < ε. We conclude that{xn} is a Cauchy sequence, and then it has a limit x due to the completeness of R.To show that x is in any B[xn, rn], we consider the sequence {xn+1, xn+2, . . . } that iscontained in B[xn, rn]. Then their limit x is also in B[xn, rn] that is a closed set.
Next we prove the “if” part by showing that if R is not complete, then there ex-ist a sequence of nested balls whose radii shrink to zero have empty intersection. Let{x1, x2, . . . } be a Cauchy sequence in R that does not converge. For each k ∈ Z+, welet Nk be a large enough such that ρ(xm, xn) ≤ 2−k−1 for all m,n > Nk. Then we con-sider the balls B[xN1 ,
12], B[xN2 ,
14], . . . . They are nested, since if m < n, then any point
x ∈ B[xNn , 2−n] satisfies
ρ(x, xNm) ≤ ρ(x, xNn) + ρ(xNn , xNm) ≤ 1
2n+
1
2m+1≤ 1
2m,
so x is also in B[xNm , 2−m]. Below we show that these closed balls have empty intersection.
Suppose otherwise x is a point in the intersections of all the closed balls. For any ε > 0,we take k to be large enough such that 2−k < ε/2, then for all n > Nk,
ρ(x, xn) ≤ ρ(x, xNk) + ρ(xNk , xn) ≤ 1
2k+
1
2k≤ ε. (9)
This implies that x is the limit of {xn}, contradictory to the assumption that {xn} hasno limit.
From the examples of Q and R, we see that the incomplete metric space (Q, ρ) is thesubspace of (R, ρ). In the remaining part of this section, we show that any incompletemetric space is a subspace of a complete metric space.
Definition 6.3. Let R = (X, ρ) be a metric space. A complete metric space R∗ =(X∗, ρ∗) is said to be the completion of R if R is isometric to an everywhere densesubspace R′ = (X ′, ρ∗) of R∗. That is, X ′ is everywhere dense in R∗ and there is anisomorphism f : X → X ′ between R and R′.
Example 6.4. R is the completion of Q, where the metric is the usual ρ(x, y) = |x− y|.
Theorem 6.2. Every metric space has a completion, and for a metric space R = (X, ρ),if R∗1 = (X∗1 , ρ
∗1) and R∗2 = (X∗2 , ρ
∗2) are both completions of R, such that the subspaces
R′1 = (X ′1, ρ∗1) and R′2 = (X ′2, ρ
∗2) of R∗1 and R∗2 respectively are the everywhere dense
subspaces with isomorphisms f1 : R→ R′1 and f2 : R→ R′2, then there is an isomorphismf ∗ : R∗1 → R∗2 such that for any x ∈ X ′1, f ∗(x) = f2(f−1
1 (x)).
To prove Theorem 6.2, we recall the concept of equivalence relation.
34
Definition 6.4. A binary relation ∼ defined on a set X is called an equivalence relationif it satisfies
1. (Reflexivity) a ∼ a.
2. (Symmetry) if a ∼ b, then b ∼ a.
3. (Transitivity) if a ∼ b and b ∼ c, then a ∼ c.
An equivalence relation ∼ on a set X divides the set into equivalent classes [a] = {x ∈X | x ∼ a}, and these equivalent classes form a set, called the quotient set X/ ∼.
We define an equivalence relation on the set of Cauchy sequences in a metric space asfollows.
Definition 6.5. The Cauchy sequences {xn} and {yn} in a metric space R are calledequivalent and denoted as {xn} ∼ {yn} if limn→∞ ρ(xn, yn) = 0.
It is straightforward to check that the relation ∼ is an equivalence relation. Thenthey define the equivalent classes of Cauchy sequences
[{xn}] = the set of Cauchy sequences {yn} such that {yn} ∼ {xn}.
Although in an incomplete metric space, a Cauchy sequence may not have a limit, wehave the following two theorems that are valid for all metric space.
Theorem 6.3. In a metric space, if two Cauchy sequences {xn} and {yn} converge tothe same limit, then they are equivalent. Conversely, if two Cauchy sequences {xn} and{yn} are equivalent and {xn} converges to x, then {yn} also converges to x.
Theorem 6.4. Let {xn} be a Cauchy sequence. Then any subsequence of it, {xnk}, is aCauchy sequence and {xn} ∼ {xnk}.
The proofs are left as an exercise.
Proof of Theorem 6.2. We first show that if R∗1 and R∗2 are both completions of R asdescribed in the theorem, then the isomorphism f ∗ exists. We denote the mappingf : R′1 → R′2 as f(x) = f2(f−1(x)). It is clear that f is an isomorphism.
Let x be any point in X∗1 . Then x is a limit point of X ′1 such that there is a sequence{x1, x2, . . . } ⊆ X ′1 that converges to x. We extend the mapping f to f ∗ on R∗1 such thatit maps x into a point in X∗2 such that
f ∗(x) = limn→∞
f(xn). (10)
Since f is an isomorphism from R′1 to R′2, we find that {f(xn)} ⊆ X ′2 is a Cauchy sequencein R′2 as well as in R∗2. Then by the completeness of R∗2, the sequence {f(xn)} has a limitin R∗2, and then f ∗(x) defined by (10) exists. To show that f ∗(x) is well-defined mapping,we need to show that it does not depend on the sequence that approaches x, that is, ifwe take another sequence {x′n} ⊆ X ′1 such that x′n → x, the definiton
f ∗(x) = limn→∞
f(x′n) (11)
35
gives the same result as (10). By Theorem 6.3, {xn} and {x′n} are equivalent Cauchysequences in R∗1 as well as in R′1. Since f is an isometry, {f(xn)} and {f(x′n)} areequivalent Cauchy sequences in R′2 as well as in R∗2. Use Theorem 6.3 again, we find that{f(x′n)} converges to the same limit in R∗2 as {f(xn)} does. Thus f ∗(x) is well-defined.By this construction, we extend the mapping f to a mapping f ∗ from R∗1 to R∗2.
We show below that f ∗ : R∗1 → R∗2 is surjective. Any y ∈ X∗2 is the limit of a Cauchysequence {yn} ⊆ X ′2. Suppose yn = f(xn), then {xn} ⊆ X ′1 is a Cauchy sequence in R′1as well as in R∗1 since f is an isomorphism between R′1 and R′2 = (X ′2, ρ
∗2). Then {xn}
has a limit in R∗1, say x. We have that
f ∗(x) = limn→∞
f(xn) = limn→∞
yn = y.
Next we show that f ∗ : R∗1 → R∗2 is isometric. Let x, y ∈ X∗1 be limits of {xn}, {yn} ⊆X ′1 respectively, then
ρ∗2(f ∗(x), f ∗(y)) = ρ∗2( limn→∞
f(xn), limn→∞
f(yn)) = limn→∞
ρ∗2(f(xn), f(yn)) = limn→∞
ρ∗1(xn, yn)
= ρ∗1(x, y).
We now finish the proof that the extended mapping f ∗ : R∗1 → R∗2 is the desired isomor-phism.
Next we show that given any metric space R = (X, ρ), we can construct a metricspace that is the completion of R. We define
S = {Cauchy sequences in R},S/ ∼= {equivalence classes of fundamental Cauchy sequences in R}.
We define a distance function ρ∗ on S/ ∼ as
ρ∗([{xn}], [{yn}] = limn→∞
ρ(xn, yn). (12)
Now we show that this function is well-defined. First, the limit in (12) is well defined,because the sequence {ρ(xn, yn)} is a Cauchy sequence in R, and it has a limit by thecompleteness of R (exercise). Next we show that the limit in (12) does not depend onthe particular choice of {xn}, {yn}, that is, if we use other sequences {x′n}, {y′n} such that{x′n} ∼ {xn}, {y′n} ∼ {yn} to represent the equivalence classes [{xn}], [{yn}], we have
limn→∞
ρ(x′n, y′n) = lim
n→∞ρ(xn, yn). (13)
To prove the identity (13), we first show that
limn→∞
ρ(x′n, y′n) ≤ lim
n→∞ρ(x′n, xn) + ρ(xn, yn) + ρ(yn, y
′n)
= limn→∞
ρ(x′n, xn) + limn→∞
ρ(xn, yn) + limn→∞
ρ(yn, y′n)
= limn→∞
ρ(xn, yn).
Similarly we can also show
limn→∞
ρ(xn, yn) ≤ limn→∞
ρ(x′n, y′n).
Then the identity (13) is proved.Furthermore, the distance function ρ∗ makes S/ ∼ a metric space. We check that it
satisfies all the four conditions:
36
1. Nonnegativity: obvious.
2. Symmetry: It is straightforward to see that ρ∗([{xn}], [{yn}]) = limn→∞ ρ(xn, yn) =limn→∞ ρ(yn, xn) = ρ∗([{yn}], [{xn}]).
3. Identity of indiscernibles: if ρ∗([{xn}], [{yn}]) = limn→∞ ρ(xn, yn) = 0, then xn ∼ ynby definition, and [{xn}] = [{yn}].
4. Triangle inequality:
ρ∗([{xn}], [{yn}]) + ρ∗([{yn}], [{zn}]) = limn→∞
ρ(xn, yn) + limn→∞
ρ(yn, zn)
≥ limn→∞
ρ(xn, yn) = ρ∗([{xn}], [{zn}]).
Thus the metric space (S/ ∼, ρ∗) is well defined, and we denote it as R∗. It containsa subspace R′ = (X ′, ρ∗) where X ′ = {[{x, x, x, . . . }] | x ∈ X}. It is straightforward tocheck that the mapping i : R→ R′ with i(x) = [{x, x, x, . . . }] is an isomorphism.
The subset X ′ is everywhere dense in R∗. For any point x = [{x1, x2, . . . }] ∈ S/ ∼,we consider the sequence {xn} ⊆ X ′ such that xn = [{xn, xn, xn, . . . }], and show that itconverges to x. Since {xn} is a Cauchy sequence, for any ε > 0, there is an Nε such thatfor all m,n ≥ Nε, ρ(xm, xn) < ε. Then for any n > Nε,
ρ∗(xn, x) = limm→∞
ρ(xm, xn) = limm>Nε,m→∞
ρ(xm, xn) ≤ ε.
Thus x is the limit of {xn}.We at last show that R∗ is a complete metric space. Let {[{x(1)
n }], [{x(2)n }], . . . } be
a Cauchy sequence in R∗. Then since X ′ is everywhere dense in R∗, there exists asequence {x1, x2, . . . } ⊆ X ′ such that xk = [{xk, xk, xk, . . . }] ∈ R, and ρ(xk, [{x(k)
n }]) <2−k. It is straightforward to check that {xn} is a Cauchy sequence in R∗, as well asin R, and then {xn} is a Cauchy sequence in R. By the argument used for the proofof the everywhere dense property of X ′, we have that the sequence {xn} converges to[{x1, x2, x3, . . . }]. Then we find that [{x1, x2, x3, . . . }] is also the limit of the Cauchy
sequence {[{x(1)n }], [{x(2)
n }], . . . }.
An important property of complete metric spaces is given by Baire’s theorem.
Definition 6.6. A set A in a metric space is nowhere dense if it is not dense in any openball.
Example 6.5. in R with the usual metric, Fq = {p/q | p ∈ Z} is nowhere dense.
Theorem 6.5 (Baire’s theorem). In a complete metric space (X, ρ), X cannot be theunion of countably many nowhere dense sets.
We are going to prove a stronger result: In a complete matric space, any nonemptyclosed ball cannot be covered by the union of countably many nowhere dense sets.
37
Proof. Let A1, A2, . . . be a collection of countably many nowhere dense sets. TakeB[x0, ε0] to be an arbitrary closed ball with ε0 > 0. We construct by induction a se-quence of nested closed balls B[x0, ε0] ⊇ B[x1, ε1] ⊇ B[x2, ε2] ⊇ · · · sucht that B[xn, εn]is disjoint to
⋃ni=1An and the radii εn → 0. Since the metric space is complete, there
exists x ∈ ∩∞n=1B[xn, εn] by the nested ball theorem. We have that x /∈⋃∞n=1An, and
then⋃∞n=1 An does not cover B[x0, ε0], let alone X.
The induction is as follows. Suppose we have already B[xk, εk]. Ak+1 is nowheredense, so Ak+1 does not contain B(xk, εk). Since B(xk, εk) is open and Ak+1 is closed, theset B(xk, εk) \ Ak+1 is a nonempty open set. We take any point in B(xk, εk) \ Ak+1 anddenote it as xk+1. Since xk+1 is an interior point of B(xk, εk) \ Ak+1, there is εk+1 > 0,that we can assume to be less than εk/2, such that B(xk+1, 2εk+1) ⊆ B(xk, εk) \ Ak+1.Then we have
B[xk+1, εk+1] ∩k+1⋃i=1
Ai =
(B[xk+1, εk+1] ∩
k⋃i=1
Ai
)∪ (B[xk+1, εk+1] ∩ Ak+1)
⊆
(B[xk, εk] ∩
k⋃i=1
Ai
)∪ (B(xk+1, 2εk+1) ∩ Ak+1) = ∅.
Also we have εn → 0.
On consequence of Baire’s theorem is that a reasonable complete space has “many”points.
Corollary 6.6. A complete metric space X with the property that any open ball hasinfinitely many points, (i.e., no point is an isolated point), has uncountably many points.
Proof. Suppose X = {x1, x2, . . . } is countable. Since any single-point set {xn} is nowheredense, the assumption X =
⋃∞n=1{xn} is contradictory to Baire’s theorem.
38
Recapitulation
• Definitions: complete metric space, completion, nowhere dense set, equivalenceclass (this is not a definition in analysis, but nevertheless useful).
• Notable results:
– R with the usual metric is complete.
– Euclidean n-space is complete.
– Q, as a subspace of R, is incomplete, with R being its completion.
• Theorems:
– Nested ball theorem: A metric space is complete if and only if any sequenceof nested, shrinking and closed balls have nonempty intersection.
– Any metric space has a completion, and all completions of the metric spaceare isometric.
– Baire’s theorem: A complete metric space is not the countable union of nowheredense subsets.
39
7 Principle of cntraction mappings
Definition 7.1. A mapping f from a metric space R to R itself is a contraction if thereexists α ∈ (0, 1) such that ρ(f(x), f(y)) ≤ αρ(x, y) for all x, y ∈ R.
Example 7.1. The function f : R+ → R+ with f(x) = x2
+ 32x
is NOT a contraction,but if we consider f on the subspace [1.5,∞) of R, then f is a contraction.
Proof. First we need to show for any x ∈ [1.5,∞), f(x) ∈ [1.5,∞) also. That is done bya simple calculation. Next, we have by the mean value theorem
|f(x)− f(y)||x− y|
= |f ′(z)|,
where z is between x and y. Then by the inequality |f ′(z)| = |1/2 − 3/(2z)| ≤ 0.5 forz ∈ [1.5,∞), we find that f is a contraction with the contraction parameter α = 1
2.
Theorem 7.1 (Principle of contraction mappings). Every contraction mapping definedin a complete metric space R has a unique fixed point, that is, the equation f(x) = x has aunique solution. Moreover, for any x ∈ R, the sequence of iterates {x, f(x), f(f(x)), . . . , fn(x), . . . }converges to the fixed point of f .
Proof. First we show the existence of the fixed point. Let x be any point in R. Thesequence {fn(x)} is a Cauchy sequence. To see it, denote ρ(x, f(x)) = M , then for anym < n,
For any ε > 0, we takeNε large enough such that αNε/(1−α) < ε/M , then ρ(fm(x), fn(x)) <ε for all m,n > Nε.
Then the sequence {fn(x)} has a limit, denoted as x0. It straightforward to verifythat a contraction mapping is a continuous mapping, so
f(x0) = limn→∞
f(fn(x)) = limn→∞
fn+1(x) = x0.
Thus we proved the existence of the fixed point.Next we show that the fixed point is unique. Suppose x0 and y0 are both fixed points
of the contraction mapping f , then the inequality
ρ(x0, y0) = ρ(f(x0), f(y0)) ≤ αρ(x0, y0)
implies that ρ(x0, y0) = 0, and then x0 = y0.
40
Remark 7.1. The completeness of the metric space is crucial. For example, th mapping fin Example 7.1 is also well defined and is a contraction with α = 1/2 if we let the spaceR′ be the subspace of R with the point set {x ∈ Q | x ≥ 1.5}. Then the contractionmapping f has no fixed point. (In the complete space R, the fixed point can be solvedexplicitly as x =
√3.)
Example 7.2. In R with the usual metric, f : R→ R is a contraction if and only if it isLipschitz continuous with the Lipschitz constant α less than 1, i.e., |f(x)−f(y)| ≤ α|x−y|.For such f , f(x) = x has a unique solution, and the iterates fn(x) for any x convergesto the fixed point. See Figure 15.
Figure 15: The intersections of the graph y = f(x) and the infinite polygonal chainrepresent the iterates.
In the following three examples we consider metric spaces with point set Rn, and themapping ~y = f(~x) = A~x+~b, that is,
yi =n∑j=1
aijxj + bi, for i = 1, . . . , n,
where ~x = (x1, . . . , xn)T , ~y = (y1, . . . , yn)T , ~b = (b1, . . . , bn)T and A = (aij)ni,j=1. We
consider under what condition the mapping is a contraction. Of course it depends on themetric space that we consider.
Example 7.3. In R1 = (Rn, ρ1), f is a contraction if and only if for any ~x, ~x′ ∈ R1
ρ1(~y, ~y′) =n∑i=1
|yi − y′i| =n∑i=1
aij(xj − x′j)| ≤ αn∑j=1
|xj − x′j| = αρ1(~x, ~x′).
Note that
n∑i=1
|n∑j=1
aij(xj − x′j)| ≤n∑i=1
(n∑j=1
|aij||xj − x′j|
)=
n∑j=1
(n∑i=1
|aij|
)|xj − x′j|.
Thus a sufficient condition for f to be a contraction is∑n
i=1|aij| ≤ α < 1 for all j =1, . . . , n.
41
Example 7.4. In R2 = (Rn, ρ2), f is a contraction if and only if for any ~x, ~x′ ∈ R2
ρ2(~y, ~y′) =
(n∑i=1
(yi − y′i)2
) 12
=
n∑i=1
(n∑j=1
aij(xj − x′j)
)2 1
2
≤ α
(n∑j=1
(xj − x′j)2
) 12
= αρ2(~x, ~x′).
Note that by the Schwarz inequality,
n∑i=1
(n∑j=1
aij(xj − x′j)
)2
≤n∑(
n∑j=1
a2ij
)(n∑j=1
(xj − x′j)2
)
=
(n∑i=1
n∑j=1
a2ij
)(n∑j=1
(xj − x′j)2
).
Thus a sufficient condition for f to be a contraction is∑n
i=1
∑nj=1 a
2ij ≤ α2 < 1.
Example 7.5. In R∞ = (Rn, ρ∞), f is a contraction if and only if for any ~x, ~x′ ∈ R∞
ρ∞(~y, ~y′) =n
maxi=1|yi − y′i| =
nmaxi=1
∣∣∣∣∣n∑j=1
aij(xj − x′j)
∣∣∣∣∣ ≤ αn
maxj=1|xj − x′j|.
Note that
nmaxi=1
∣∣∣∣∣n∑j=1
aij(xj − x′j)
∣∣∣∣∣ ≤ nmaxi=1
n∑j=1
|aij|(
nmaxj=1|xj − x′j|
)=
(n
maxi=1
n∑j=1
|aij|
)(n
maxj=1|xj − x′j|
).
Thus a sufficient condition for f to be a contraction is∑n
j=1|aij| ≤ α < 1 for all i =1, . . . , n.
The three conditions in Examples 7.3–7.5 are not equivalent. The three examplesshow that different metrics can be used for various fixed point problem.
42
8 Application of the principle of contraction map-
pings: Picard’s theorem
(Note: In this section, we consider the metric space C∞(I) where I is a closed interval,and abbreviate it as C(I). Also we use the result of a tutorial problem that C(I) is acomplete metric space.)
One important application of the principle of contraction mappings is the proof ofPicard’s theorem on the existence and uniqueness of the local solution of the initial valueproblem of ordinary differential equation (ODE)
dy
dx= f(x, y), where y(x0) = y0, (14)
such that f is a continuous function on an open set G ⊆ R2 that contains (x0, y0), andsatisfies the Lipschitz condition with respect to y with Lipschitz constant M such that
|f(x, y1)− f(x, y2)| ≤M |y1 − y2| for all (x, y1), (x, y2) ∈ G. (15)
Note that the Lipschitz condition is important for the uniqueness.
Example 8.1. Consider the ODE
dy
dx= 2√|y|,
where the function f is well defined on the whole R2, but the Lipschitz condition is notsatisfied when around the axis y = 0. The ODE with initial condition y(0) = 0 has twosoluitons:
y(x) = 0,
and
y(x) =
{x2 x ≥ 0,
−x2 x < 0.
Later we will see that the existence does not depend on the Lipschitz condition.To apply the principle of contraction mappings, we consider not the initial value of
ODE directly, but an equivalent integral equation problem.
Lemma 8.1. A continuous function y(x) ∈ C(I) where I = [x1, x2] containing x0 satisfiesthe initial value problem of the ODE (14) if and only if it satisfies the integral equation
y(x) = y0 +
∫ x
x0
f(t, y(t))dt. (16)
Proof. Suppose y(x) ∈ C(I) satisfied (14), then y(x) is a differentiable function in x withcontinuous derivative. Taking integral on both sides of (14), we obtain (16). On the otherhand, suppose y(x) ∈ C(I) satisfies (16). The right-hand side of (16) is an integral withcontinuous integrand for all x ∈ I, so y(x) is differentiable with continuous derivative.Taking derivative on both sides of (16), we obtain (14).
43
The main result of this section is
Theorem 8.2 (Picard). Suppose f is a continuous function on G ⊆ R2 and satisfiesthe Lipschitz condition (15). For any (x0, y0) ∈ G, there is an interval I = [x1, x2] thatcontains x0 such that there exists a unique y ∈ C(I) that satisfies the integral equation(16).
This is equivalent to the Picard theorem in the more convenient form:
Theorem 8.3 (Picard). Suppose f is a continuous function on an open set G ⊆ R2
and satisfies the Lipschitz condition (15). For any (x0, y0) ∈ G, there is an intervalI = [x1, x2] that contains x0 such that there exists a unique y ∈ C(I) that satisfies theinitial value problem of ordinary differential equation (14).
The proof of Theorem (8.2) requires a technical result:
Theorem 8.4. Suppose (X, ρ) is a complete metric space and A is a closed set in X.Then the subspace (A, ρ) is also a complete metric space.
Proof. Let {xn} ⊆ A be any Cauchy sequence in the subspace A (and hence a Cauchysequence in X). Then by the completeness of X, it converges to x ∈ X. By the closednessof A, we have that x ∈ A. Thus A is complete.
Proof of Theroem 8.2. Since f is continuous at (x0, y0) ∈ G, there exists K > 0 andd ∈ (0, 1/M) such that the rectangle [x0 − d, x0 + d] × [y0 − Kd, y0 + Kd] ⊆ G and|f(x, y)| < K for all (x, y) in the rectangle, see Figure 16. We prove that it suffices to letI = [x0 − d, x0 + d].
Figure 16: The rectangle and the interval I in the proof of Picard’s theorem.
The constant function y(x) = y0 is a point in C(I), and the closed ball B[y0, Kd] thatcontains all continuous functions ϕ on I such that |ϕ(x) − y0| ≤ Kd for all x ∈ I is aclosed set in C(I). The completeness of C(I) and Theorem 8.4 imply that the subspaceof C(I) consisting of points in B[y0, Kd] is a complete metric space. We denote it C∗
below.Consider the mapping F : C∗ → C∗ such that
(Fϕ)(x) = y0 +
∫ x
x0
f(t, ϕ(t))dt, for x ∈ I.
44
First we show that F is well defined. For any x ∈ I and ϕ ∈ C∗,
|(Fϕ)(x)− y0| =∣∣∣∣∫ x
x0
f(t, ϕ(t))dt
∣∣∣∣≤∣∣∣∣∫ x
x0
|f(t, ϕ(t))|dt∣∣∣∣
≤∣∣∣∣∫ x
x0
Kdt
∣∣∣∣ = K|x− x0| ≤ Kd.
Next we show that F is a contraction. For any ϕ1, ϕ2 ∈ C∗,
ρ(Fϕ1, Fϕ2) = maxx∈I|(Fϕ1)(x)− (Fϕ2)(x)|
= maxx∈I
∣∣∣∣∫ x
x0
f(t, ϕ1(t))dt−∫ x
x0
f(t, ϕ2(t))dt
∣∣∣∣≤ max
x∈I
∣∣∣∣∫ x
x0
|f(t, ϕ1(t))− f(t, ϕ2(t))|dt∣∣∣∣
≤ maxx∈I
∣∣∣∣∫ x
x0
M |ϕ1(t)− ϕ2(t)|dt∣∣∣∣
≤ dMρ(ϕ1(t), ϕ2(t)).
Thus F is a contraction with the contraction factor α = dM < 1.By the principle of contraction mappings, F has a unique fixed point in C∗, and it
is a solution to (16). Now we prove the existence of the solution to (16). But for theuniqueness, we need to show that the solution we have found is unique not only in C∗
but in C(I), that is, no function in C(I) \ C∗ can be a solution to (16).Suppose to the contrary φ(x) ∈ C(I) \C∗ is a solution to (16). Then |φ(x′)−φ(x0) =
y0| > Kd for some x′ ∈ I. Without loss of generality, suppose x′ ∈ (x0, x0 + d]. Thenby the continuity of φ, there exists x′′ ∈ (x0, x
′) such that |φ(x′′) − φ(x0)| = Kd and|φ(x) − φ(x0)| < Kd for all x ∈ (x0, x
′′). See Figure 17. Note that φ(x) is differentiabledue to the assumption that φ(x) satisfies (16). Then by the mean value theorem, thereis x ∈ (x, x′′) such that
|φ′(x)| = |φ(x′′)− φ(x0)|x′′ − x0
>Kd
d= K.
Since (x, φ(x)) is in the rectangle [x0 − d, x0 + d]× [y0 −Kd, y0 +Kd], |f(x, φ(x))| < K.Thus φ cannot satisfy (16). We finish the proof of the uniqueness.
45
Figure 17: The choice of x′ and x′′.
9 Sequentially compact sets in metric spaces
Definition 9.1. A set M in the metric space R is relatively sequencially compact if everysequence of elements in M contains a subsequence that converges to some point x ∈ R.M is sequentially compact if it is relatively sequentially compact and closed, that is, everysequence of elements in M has a subsequence that converges to some point x ∈M .
Remark 9.1. Another definition of sequentially compact set is that M is sequentiallycompact if every sequence of elements in M has a subsequence that converges to somepoint x ∈M . This definition looks weaker, since it does not require M to be closed, andthen a sequence in M may have two kinds of convergent subsequences, ones with limitsin M and others with limits outside of M . However, these two definitions are equivalent.Suppose M is sequentially compact by the second definition. Suppose {xn} ⊆ M is anyconvergent sequence, but its limit x may be or may not be in M . By assumption, {xn}has a convergent subsequence, say {xnk} whose limit is in M . But the limit of {xnk}has to be identical to the limit of {xn}, so we have that x ∈M , and conclude that M isclosed.
Example 9.1 (Bolzano-Weierstrass). A bounded set in the Eclidean n space is rela-tively sequentially compact; a bounded closed set in the Euclidean n-space is sequentiallycompact.
Example 9.2. The closed ball M = B[(0, 0, 0, . . . ), 1] in the `2 space is bounded inthe usual sense that supx,y∈M ρ2(x, y) = 2 < ∞, but is not sequentially compact. Thesequence {xn} where
xn = (0, 0, . . . , 0︸ ︷︷ ︸n−1
, 1, 0, 0, . . . )
does not have any convergent subsequence, since ρ2(xm, xn) =√
2 for any m 6= n.
Definition 9.2. For any set M in the metric space R, the set A is an ε-net with respectto M if for any point x ∈ M , there exists an a ∈ A such that ρ(a, x) < ε. The set M istotally bounded if for any ε > 0, there exists a finite ε-net with respect to M .
46
Note that a totally bounded set is bounded in the usual sense, but generally a boundedset may not be totally bounded, as shown in Example 9.2.
Example 9.3. In the Euclidean n-space, a set is totally bounded if and only if it isbounded.
Proof. Suppose M is a bounded set in the Euclidean n-space. We cover M in a largecube. For any ε > 0, we divide the cube into small enough cubicles, such that the sideof each cubicle is less than ε/n, and then the vertices of the cubicles form an ε-net forM .
Theorem 9.1. A set M in a complete metric space R is relatively sequentially compactif and only if M is totally bounded.
Proof. First we prove the “only if” part by contradiction. Suppose M is not totallybounded. Then there exists ε > 0 such that no finite set is an ε-net for M . We constructa sequence as follows. First take any x1 ∈ M , and then choose xn recursively such thatxn ∈ M and ρ(xn, xi) ≥ ε for all i = 1, . . . , n − 1. This is possible since {x1, . . . , xn−1}do not form an ε-net. Then by the construction of xn, we have ρ(xm, xn) ≥ ε for allm 6= n, and then no subsequence of {xn} is Cauchy, let along convergent. Thus M is notrelatively sequentially compact.
Then we prove the “if” part. Suppose M is totally bounded such that for any k, it hasa 2−k-net Ak = {a(k)
1 , . . . , a(k)nk }, and {xn} is a sequence in M . We construct a subsequence
{x′n} of {xn} that converges. Without loss of generality, we assume xn are distinct.
Since the closed balls B[a(1)i , 1/2] (i = 1, . . . , n1) cover M , at least one of them contains
infinitely many terms of {xn}, that is, a subsequence of {xn}. We denote the subsequence
as {x(1)n }, and let x′1 = x
(1)1 . Then we find other x′n recursively. Suppose we have {x(k)
n },a subsequence of {xn}, we can select a subsequence of it, say {x(k+1)
n }, that is covered
by a ball B[a(k)i , 1/2k]. We then let x′k+1 be the first element in {x(k+1)
n } that is distinctto x′1, . . . , x
′k. By the construction, for any m < n, x′m and x′n are both covered by a
closed ball B[a(m)i , 1/2m], and ρ(x′m, x
′n) ≤ 1/2m−1. Thus {x′m} is a Cauchy sequence and
it converges.
Remark 9.2. From the proof of Theorem 9.1, we find that a relatively sequentially compactset is totally bounded even if the metric space is not complete.
Corollary 9.2. A set M in a complete metric space is relatively sequentially compact ifand only if for any ε > 0, there exists a sequentially compact ε-net for M .
Proof. If M is relatively sequentially compact, then for any ε > 0 it has a finite ε-netthat is sequentially compact (exercise). On the other hand, suppose for any ε > 0, Mhas a sequentially compact ε/2-net A. Then A has a finite ε/2-net B by Theorem 9.1.We find that B is an ε-net for M . Thus M is relatively sequentially compact.
47
10 Arzela-Ascoli theorem and its applications
For the important metric space C([a, b]) = C∞([a, b]), we have a criterion of relativelysequentially compact sets.
Definition 10.1. A family of functions {ϕ(x)} ⊆ C([a, b]) are uniformly bounded if thereexists a number M such that |ϕ(x)| < M for all x ∈ [a, b] and for all ϕ in this family.
Definition 10.2. A family of functions {ϕ(x)} ⊆ C([a, b]) are equicontinuous at x ∈ [a, b]if for every ε > 0, there is a δ > 0 such that |ϕ(x′) − ϕ(x)| < ε for all x′ ∈ [a, b] suchthat |x′ − x| < δ and for all ϕ in this family. If the family of functions is equicontinuousat all points in [a, b], then the family of functions are equicontinuous. We say the familyof functions are uniformly equicontinuous if for every ε > 0, there is a δ > 0 such that|ϕ(x1) − ϕ(x2)| < ε for all x1, x2 ∈ [a, b] such that |x1 − x2| < δ and for all ϕ in thisfamily.
Lemma 10.1. A family of functions in C([a, b]) where [a, b] is a bounded closed intervalare uniformly equicontinuous if and only if they are equicontinuous.
The proof of the lemma is left for exercise. It helpful to review the proof of thetheorem that a function is uniformly continuous in a bounded closed interval if and onlyif it is continuous.
Theorem 10.2 (Arzela-Ascoli). A family {ϕ(x)} of functions in C([a, b]) is relativelysequentially compact, that is, any sequence of functions in this family has a subsequencethat converges uniformly to a continuous function, if and only if this family of functionsare uniformly bounded and equicontinuous.
Example 10.1. The family of function {f(x) ∈ C([a, b]) | f(a) = 0 and |f(x1)−f(x2)| ≤K|x1 − x2|} is relatively sequentially compact.
Proof. The family of functions are uniformly bounded since each function f satisfies|f(x)| = |f(x)−f(a)| ≤ K(x−a) ≤ K(b−a). The family of functions are equicontinuous,since for each ε we can choose δ = ε/K to satisfy the condition for uniform equicontinuity.So by the Arzela-Ascoli theorem, the family of functions is relatively sequentially compact.
Proof of Theorem 10.2. First we show that if {ϕ(x)} is uniformly bounded and uniformlyequicontinuous, then given any ε > 0, there is a finite ε-net for {ϕ(x)} and then {ϕ(x)}is relatively sequentially compact.
Suppose for any ϕ in this family and any x, x1, x2 ∈ [a, b], |ϕ(x)| < M and |ϕ(x1) −ϕ(x2)| < ε/2 if |x1−x2| < (b−a)/N . Denoting ai = a+(b−a)i/N where i = 0, 1, . . . , N ,we consider the set of piecewise linear functions
Xε = {f(x) ∈ C([a, b]) | |f(ai)| ≤M and f(ai) is an integer multiple ofε
2,
and f(x) is linear on [ai−1, ai] for all i = 0, . . . , N}.
Clearly Xε is a finite set. We show that it is an ε-net of {ϕ(x)}.
48
For any ϕ(x) in the family, there is an f(x) ∈ Xε such that |ϕ(ai)− f(ai)| ≤ ε/2 forall i = 0, . . . , N . For any x ∈ [a, b], suppose x ∈ [ai, ai+1], then
|ϕ(x)− f(ai)| ≤ |ϕ(x)− ϕ(ai)|+ |ϕ(ai)− f(ai)| <ε
2+ε
2= ε.
Similarly, |ϕ(x) − f(ai+1)| < ε. Since f is linear on [ai, ai+1], the value f(x) is betweenf(ai) and f(ai+1). So |ϕ(x)− f(x)| < ε. Since x is arbitrary, ρ(ϕ, f) ≤ ε in C([a, b]) andXε is an ε-net for {ϕ}.
Next suppose {ϕ(x)} is relatively sequentially compact. Then for any ε > 0, thefamily has a finite ε
3-net, say, f ε1(x), . . . , f εNε(x) ∈ C([a, b]). Taking ε = 1, we have
that any ϕ in the family satisfies maxx∈[a,b]|ϕ(x) − f 1i (x)|< 1
3for some i. Denote M =
maxi=1,...,N1 maxt∈[a,b]|f 1i (x)|, we have that |ϕ(x) < M + 1
3for all x ∈ [a, b]. Thus the
family is uniformly bounded.To prove that the family is equicontinuous at any point, we let ε be an arbitrary
positive number. At any point x, for each i = 1, . . . , Nε, there is a constant δi > 0 suchthat |f εi (x′)− f εi (x)| < ε
3for all x′ ∈ [a, b] and |x′ − x| < δi. Let δ = mini=1,...,Nε δi. Then
for any x′ ∈ [a, b] such that |x′ − x| < δ, and for any ϕ in the family, we take f εk(x) suchthat ρ(ϕ, f εk) <
ε3, and then
|ϕ(x′)− ϕ(x)| ≤ |ϕ(x′)− f εk(x′)|+ |f εk(x′)− f εk(x)|+ |f εk(x)− ϕ(x)| < ε
3+ε
3+ε
3= ε.
Therefore we prove the equicontinuity at x.
One important application of the Arzela-Ascoli theorem is the existence of local so-lutions to an initial value problem of ordinary differential equation (ODE). Recall that anatural setting of ODE is that f is a continuous function on an open set G ⊆ R2, and apoint (x0, y0) ∈ G. We look for a function y(x) that is defined on an (closed) interval Icontaining x0 such that y(x) is differentiable and satisfies the condition
dy
dx= f(x, y), where y(x0) = y0. (17)
In Section 8 we proved Lemma 8.1 that confirms that the differential equation above isequivalent to the integral equation
y(x) = y0 +
∫ x
x0
f(t, y(t))dt (18)
on the same interval. Picard’s theorem (Theorem 8.2 and equivalently Theorem 8.3)shows that if we impose a Lipschitz condition of f with respect to the y variable, thereexists an I 3 x0 such that on the interval I, the initial value problem of the ODE, orequivalently the integral equation, has a unique solution. Furthermore, by Example 8.1,we showed that if the Lipschitz condition is removed, the uniqueness fails. A naturalquestion is: Without the Lipschitz condition, is there always a local solution (i.e., asolution on an interval that t containing x0) to the initial value problem of ODE, orequivalently the integral equation, even if it may not be the unique solution? The questionis answered by Peano’s theorem positively.
49
Theorem 10.3 (Peano). Suppose f is a continuous function on an open set G ⊇ R2.For any (x0, y0) ∈ G, there is an interval I = [x1, x2] that contains x0 such that thereexists a y ∈ C(I) that satisfies the initial value problem of ODE (17), or equivalently,there exists a y ∈ C(I) that satisfies the integral equation (18).
Proof. We consider the integral equation (18). Without loss of generality, we assume(x0, y0) = (0, 0). Since f(x, y) is continuous in G, shrinking the region G if necessary, weassume that |f(x, y)| < M for all (x, y) ∈ G. Since (0, 0) is an interior point of G, thereis an r > 0 such that the rectangle [−r, r]× [−Mr,Mr] is contained in g. We prove thatthere exists a solution to the integral equation on the interval I = [−r, r].
For any n ∈ Z+, we define the piecewise linear function ϕn(x) ∈ C([−r, r]) by theconditions
ϕn(0) = 0,
ϕn
(k + 1
nr
)= ϕn
(k
nr
)+ f
(k
nr, ϕn
(k
nr
))r
n, for k = 0, 1, . . . , n− 1,
ϕn
(−k + 1
nr
)= ϕn
(k
nr
)− f
(−knr, ϕn
(−knr
))r
n, for k = 0, 1, . . . , n− 1,
ϕn(x) is linear on [k
nr,k + 1
nr] for k = −n,−n+ 1, . . . , n− 1.
We can show that ϕn(x) is well defined and |ϕn(x)| < Mr for all x ∈ [−r, r]. (Exercise.Hint: Show |ϕn(± k
nr)| < k
nMr iteratively.) Thus the family of functions {ϕn(x)} are
uniformly bounded.Also we can show that {ϕn(x)} are equicontinuous. First suppose x′1 and x′2 are two
points such that knr ≤ x′1 < x′2 ≤ k+1
nr. Then On the interval [ k
nr, k+1
nr], ϕn(x) is linear,
with slope equal to either f( knr, ϕn( k
nr)) or f(k+1
nr, ϕn(k+1
nr)). Hence
|ϕn(x′2)− ϕn(x′1)| ≤ max
(∣∣∣∣f (knr, ϕn(k
nr
))∣∣∣∣ , ∣∣∣∣f (k + 1
nr, ϕn
(k + 1
nr
))∣∣∣∣) |x′2 − x′1|< M |x′2 − x′1|.
More generally, suppose x′1 ∈ [k1nr, k1+1
nr) and x′2 ∈ (k2−1
nr, k2
nr] where k1 < k2. Then
|ϕn(x′2)− ϕn(x′1)| ≤∣∣∣∣ϕn(x′2)− ϕn
(k2 − 1
nr
)∣∣∣∣+
∣∣∣∣ϕn(k2 − 1
nr
)− ϕn
(k2 − 2
nr
)∣∣∣∣+ · · ·+
∣∣∣∣ϕn(k1 + 1
nr
)− ϕn(x′1)
∣∣∣∣< M
∣∣∣∣x′2 − k2 − 1
nr
∣∣∣∣+M
∣∣∣∣k2 − 1
nr − k2 − 2
nr
∣∣∣∣+ · · ·+M
∣∣∣∣k1 + 1
nr − x′
∣∣∣∣= M |x′2 − x′1|.
(19)
Thus we prove the equicontinuity. By Arzela-Ascoli theorem, we find that there is asubsequence {ϕnk(x)} that converges to a limit, say ϕ(x).
50
Next we prove that ϕ(x) satisfies the integral equation (18). Since ϕnk(x) convergeuniformly to ϕ(x), by the continuity of f ,∫ x
0
f(t, ϕnk(t))dt→∫ x
0
f(t, ϕ(t))dt uniformly on I.
Thus we need only to show that
ϕnk(x)−∫ x
0
f(t, ϕnk(t))dt→ 0 uniformly on I (20)
to prove that the limit ϕ(x) satisfies (18).Our proof needs a technical result that there is a sequence of positive numbers ε(n)
that converges to 0 such that
|f(a, b)− f(a′, b′)| < ε(n) (21)
for all (a, b), (a′, b′) ∈ [−r, r]× [−Mr,Mr] with |a−a′| ≤ rn
and |b− b′| ≤ Mrn
. This resultis a direct consequence of the (uniform) continuity of f on [−r, r]× [−Mr,Mr].
Consider any 0 ≤ k ≤ n− 1 and x ∈ ( knr, k+1
nr]. We have∣∣∣∣ϕn(x)−
∫ x
0
f(t, ϕn(t))dt
∣∣∣∣−∣∣∣∣∣ϕn
(k
nr
)−∫ k
nr
0
f(t, ϕn(t))dt
∣∣∣∣∣≤
∣∣∣∣∣ϕn(x)− ϕn(k
nr
)−∫ x
knr
f(t, ϕn(t))dt
∣∣∣∣∣=
∣∣∣∣∣∫ x
knr
f
(k
nr, ϕn
(k
nr
))− f(t, ϕn(t))dt
∣∣∣∣∣=
∫ x
knr
∣∣∣∣f (knr, ϕn(k
nr
))− f(t, ϕn(t))
∣∣∣∣ dt.
(22)
We have | knr − t| ≤ r
nfor all t in the integrand on the right-hand side of (22), and hence
|ϕn( knr)− ϕn(t)| ≤ Mr
nby (19). Using (21), we have that∣∣∣∣ϕn(x)−
∫ x
0
f(t, ϕn(t))dt
∣∣∣∣−∣∣∣∣∣ϕn
(k
nr
)−∫ k
nr
0
f(t, ϕn(t))dt
∣∣∣∣∣ ≤ ε(n)
(x− k
nr
). (23)
Using the inequality iteratively to x′1 = 1nr, x′2 = 2
nr, . . . , x′k = k
nr, we obtain that for
any x ∈ [0, r], ∣∣∣∣ϕn(x)−∫ x
0
f(t, ϕn(t))dt
∣∣∣∣ ≤ ε(n)|x|. (24)
Analogously we check that the estimate (24) is true for x ∈ [−r, 0] as well. Thus for alln,
maxx∈I
∣∣∣∣ϕn(x)−∫ x
0
f(t, ϕn(t))dt
∣∣∣∣ ≤ ε(n)r.
Thus we prove the uniform convergence result (20) and finish the proof.
51
11 Compact sets and compact metric spaces
Definition 11.1. In a metric space, the set M is compact if any open covering of M hasa finite subcovering, that is, if open sets {Gα}α∈I satisfies
Theorem 11.1. A set M in a metric space is compact if and only if M is sequentiallycompact.
Proof. First we prove the “if” part by contradiction. Suppose M is a sequentially compactset but not a compact set. Then there is an open covering {Gα}α∈I of M that has nofinite subcovering. Since M is sequentially compact, for any n = 1, 2, 3, . . . , there is an1/n-net of M , denoted as A(n) = {a(n)
1 , . . . , a(n)N(n)}. For any n, consider the subsets of M
B[a(n)1 ,
1
n] ∩M, . . . , B[a
(n)N(n),
1
n] ∩M.
These subsets cover M , and {Gα}α∈I is an open covering for each of them. If for every i =
1, . . . , N(n) there are a finite number of open sets in {Gα}α∈I that cover B[a(n)i , 1/n]∩M ,
then the collection of these finite sets give rise to a finite subcovering of M , contradictoryto that {Gα}α∈I has no finite subcovering of M . Thus at least one of these subsets doesnot have a finite subcovering of {Gα}α∈I , and without loss of generality we assume it is
B[a(n)1 , 1/n] ∩M . We take a(n) ∈ B[a
(n)1 , 1/n] ∩M .
By the sequential compactness of M , the sequence {a(n)} has a subsequence, say{a(nk)}, that converges to a ∈ M . Note that a is contained in a Gα′ ∈ {Gα}α∈I . SinceGα′ is open, there is a neighbourhood Nε(a) ⊆ Gα′ . Take a large enough k such that
ρ(a(nk), a) < ε/3 and 1/nk < ε/3. Then B[a(nk)1 , 1/nk] ⊆ Nε(a) ⊆ Gα′ , and B[a
(nk)1 , 1/nk]
is covered by a single Gα′ , a contradiction to that it cannot be covered by finitely manyopen sets in {Gα}α∈I . Therefore we conclude the proof of the “if” part.
Next we prove the “only if” part by proving its contraposition. Suppose M is notsequentially compact, then there is a sequence {xn} ⊆ M with distinct (think: why wecan assume them to be distinct?) elements such that none of its subsequence convergesto a point in M . Equivalently, no point in M is a limit point of {xn}. Thus for all a ∈M ,there is a neighbourhood Nε(a)(a) that contains only finitely many points in {xn}. It isobvious that any finitely collection these neighbourhoods cover any finitely many pointsin {xn} ⊆M , so cannot cover M . Therefore the open covering {Nε(a)(a)}a∈M of M has nofinite subcovering of M , and M is not compact. Hence the “only if” part is proved.
Theorem 11.1 shows that for a set in a metric space, sequential compactness is equiv-alent to compactness. Later we do not distinguish these two concepts if there is nopossibility of confusion.
Definition 11.2. A metric space (X, ρ) is a compact metric space if X is compact inthis space.
Theorem 11.2. Let (X, ρ) be a metric space and M ⊆ X. Then the subspace (M,ρ) isa compact metric space if and only if M is a compact set of M .
52
Proof. M is a compact set of M if and only if any sequence {xn} ⊆M has a subsequencethat converges to a point x ∈ M with respect to the metric ρ. The subspace (M,ρ) isa compact metric space if and only if any sequence {xn} ⊆ M has a subsequence thatconverges to a point x ∈M also. Thus they are equivalent.
Theorem 11.3. A metric space (X, ρ) is a compact metric space if and only if it iscomplete and X is totally bounded.
Proof. First we prove the “if” part. If X us a totally bounded set in a complete metricspace, then X is relatively sequentially compact. Since X is closed, X is sequentiallycompact, and then compact.
To prove that “only if” part, we first show that (X, ρ) is complete if it is compact.Let {xn} be any Cauchy sequence in this space. The compactness of X implies thata subsequence of {xn}, say {xnk}, converges to x ∈ X. Then we have that {xn} alsoconverges to x by a tutorial problem. Thus (X, ρ) is complete. Next, since X is a(sequentially) compact set in a complete metric space, X is totally bounded. We completethe proof.
The compact metric spaces can be characterized by open sets as well as closed sets.
Corollary 11.4. A metric spce (X, ρ) is compact if and only if either of the followingtwo conditons holds:
1. Any open covering {Gα}α∈I of the space has a finite subcovering.
2. Any collection of the closed sets {Fα}α∈I , such that any finite number of thesesets have nonempty intersection,(we say that {Fα}α∈I has the finite intersectionproperty,) has nonempty intersection.
Proof. The equivalence between the compactness of the space and Condition 1 is directlyfrom the definition. Conditions 1 and 2 are equivalent to each other. To see it, we takeGα = F c
α, and find
•⋃α∈I Gα = X if and only if
⋂α∈I Fα = ∅.
• Gα1 , . . . , Gαn cover X if and only if Fα1 , . . . , Fαn has empty intersection.
Thus Condition 2 is the contraposition of Condition 1.
Theorem 11.5. Let (X, ρ) and (Y, ρ′) be two metric spaces and f : X → Y be a contin-uous mapping. Suppose M ⊆ X is a compact set in (X, ρ). Then f(M) is a compact setin (Y, ρ′).
Proof. Let {Gα}α∈I be any open covering of f(M). Then every f−1(Gα) is open and{f−1(Gα)}α∈I is an open covering ofM . SinceM is compact, there are f−1(Gα1), . . . , f
−1(Gαn)that cover M . Then Gα1 , . . . , Gαn form a subcovering of the covering {Gα}α∈I of f(M).Thus f(M) is compact.
Theorem 11.6. Suppose (X, ρ) is a compact metric space and Y ⊆ X. Then
1. Y is a compact set if and only if Y is a closed set.
53
2. The subspace (Y, ρ) is a compact metric space if and only if Y is a closed set in X.
Proof. It is clear by Theorem 11.2 that the two parts of the theorem are equivalent. Weprove Part 1 below.
We first prove the “if” part. Since (X, ρ) is compact, the space is complete and Xis totally bounded by Theorem 11.3. We have that Y , a subset of X, is also totallybounded. The reason is that for any ε > 0, the finite ε-net of X is also a finite ε-net ofY . Then Y is relatively sequentially compact since it is totally bounded in a completespace. Since Y is closed by assumption, we have that Y is (sequentially) compact, andthen by Theorem 11.2, Y is a compact set.
Next we prove the “only if” part. If Y is a compact set in (X, ρ), Then Y is sequentiallycompact, and by definition Y is closed.
Theorem 11.7. Suppose (X, ρ) and (Y, ρ′) are metric spaces and (X, ρ) is compact. Letf : X → Y be a bijective continuous mapping. Then f−1 : Y → X is also continuous.
Proof. Since (X, ρ) is a compact metric space, X is a compact set in the space, and thenby Theorem 11.5, Y = f(X) is a compact set in (Y, ρ′), so that (Y, ρ′) is also a compactmetric space.
To show that f−1 is continuous, we show that the inverse image by f−1 of any closedset is closed. That is, for any closed set M ⊇ X, (f−1)−1(M) = f(M) is also closed in(Y, ρ′).
If M is closed in (X, ρ), then by Theorem 11.6, it is a compact set. By Theorem 11.5,we have that f(M) is also a compact set in (Y, ρ′), and then by Theorem 11.6 again wehave that f(M) is a compact set. Thus we prove the theorem.
Below we consider the set of continuous mappings between two metric spaces (X, ρX)and (Y, ρY ), denoted as
CXY = {f : X → Y | f is continuous}. (25)
It is desirable to define distance function ρ on a pair of continuous mappings to CXY ametric space, such that
ρ(f, g) = supx∈X
(ρY (f(x), g(x))). (26)
But this distance function is not even well defined. However, if we require X is compact,the definition is valid.
Theorem 11.8. Let (X, ρX) be a compact metric space and (Y, ρY ) be any metric space.Then the set of mappings CXY defined in (25) with the distance ρ defined in (26) is ametric space.
Proof. First we show that ρ is well defined, that is, for any f, g ∈ CXY , the distanceρ(f, g) is a finite number. Otherwise, suppose ρ(f, g) = supx∈X ρY (f(x), g(x)) is notfinite. Then there is a sequence {xn} ⊆ X such that ρY (f(xn), g(xn)) > n for all n. Bythe compactness of X, we have that {xn} has a subsequence, say {xnk}, that convergesto x ∈ X. Hence we have that as k → ∞, by the continuity of f and g, {f(xnk)}converges to f(x) and {g(xn)} converges to g(x), and ρY (f(xnk), g(xnk)) converges to
54
ρY (f(x), g(x)) <∞. That is contradictory to the assumption that ρY (f(xnk), g(xnk)) goto infinity.
Next we show that ρ is a well defined distance function. We only check that ρ satisfiesthe triangle inequality, since the other three conditions are easier to check. To show thatfor any f, g, h ∈ CXY , ρ(f, g) + ρ(g, h) ≥ ρ(f, h), it suffices to show that for any ε > 0,
ρ(f, g) + ρ(g, h) > ρ(f, h)− ε. (27)
Take x0 ∈ X such that
ρ(f, h) = supx∈X
ρY (f(x), h(x)) < ρY (f(x0), h(x0)) + ε.
Then we have that
ρ(f, g) + ρ(g, h) = supx∈X
ρY (f(x), g(x)) + supx∈X
(g(x), h(x))
≥ ρY (f(x0), g(x0)) + ρY (g(x0), h(x0))
≥ ρY (f(x0), h(x0))
> ρ(f, h)− ε.
Thus we prove (27) and then the triangle inequality of ρ.
Furthermore, we have that
Theorem 11.9. Suppose (X, ρX) is a compact metric space and (Y, ρY ) is a completemetric space, then CXY defined by (25) and (26) is a complete metric space.
Proof. Let {fn} be a Cauchy sequence in CXY , such that given any ε > 0, there is an Nε
such that for all m,n > Nε, ρ(fm, fn) < ε. Then for any x ∈ X, we have that {fn(x)},a sequence in (Y, ρY ), is a Cauchy sequence. Since (Y, ρY ) is complete, for each x, theCauchy sequence {fn(x)} has a limit, denoted as f(x).
We claim that the mapping f : X → Y that assign each x the value f(x) is acontinuous mapping. To see that f is continuous, we use the inequality that for allx ∈ X and ε > 0, we have
ρY (fn(x), f(x)) ≤ ε, for all n > Nε. (28)
To check inequality (28), we note that for all m,n > Nε, ρY (fm(x), fn(x)) ≤ ρ(fm, fn) < ε,that is, fm(x) is in the closed ball B[fn(x), ε]. Then the limit f(x) = limm→∞ fm(x) is alsoin the closed ballB[fn(x), ε], and we prove (28). We take any x0 ∈ X and n > Nε/3, and bythe continuity of fn, we have a δ > 0 such that for all x ∈ Nε(x0), ρY (fn(x), fn(x0)) < ε/3.Then for any x ∈ Nδ(x0),
Thus f is a continuous mapping at x0, and then it is a continuous mapping since x0 isarbitrary.
To see that {fn} converges to f , we use inequality (28) again. Since the ε in (28) doesnot depend on x, it is equivalent to that ρ(fn, f) ≤ ε for all n > Nε, and it means that fis the limit of {fn}.
55
Now we consider a generalization of Arzela-Ascoli theorem. The concept of equicon-tinuity can be defined for set of mappings between any metric spaces.
Definition 11.3. Let (X, ρX) and (Y, ρY ) be metric spaces, and {fα}α∈I be a family ofcontinuous mappings from X to Y . We say that they are equicontinuous at point x ∈ Xif for any ε > 0, there is a δx > 0 such that
ρY (fα(x′), fα(x)) < ε for all fα and x′ ∈ Nδ(x).
If the set of mappings are equicontinuous at any point of X, we say that they are equicon-tinuous. If for the set of mappings and any ε > 0, there is a δ > 0 independent of ε suchthat
ρY (fα(x′), fα(x)) < ε for all fα, x ∈ X and x′ ∈ Nδ(x),
we say that {fα}α∈I are uniformly equicontinuous.
Lemma 11.10. Suppose (X, ρX) is a compact metric space and (Y, ρY ) is any metricspace. Then a set of continuous mappings {fα}α∈I from X to Y are uniformly equicon-tinuous if and only if they are equicontinuous.
The proof is left as an exercise. By this lemma, we do not need to distinguish equicon-tinuity and uniform equicontinuity if the space X is compact.
Theorem 11.11 (Generalized Arzela-Ascoli theorem). Let (X, ρX) and (Y, ρY ) be bothcompact metric spaces, and D = {fα}α∈I ⊆ CXY be a set of continuous mappings from Xto Y . Then D is a relatively sequentially compact set if and only if D is equicontinuous.
Proof. First we give the proof of the “if” part. Suppose D is equicontinuous. Thenfor each ε > 0, we construct an ε-net for D, to show that D is relatively sequentiallycompact. By the uniform equicontinuity of D, for any ε > 0, there exists δ > 0 suchthat ρY (f(x), f(x′)) < ε/3 for all f ∈ D and all x, x′ ∈ X with ρX(x, x′) < δ. Since Xis compact, we choose a finite δ-net {x1, . . . , xn} ⊆ X. Then B(x1, δ), . . . , B(xn, δ) coverX. Also since Y is compact, it has a finite ε/7-net {y1, . . . , ym}. We consider the finiteindex set
I = {(i1, . . . , in) | 1≤ ij ≤ m for all j = 1, . . . , n},and for each (i1, . . . , in) ∈ I, define the (possibly empty) set
D(11,...,in) = {f ∈ D | ρY (f(xj), yij) <ε
3for all j = 1, . . . , n}.
Then we have that⋃
(i1,...,in)∈I D(i1,...,in) = D. In each nonempty D(i1,...,in), we pick up oneelement f(i1,...,in) ∈ D(i1,...,in), and we claim that these f(i1,...,in) constitute an ε-net of D.To see it, we take any fε ∈ D, and it belongs to a nonempty D(i1,...,in). Take any x ∈ X,there is an xi such that ρX(x, xk) < δ, so
Thus ρ(fα, ρ(i1,...,in)) < ε and we finish the proof of the “if” part.
56
Next we give the proof of the “only if” part. Take any x ∈ X, we show that D isequicontinuous at x. Suppose {f1, . . . , fn} ⊆ CXY for a ε/3-net of D. For each fi, since{fi} is continuous at x, for any ε > 0, there is an δi such that ρY (fi(x
′), fi(x)) < ε/3for all x′ ∈ Nδi(x). Then take δ = minni=1 δi, we have that ρY (fi(x
′), fi(x)) < ε3
for alli = 1, . . . , n and for all x′ ∈ Nδ(x).
For any f ∈ D, there is a fi such that ρ(f, fi) <ε3. Then for any x′ ∈ Nδ(x),
ρY (f(x′), f(x)) ≤ ρY (f(x′), fi(x′)) + ρY (fi(x
′), fi(x)) + ρY (fi(x), f(x)) <ε
3+ε
3+ε
3= ε.
Thus we prove that D is equicontinuous at x, and then D is an equicontinuous.
The mappings from a metric space to the real line are naturally important. Usuallywe call these mappings functions. If the metric space consists of functions, then we callthese mappings functionals. If the metric space X is compact, then the set of mappingsCXR defined by (25) together with the distance function defined by (26) form a completemetric space. For this metric space, although Theorem 11.11 does not apply directly,we also have a generalized Arzela-Ascoli theorem. To state the theorem, we need thegeneralized concept of uniform boundedness.
Definition 11.4. A set of functions {fα}α∈I from a metric space X to R with the usualmetric are uniformly bounded if there is an M > 0 such that |fα(x)| < M for all fα andall x ∈ X.
Theorem 11.12 (Generalized Arzela-Ascoli theorem). Let X be a compact metric space,and D = {fα}α∈I ⊆ CXR be a set of continuous functions from X to R with the usualmetric. Then D is relatively sequentially compact if and only if D is equicontinuous anduniformly bounded.
Proof. The “only if” part of the proof is the same of that of the classical Arzela-Ascolitheorem. For the “if” part, note that if D is uniformly bounded, then there is an M suchthat fα(x) ∈ (−M,M) for all fα ∈ D and x ∈ X. We view D as a set in CX,[−M,M ],and use Theorem 11.11 we see that D is relatively sequentially compact as a set inCX,[−M,M ]. It means that any sequence in D has a subsequence that converges uniformlyto a continuous function from X to [−M,M ]. But this means also that D is relativelysequentially compact in CXR.
57
A Appendix
Proof of Theorem 4.2, given in class. Let G be an open set, and {x1, x2, . . . } be the ra-tional numbers in G. Define the family of open intervals
Ai = {(a, b) | a, b are rational numbers, and xi ∈ (a, b) ⊆ G}.
Then each Ai is not empty. Let
Ii =⋃
(a,b)∈Ai
(a, b).
Then each Ii is an open interval, and
Ii = ( inf(a,b)∈Ai
a, sup(a,b)∈Ai
b).
To see it, we first check that Ii ⊆ (inf a, sup b) (We omit the subscripts for notationalsimplicity). On the other hand, we consider an arbitrary y ∈ (inf a, sup b), and withoutloss of generality assume that inf a < y < x. Since y > inf a, there exists (a′, b′) such thatinf a < a′ < y < x < b′, and then y ∈ (a′, b′) ⊆ Ii. thus we show that (inf a, sup b) ⊆ Ii.See Figure 9 for the formation of Ii. We have
∞⋃i=1
Ii =⋃
a,b∈Q, (a,b)⊆G
(a, b) = G. (29)
Below we show that if Ii 6= Ij, then they are disjoint. Thus if we count identical Ii as oneinterval, the decomposition (29) gives the desired disjoint open interval decomposition ofG.
We note that if we consider a larger family
A′i = {(a, b) | a, b are real numbers, and xi ∈ (a, b) ⊆ G},
then the interval( inf(a,b)∈A′i
a, sup(a,b)∈A′i
b) = ( inf(a,b)∈Ai
a, sup(a,b)∈Ai
b) = Ii,
because the right-hand side is obviously no smaller than Ii while each (a, b) ∈ Ai can beshown to be in Ii. Suppose Ii and Ij are not disjoint, so that Ii ∩ Ij is an open interval,and xk is a rational number in their intersection, see Figure 10. Since the intervals Ii, Ijcontain xk, we have Ii ∈ A′k and Ij ∈ A′k, and then Ii ⊆ Ik and Ij ⊆ Ik. On the otherhand, Ik is an interval such that Ik ⊇ Ii 3 xi, so Ik ∈ A′i and Ik ⊆ Ii. Similarly Ik ⊆ Ij.Therefore Ii = Ik = Ij. We complete the proof.
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