3.8. Compactness#

The material in this section is primarily based on [2, 41].

Recall from Definition 1.21 that for a subset AX, a cover is a family {Ai}iI of subsets of X such that

AiIAi.

3.8.1. Open Covers#

Definition 3.54 (Open cover)

A family of open subsets {Ai}iI of subsets of (X,d) is an open cover of A if it covers A.

Theorem 3.74 (Lindelöf)

Every open cover of a subset of Rm can be reduced to an at-most countable subcover.

Proof. We call a point a=(a1,,am)Rm a rational point if every component of a is a rational number.

  1. Let A be a subset of Rm.

  2. Let {Oi}iI be an open cover of A (possibly uncountable).

  3. Thus, AiIOi.

  4. For each xA,

    1. Choose an index ixI such that xOix.

    2. Pick a rational point axRm and a rational positive number rx such that xB(ax,rx)Oix.

  5. Consider the collection C={B(ax,rx)|xA}.

  6. Since the set of rational points is countable and the set of rational numbers is countable, hence the set of all open balls centered at rational points with rational radii is countable.

  7. Hence, C which is a subset of open balls with rational points as centers and rational radii, is at most countable.

  8. Thus, C is an at most countable open cover of A.

  9. Since each B(ax,rx) is a subset of an Oix, hence there exists an at most countable subcover of {Oi}iI for A.

3.8.2. Compact Sets#

Definition 3.55 (Compact set)

Let (X,d) be a metric space. A subset A of X is called compact if every open cover of A can be reduced to a finite subcover.

Definition 3.56 (Compact metric space)

Let (X,d) be a metric space. If X is itself a compact set, then (X,d) is called a compact metric space.

Example 3.24 ((0,1) is not compact)

The set (0,1) is not compact in R. We first show this the hard way by picking an open cover for (0,1) which cannot be reduced to a finite subcover. Later, we discuss how to verify compactness through easy checks.

  1. Consider the family of open intervals:

    C={(1n,1)|n2}.
  2. For every x(0,1), there is a natural number n such that x>1n.

  3. Thus, x(1n,1).

  4. Thus,

    (0,1)n=2(1n,1)

    implying that C is an open cover of (0,1).

  5. At the same time for every n2, we have:

    (1n,1)(0,1)

    since 1n>0.

  6. Thus,

    (0,1)=n=2(1n,1).
  7. But there is no finite subcover of (0,1) in C.

  8. If there was a finite subcover, we could pick a maximum n among those intervals.

  9. But then x=1n won’t belong to any of those intervals in the finite subcover.

  10. Hence (0,1) is not compact.

We later show in Theorem 3.76 that every compact set is closed and bounded. Hence, an easy way to say that (0,1) is not compact is by noticing that it is not closed.

3.8.3. Characterization of Compactness#

We have defined compactness as a property where every open cover can be reduced to a finite subcover. The characterization of a property involves identifying other properties which are equivalent in the sense that property A property B. If one is true then the other must be true and vice versa.

We will see later that compactness of a set A is equivalent to the property that every sequence of A has a subsequence that converges to a point in A. However, before we go there, let us examine some implications of this property that every sequence of a set A has a subsequence that converges within the set A. These results will be useful later in the characterization of compact sets.

Lemma 3.1 (Lebesgue number)

Let AiIOi be an open cover of A.

If every sequence in A has a subsequence which converges to a point of A, then, there exists a number δ>0 such that for each xA, we have B(x,δ)Oi for at least one iI.

Any such number δ>0 is called a Lebesgue number of A for the open cover {Oi}iI.

Proof. Assume the claim is false.

  1. Then, for each δ>0, there exists some xA such that B(x,δ) is not a subset of any Oi.

  2. In particular, for each n, there exists some xnA such that B(xn,1n)(XOi) holds for each iI.

  3. Consider the sequence {xn}.

  4. By the hypothesis, every sequence has a subsequence that converges to a point of A.

  5. Let xA be the limit of such a subsequence of {xn}.

  6. Since xA, xOi for at least one iI.

  7. Pick some iI such that xOi.

  8. Choose some r>0 such that xB(x,r)Oi.

    1. Recall that Oi is open and x is its interior point. So such an r>0 can be chosen.

  9. Now, select some n such that 1n<r2 and d(x,xn)<r2.

  10. It follows that B(xn,1n)B(x,r)Oi.

  11. But this contradicts with the selection of xn such that B(xn,1n) is not contained in any Oi.

  12. Thus, a δ>0 must exist satisfying the condition that for each xA, B(x,δ)Oi for at least one iI.

Lemma 3.2 (Existence of finite cover of open balls)

Let AX of a metric space (X,d). If every sequence in A has a subsequence which converges to a point of A, then, for every r>0 there exist x1,,xnA such that

Aj=1nB(xj,r).

This lemma simply claims that we can construct a finite open cover of open balls for A for every r>0.

Proof. Assume the claim to be false. Choose r>0 such that it is not possible to select a finite number of points from A such that

Aj=1nB(xj,r).
  1. Pick some x1A.

  2. Choose some x2AB(x1,r). We can choose x2 since A doesn’t have a finite open ball cover.

  3. Assuming that x1,x2,,xn have been chosen inductively, choose xn+1 from the set Ai=1nB(xi,r).

  4. By design, d(xn,xm)r holds for every nm.

  5. Then, no subsequence of {xn} can converge.

  6. This contradicts with the hypothesis that every sequence has a subsequence that converges in A.

  7. Hence, the claim must be true.

Theorem 3.75 (Characterization of compactness)

Let A be a subset of a metric space (X,d). The following statements are equivalent.

  1. A is compact.

  2. Every infinite subset of A has an accumulation point in A.

  3. Every sequence in A has a subsequence which converges to a point of A.

Proof. (1) (2)

  1. Let S be an infinite subset of A.

  2. Assume, by way of contradiction, that S has no accumulation point in A.

  3. Then for every xA, there exists a deleted neighborhood of x that is disjoint with S.

  4. That is, xA,rx>0|B(x,rx)(S{x})=.

  5. Then, B(x,rx)S can either be empty or it can at most contain x.

  6. Thus, B(x,rx)S{x}.

  7. Consider the open cover C=xAB(x,rx).

  8. It is easy to see that AC.

  9. Since A is compact (by hypothesis), there exists a finite set {x1,,xn}A such that Ai=1nB(xi,rxi).

  10. But then

    S=AS=i=1n[B(xi,rxi)S]{x1,,xn}.
  11. Thus, S must be a finite set containing at most n elements.

  12. We arrive at a contradiction as S is infinite.

  13. Hence, S has an accumulation point in A.

(2) (3)

  1. We assume that every infinite subset of A has an accumulation point in A.

  2. Let {xn} be an arbitrary sequence of A.

  3. If {xn} is constant, then every subsequence is constant and convergent.

  4. If {xn} takes a finite number of distinct values, then there is at least one value which must occur infinite times. Thus, at least one subsequence is a constant sequence and hence convergent.

  5. We are left with the case where {xn} contains infinite distinct values.

  6. Then, we can choose a subsequence {yn} of {xn} which consists of all distinct values; i.e., ynym whenever nm.

    1. Let k1=1.

    2. Assuming k1,,kn have been selected, choose kn+1>kn such that xkn+1xki for 1in.

    3. This is possible since {xn} has infinite distinct values and so far only n distinct values have been chosen.

    4. Now, let yn=xkn.

    5. It is clear that {yn} is a subsequence of {xn} consisting of all distinct values.

  7. Consider the set Y={y1,y2,} (not the sequence but the set).

  8. By our hypothesis (2), since Y is an infinite subset of A, hence it must have an accumulation point in A.

  9. Let x be an accumulation point of Y.

  10. Assume that xyn for each n. If x=yk for some k, then drop the first k elements of {yn}.

  11. This ensures that d(x,yn)>0 for every n.

  12. We will now select a subsequence of {yn} that converges to x.

  13. Towards this, choose m1 such that d(ym1,x)<1.

  14. Now, inductively, assuming that m1<m2,<mn have been chosen, choose mn+1 such that

    d(ymn+1,x)<rn+1=min{1n+1,d(y1,x),d(y2,x),,d(ymn,x)}.
  15. Since d(x,yn)>0, hence rn+1>0.

  16. Since x is an accumulation point of Y, hence for every r>0, there exists a point yY such that d(x,y)<r.

  17. Thus, it is possible to pick a suitable mn+1 such that d(ymn+1,x)<rn+1.

  18. Also, by design, mn+1 must be greater than mn as rn+1<d(yi,x)1imn.

  19. Thus, {ymn} is a subsequence of {yn}.

  20. Hence, {ymn} is a subsequence of {xn} too.

  21. Since d(x,ymn)<1n, it follows that limymn=x.

  22. Thus, {xn} has a convergent subsequence which converges to a point of A.

(3) (1)

  1. Let AiIOi be an arbitrary open cover of A.

  2. We can pick a Lebesgue number δ>0 such that for each xA, we have B(x,δ)Oi for at least one iI thanks to Lemma 3.1.

  3. Now, thanks to Lemma 3.2, we can pick x1,,xnA such that:

    Aj=1nB(xj,δ).
  4. Now, for each j pick some ijI such that B(xj,δ)Oij.

  5. Then,

    Aj=1nB(xj,δ)j=1nOij.
  6. Thus, the open cover of A has a finite subcover.

  7. Thus, A is compact.

Definition 3.57 (Bolzano-Weierstrass property)

A set A in a metric space has the Bolzano-Weierstrass property if every sequence in A has a convergent subsequence that converges to a point in A.

Observation 3.1

A compact set has the Bolzano-Weierstrass property.

3.8.4. Closedness and Boundedness#

Theorem 3.76 (Compact sets are closed and bounded)

A compact set is closed and bounded.

Proof. Assume A to be compact. We shall show that A must be bounded.

  1. Choose an open cover for A as AxAB(x,1).

  2. Since A is compact, hence there exist finite set of points {x1,xn}A such that Ai=1nB(xi,1).

  3. Let M=max{d(xi,xj)|1i,jn}.

  4. For any x,yA, choose i,j such that xB(xi,1) and yB(xj,1).

  5. Then, by triangle inequality, we have:

    d(x,y)d(x,xi)+d(xi,xj)+d(xj,y)<M+2<.
  6. Thus, diamA=supd(x,y) is finite and A is bounded.

We now show that if A is compact then A must be closed too. Towards this, we show that A contains all its closure points.

  1. Let xclA.

  2. Due to Theorem 3.31, there exists a sequence {xn} of A with limxn=x.

  3. Due to Theorem 3.75, {xn} has a subsequence that converges to a point of A (since A is compact by hypothesis).

  4. As per Theorem 3.35, subsequences of a convergent sequence converge to the same limit.

  5. Thus, x must be in A.

  6. Thus, clAA. Thus, clA=A. Thus, A is closed.

Although every compact set is closed and bounded, the converse need not be true. See Example 3.29 for an example of closed and bounded set (in discrete space) which is not compact. In fact, discrete space is a complete metric space. Yet, it has closed and bounded sets which are not compact.

In the specific case of Euclidean spaces, all closed and bounded sets are compact too. See Heine-Borel theorem below.

3.8.5. Nested Sequence of Compact Sets#

Theorem 3.77 (Nonempty nested intersection property)

Let {An} be a nested sequence of nonempty compact subsets of (X,d) such that

S1S2S3

Then their intersection is nonempty; i.e.,

n=1Sn.

Proof. We prove this result by way of contradiction.

  1. Assume for contradiction that n=1Sn=.

  2. Let Vn=XSn for every nN.

  3. Since Sn+1Sn for every n hence VnVn+1 for every n.

  4. Then V={Vn|nN} is a collection of open sets.

  5. Since n=1Sn=, hence

    X=X=X(n=1Sn)=n=1(XSn)=n=1Vn.
  6. Hence V is an open cover for every Sn as SnX=n=1Vn.

  7. In particular, V is an open cover for S1.

  8. Since S1 is compact, hence there exists a finite subcover of S1 indexed by a finite set of natural numbers I.

  9. Let I={i1,i2,,ik} and the finite subcover be Fk={ViV|iI}.

  10. Since I is a finite set, it has a maximum element.

  11. Let m=maxI=max{i1,i2,,ik}.

  12. Then Fm={V1,,Vm} is also a finite subcover of S1 since FkFm.

  13. But then k=1mVk=Vm since VnVn+1 for every n.

  14. Hence S1Vm.

  15. Then SmS1Vm.

  16. But Vm=XSm, hence VmSm= a contradiction.

  17. Hence, the intersection of the nested compact sets must be nonempty.

3.8.6. Continuity#

Theorem 3.78 (Continuous images of compact sets are compact)

Let f:(X,d)(Y,ρ) be a continuous function. Let A be a compact subset of X with Adomf. Then f(A) is a compact subset of Y.

Proof. We prove this by showing that any open cover of f(A) can be reduced to a finite subcover.

  1. Let f(A)iIOi be an open cover for f(A).

  2. Then AiIf1(Oi).

  3. Since f is continuous, hence f1(Oi) is an open subset of X for every iI.

  4. Since A is compact, there exist indices i1,,in such that Aj=1nf1(Oij).

  5. Then

    f(A)f(j=1nf1(Oij))=j=1nf(f1(Oij))j=1nOij.
  6. Thus, A is compact.

3.8.7. Lipschitz Continuity#

Theorem 3.79

Let f:(X,d)(Y,ρ) be a (partial) function with S=domf. Assume that f is locally Lipschitz continuous at every xS. Let AS be a compact subset of S. Then, f is Lipschitz function on A. In other words, there exists a constant L>0 such that

ρ(f(x),f(y))Ld(x,y)

for every x,yA.

Proof. We proceed as follows.

  1. Since f is locally Lipschitz continuous on S hence f is continuous by Theorem 3.57.

  2. Thus, by Theorem 3.78, f(A) is compact.

  3. Hence, f(A) is closed and bounded.

  4. Thus, there exists M>0 such that for any x,yA, ρ(f(x),f(y))M.

  5. For contradiction, assume that f is not Lipschitz on A.

  6. Then, there is no L>0 such that

    ρ(f(x),f(y))Ld(x,y)x,yA.
  7. Then, there exist two sequences {xn} and {yn} of A such that

    limnρ(f(xn),f(yn))d(xn,yn)=.
  8. But ρ(f(xn),f(yn))M.

  9. Thus,

    limd(xn,yn)=0.
  10. Since A is compact, hence {xn} has a convergent subsequence.

  11. Let {xnk} be a convergent subsequence with x=limkxnk.

  12. By compactness of A, xA.

  13. Then, f cannot be not locally Lipschitz continuous at x.

  14. This contradicts our hypothesis.

  15. Thus, f must be Lipschitz continuous on A.

3.8.8. Homeomorphism#

Theorem 3.80 (Homeomorphism preserves compactness)

Let f:(X,d)(Y,ρ) be a homeomorphism. Then, A is a compact subset of (X,d) is and only if f(A) is a compact subset of (Y,ρ).

Proof. Let AX be compact. Then, f(A) is compact since f is continuous due to Theorem 3.78.

Let f(A) be compact. Since f is a homeomorphism, hence f1 is continuous and bijective. Hence, f1(f(A))=A is compact due to Theorem 3.78.

3.8.9. Compact Spaces#

Theorem 3.81

Every closed subset of a compact space is compact.

Proof. Let (X,d) be a compact metric space and let A be a closed subset of X.

  1. Let {Oi}iI be an open cover of A. We have, AiIOi.

  2. X=A(XA).

  3. Then, X(XA)iIOi.

  4. Since all Oi are subsets of X, hence we can write it as: X=(XA)iIOi.

  5. Since A is closed, hence XA is open.

  6. Thus, (XA)iIOi is an open cover of X.

  7. But X is compact. Hence, there exist finite indices i1,,in such that X=(XA)Oi1Oin.

  8. But then AX and A(XA)= imply that: AOi1Oin.

  9. Thus, A is compact.

Theorem 3.82 (Continuous maps are closed)

Let (X,d) be a compact metric space and suppose that f:(X,d)(Y,ρ) is a (total) continuous function. Then f is a closed mapping. If f is bijective, then f is a homeomorphism.

Proof. Let C be a closed subset of X.

  1. Due to Theorem 3.81, C is compact.

  2. Since f is continuous, hence, due to Theorem 3.78, f(A) is compact.

  3. As per Theorem 3.76, since f(A) is compact, hence f(A) is closed.

  4. Thus, f maps every closed set to a closed set.

  5. Thus, f is a closed mapping.

Now, assume that f is bijective too.

  1. Thus, f1 exists. Let g=f1.

  2. Then, due to bijection property g1(A)=f(A) holds for every subset A of X.

  3. Thus, g1(A)=f(A) is a closed subset of Y whenever A is a closed subset of X.

  4. Thus, as per Theorem 3.42 [(5)(1)], g is continuous.

  5. Thus, both f and f1=g are continuous.

  6. Thus, f is a homeomorphism.

Theorem 3.83 (Compact domain + Continuity = Uniform continuity)

Let f:(X,d)(Y,ρ) be continuous on X. If X is compact, then f is uniformly continuous.

Proof. We proceed as follows.

  1. Let ϵ>0.

  2. Since f is continuous on X, hence for every xX, there exists rx>0 such that ρ(f(y),f(x))<ϵ holds whenever d(x,y)<2rx.

  3. The collection of open balls B(x,rx) covers X; i.e., X=xXB(x,rx).

  4. Since X is compact, there exists a set of finite number of points x1,,xn such that X=i=1nB(xi,rxi).

  5. Now, let δ=min{rx1,,rxn}.

  6. Since δ is the minimum of a finite number of positive numbers, hence δ>0.

  7. Now, pick any x,yX that satisfy d(x,y)<δ.

  8. There exists an integer i such that d(x,xi)<rxi (due to the finite open cover).

  9. Therefore, ρ(f(x),f(xi))<ϵ.

  10. Now, by triangle inequality:

    d(y,xi)d(y,x)+d(x,xi)<δ+rxi2rxi

    holds true.

  11. Thus, ρ(f(xi),f(y))<ϵ, since d(y,xi)<2rxi.

  12. Thus,

    ρ(f(x),f(y))ρ(f(x),f(xi))+ρ(f(xi),f(y))<ϵ+ϵ=2ϵ.
  13. Thus, f is uniformly continuous.

3.8.10. Euclidean Spaces#

Recall that Rm are called Euclidean spaces with the standard metric:

d(x,y)(i=1m|xiyi|2)12.

By 0Rm we shall mean the vector (0,,0).

The compact subsets of a Euclidean space are precisely those sets which are closed and bounded.

Theorem 3.84 (Heine-Borel theorem)

A subset of a Euclidean space is compact if and only if it is closed and bounded.

Compare this to Heine-Borel theorem for the real line. We established there that for a closed and bounded subset of R, any open cover can be reduced to a finite subcover. There, we defined closed and bounded sets as compact sets. Our treatment of compactness in this section is more general.

We start here with the definition that a compact set is one for which any open cover can be reduced to a finite subcover. We then show in this theorem that in the special case of Euclidean spaces Rm, the compact subsets are identical to the closed and bounded subsets of Rm.

Proof. We have shown in Theorem 3.76 that every compact set is closed and bounded.

For the converse, we assume A is a closed and bounded subset of Rm. We will show that every sequence of A has a subsequence converging in A.

  1. Since A is bounded, we can pick M>0 such that d(x,y)M for all x,yA.

  2. Fix an element yA.

  3. Let a=(a1,,am)A be some arbitrary point of A. Then

    |ai|d(a,0)d(a,y)+d(y,0)M+d(y,0)

    holds for every 1im.

  4. Thus, the set of real numbers consisting of the i-th coordinates of the elements of A is a bounded set.

  5. Choose an arbitrary sequence {xn} of A.

  6. Recall from Bolzano Weierstrass theorem for real numbers that every bounded sequence of real numbers has a convergent subsequence.

  7. Note that if we form the sequence of real numbers from any particular coordinate of {xn}, (say first coordinates or second coordinates) then the sequence is bounded by M+d(y,0).

  8. Every such sequence of real numbers (from a fixed coordinate of {xn}) will have a convergent subsequence.

  9. Thus, there is a subsequence {xn1} of {xn} whose first coordinates form a sequence in R that converges in R.

  10. Now, we choose {xn2} as a subsequence of {xn1} so that the corresponding sequence of second coordinates of {xn2} converge in R.

  11. Proceeding in this manner, after m steps, we have a subsequence {xnm} of {xn} with the property that for each 1im, the sequence of its i-th coordinates forms a convergent subsequence in R.

  12. Since each of the coordinates of {xnm} converges in R, hence, {xnm} converges in Rm.

  13. But since A is closed, hence {xnm} converges to a point of A.

  14. Thus, every sequence of A has a convergent subsequence in A.

  15. Thus, by Theorem 3.75, A is compact.

Note that the property convergence in individual coordinates implies convergence in Rm is due to the specific choice of Euclidean metric.

Theorem 3.85 (Attainment of minimum and maximum values)

Let f:(X,d)R be a real valued function. If f is continuous then f attains a maximum and minimum value on every compact subset of domf.

Proof. Let A be a compact subset of domf.

  1. Then, f(A) is compact in R due to Theorem 3.78.

  2. Then, f(A) is closed and bounded due to Heine-Borel theorem.

  3. Since f(A) is bounded, hence it has an infimum and supremum.

  4. Since f(A) is closed, hence its infimum and supremum lie inside f(A) itself.

  5. Thus, f attains a maximum and minimum value in A.

Theorem 3.86 (Bolzano Weierstrass theorem for bounded subsets of Rm)

Every sequence in a closed and bounded set in Rm has a convergent subsequence.

Proof. Let A be a closed and bounded subset of Rn.

  1. By Theorem 3.84, A is compact.

  2. By Theorem 3.75, every sequence in A has a subsequence which converges to a point of A.

Theorem 3.87 (Bolzano Weierstrass theorem for bounded sequences of Rm)

Every bounded sequence in Rm has a convergent subsequence.

Proof. Let {xm} be a bounded sequence of Rm. Then there exists a closed ball B[0,M] such that {xm}B[0,M].

B[0,M] is closed and bounded. From Theorem 3.86, every sequence in a closed and bounded subset of Rn has a convergent subsequence.

3.8.11. Totally Bounded Metric Spaces#

Definition 3.58 (Totally bounded space)

A metric space (X,d) is called totally bounded if for each r>0, there exists a finite number of points x1,,xnX such that

X=i=1nB(xi,r).

Theorem 3.88 (Compactness implies total boundedness)

A compact metric space is totally bounded.

Proof. Let (X,d) be a compact metric space.

  1. Let r>0.

  2. Consider the family of open balls {B(x,r)}xX.

  3. Then X=xXB(x,r).

  4. Since X is compact, there is a finite subcover of open balls.

  5. Thus, for every r>0, there X is a union of finite open balls.

  6. Thus, X is totally bounded.

Example 3.25

We showed earlier in Example 3.24 that (0,1) is not compact.

However (0,1) is totally bounded.

  1. Let r>0.

  2. Pick any n>1r. Thus, r>1n.

  3. Consider the points 1n,2n,,n1n.

  4. kn+r>kn+1n=k+1n.

  5. knr<kn1n=k1n.

  6. Thus,

    (k1n,k+1n)B(kn,r)
  7. Thus,

    (0,1)=k=1n1B(kn,r).

    with the caveat that the first and last balls are restricted within the set (0,1).

  8. Thus, we have a finite union of open balls.

  9. Thus, (0,1) is totally bounded.

Theorem 3.89 (Completeness and Compactness)

A metric space is compact if and only if it is complete and totally bounded.

Proof. Let (X,d) be a metric space. Assume that (X,d) is compact.

  1. (X,d) is totally bounded (Theorem 3.88).

  2. Since (X,d) is compact, hence every sequence has a convergent subsequence that converges to a point in X (Theorem 3.75 (3)).

  3. Thus, if {xn} is a Cauchy subsequence of X, then it has a subsequence with limit xX leading to limxn=x.

  4. Thus, every Cauchy sequence of X converges in X.

  5. Thus, (X,d) is complete.

For the converse, assume that (X,d) is complete and totally bounded. To show that (X,d) is compact, we will show that every infinite subset of X has an accumulation point in X (Theorem 3.75 (2)).

  1. Let A be an infinite subset of X.

  2. Since X is totally bounded, there exists a finite subset F1X such that

    X=xF1B(x,1).
  3. We can extend the open balls with closed balls without problem:

    X=xF1B[x,1].
  4. Thus:

    A=AX=xF1(AB[x,1]).
  5. Since A is infinite, there is some x1F1 such that A1=AB[x1,1] is an infinite set.

    1. If AB[x,1] were finite for each xF, then A would be finite as a finite union of finite sets.

  6. Since X is totally bounded, we can again find a finite subset F2X such that

    X=xF2B[x,12].
  7. Since A1 is infinite, there is some x2F2 such that A2=A1B[x2,12] is an infinite set.

  8. Proceeding inductively, if x1,x2,,xn have been chosen, we can choose xn+1 such that the set

    AB[x1,1]B[x2,12]B[xn,1n]B[xn+1,1n+1]

    is infinite.

  9. Define

    En=B[x1,1]B[x2,12]B[xn,1n]

    for each n.

  10. Then for each n:

    1. En is nonempty and closed.

    2. AEn is infinite.

    3. En+1En.

    4. diamEn2n. limdiamEn=0.

  11. Due to Theorem 3.59, there exists aX, such that aEn for each n.

  12. Now, if yAEn, then

    d(a,y)d(a,xn)+d(xn,y)<1n+1n=2n.
  13. Thus, for every r>0, we can pick n>2r such that for every yAEn, yB(a,r).

  14. Thus, a is an accumulation point of A.

  15. Thus, every infinite subset of X has an accumulation point in X.

  16. Thus, X is compact.

3.8.12. Equivalent Metrics#

Theorem 3.90 (Metric equivalence and compactness)

Let da and db be two different metrics on the same set X that are equivalent.

Then, a set AX is compact in (X,da) if and only if A is compact in (X,db).

In other words, the compact sets in the two metric spaces are identical.

Proof. Assume A to be compact in (X,da).

  1. Let AiIOi be an open cover of A in (X,db); i.e. Oi are open in (X,db).

  2. Since da and db are equivalent, hence Oi are open in (X,da) too.

  3. Thus, iIOi is an open cover for A in (X,da) too.

  4. Since A is compact in (X,da), hence, there exist finite indices i1,,in such that AOi1Oin.

  5. But then, Oi1,,Oin are open in (X,db) too.

  6. Thus, AOi1Oin is a finite open subcover of A in (X,db).

  7. Thus, every open cover of A in (X,db) can be reduced to a finite subcover.

  8. Thus, A is compact in (X,db).

A similar reasoning establishes that if A is compact in (X,db) then A is compact in (X,da) too.