3.8. CompactnessΒΆ

The material in this section is primarily based on [2, 23].

Recall from Definition 1.21 that for a subset AβŠ†X, a cover is a family {Ai}i∈I of subsets of X such that

AβŠ†β‹ƒi∈IAi.

3.8.1. Open CoversΒΆ

Definition 3.54 (Open cover)

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

Theorem 3.74 (LindelΓΆf)

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}i∈I be an open cover of A (possibly uncountable).

  3. Thus, AβŠ†β‹ƒi∈IOi.

  4. For each x∈A,

    1. Choose an index ix∈I such that x∈Oix.

    2. Pick a rational point ax∈Rm and a rational positive number rx such that x∈B(ax,rx)βŠ†Oix.

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

  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}i∈I for A.

3.8.2. Compact SetsΒΆ

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)|nβ‰₯2}.
  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)βŠ†β‹ƒn=2∞(1n,1)

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

  5. At the same time for every nβ‰₯2, we have:

    (1n,1)βŠ†(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 CompactnessΒΆ

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 AβŠ†β‹ƒi∈IOi 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 Ξ΄>0 such that for each x∈A, we have B(x,Ξ΄)βŠ†Oi for at least one i∈I.

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

Proof. Assume the claim is false.

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

  2. In particular, for each n, there exists some xn∈A such that B(xn,1n)∩(Xβˆ–Oi)β‰ βˆ… holds for each i∈I.

  3. Consider the sequence {xn}.

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

  5. Let x∈A be the limit of such a subsequence of {xn}.

  6. Since x∈A, x∈Oi for at least one i∈I.

  7. Pick some i∈I such that x∈Oi.

  8. Choose some r>0 such that x∈B(x,r)βŠ†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)βŠ†B(x,r)βŠ†Oi.

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

  12. Thus, a Ξ΄>0 must exist satisfying the condition that for each x∈A, B(x,Ξ΄)βŠ†Oi for at least one i∈I.

Lemma 3.2 (Existence of finite cover of open balls)

Let AβŠ†X 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,…,xn∈A such that

AβŠ†β‹ƒj=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

AβŠ†β‹ƒj=1nB(xj,r).
  1. Pick some x1∈A.

  2. Choose some x2∈Aβˆ–B(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 Aβˆ–β‹ƒi=1nB(xi,r).

  4. By design, d(xn,xm)β‰₯r holds for every nβ‰ m.

  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 x∈A, there exists a deleted neighborhood of x that is disjoint with S.

  4. That is, βˆ€x∈A,βˆƒrx>0|B(x,rx)∩(Sβˆ–{x})=βˆ….

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

  6. Thus, B(x,rx)∩SβŠ†{x}.

  7. Consider the open cover C=⋃x∈AB(x,rx).

  8. It is easy to see that AβŠ†C.

  9. Since A is compact (by hypothesis), there exists a finite set {x1,…,xn}βŠ†A such that AβŠ†β‹ƒi=1nB(xi,rxi).

  10. But then

    S=A∩S=⋃i=1n[B(xi,rxi)∩S]βŠ†{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., yn≠ym whenever n≠m.

    1. Let k1=1.

    2. Assuming k1,…,kn have been selected, choose kn+1>kn such that xkn+1β‰ xki for 1≀i≀n.

    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 x≠yn 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,β‹―<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 y∈Y 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)βˆ€1≀i≀mn.

  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 AβŠ†β‹ƒi∈IOi be an arbitrary open cover of A.

  2. We can pick a Lebesgue number Ξ΄>0 such that for each x∈A, we have B(x,Ξ΄)βŠ†Oi for at least one i∈I thanks to Lemma 3.1.

  3. Now, thanks to Lemma 3.2, we can pick x1,…,xn∈A such that:

    AβŠ†β‹ƒj=1nB(xj,Ξ΄).
  4. Now, for each j pick some ij∈I such that B(xj,Ξ΄)βŠ†Oij.

  5. Then,

    AβŠ†β‹ƒj=1nB(xj,Ξ΄)βŠ†β‹ƒj=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 BoundednessΒΆ

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 AβŠ†β‹ƒx∈AB(x,1).

  2. Since A is compact, hence there exist finite set of points {x1,…xn}βŠ†A such that AβŠ†β‹ƒi=1nB(xi,1).

  3. Let M=max{d(xi,xj)|1≀i,j≀n}.

  4. For any x,y∈A, choose i,j such that x∈B(xi,1) and y∈B(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 x∈clA.

  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, clAβŠ†A. 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 SetsΒΆ

Theorem 3.77 (Nonempty nested intersection property)

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

S1βŠ‡S2βŠ‡S3βŠ‡β€¦

Then their intersection is nonempty; i.e.,

β‹‚n=1∞Snβ‰ βˆ….

Proof. We prove this result by way of contradiction.

  1. Assume for contradiction that β‹‚n=1∞Sn=βˆ….

  2. Let Vn=Xβˆ–Sn for every n∈N.

  3. Since Sn+1βŠ†Sn for every n hence VnβŠ†Vn+1 for every n.

  4. Then V={Vn|n∈N} is a collection of open sets.

  5. Since β‹‚n=1∞Sn=βˆ…, hence

    X=Xβˆ–βˆ…=Xβˆ–(β‹‚n=1∞Sn)=⋃n=1∞(Xβˆ–Sn)=⋃n=1∞Vn.
  6. Hence V is an open cover for every Sn as SnβŠ†X=⋃n=1∞Vn.

  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={Vi∈V|i∈I}.

  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 FkβŠ†Fm.

  13. But then ⋃k=1mVk=Vm since VnβŠ†Vn+1 for every n.

  14. Hence S1βŠ†Vm.

  15. Then SmβŠ†S1βŠ†Vm.

  16. But Vm=Xβˆ–Sm, hence Vm∩Sm=βˆ… a contradiction.

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

3.8.6. ContinuityΒΆ

Theorem 3.78 (Continuous images of compact sets are compact)

Let f:(X,d)β†’(Y,ρ) be a continuous function. Let A be a compact subset of X with AβŠ†domf. 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)βŠ†β‹ƒi∈IOi be an open cover for f(A).

  2. Then AβŠ‚β‹ƒi∈Ifβˆ’1(Oi).

  3. Since f is continuous, hence fβˆ’1(Oi) is an open subset of X for every i∈I.

  4. Since A is compact, there exist indices i1,…,in such that AβŠ†β‹ƒj=1nfβˆ’1(Oij).

  5. Then

    f(A)βŠ†f(⋃j=1nfβˆ’1(Oij))=⋃j=1nf(fβˆ’1(Oij))βŠ†β‹ƒj=1nOij.
  6. Thus, A is compact.

3.8.7. Lipschitz ContinuityΒΆ

Theorem 3.79

Let f:(X,d)β†’(Y,ρ) be a (partial) function with S=domf. Assume that f is locally Lipschitz continuous at every x∈S. Let AβŠ†S 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,y∈A.

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,y∈A, 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)βˆ€x,y∈A.
  7. Then, there exist two sequences {xn} and {yn} of A such that

    limnβ†’βˆžΟ(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=limkβ†’βˆžxnk.

  12. By compactness of A, x∈A.

  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. HomeomorphismΒΆ

Theorem 3.80 (Homeomorphism preserves compactness)

Let f:(X,d)β†’(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 AβŠ†X 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 fβˆ’1 is continuous and bijective. Hence, fβˆ’1(f(A))=A is compact due to Theorem 3.78.

3.8.9. Compact SpacesΒΆ

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}i∈I be an open cover of A. We have, AβŠ†β‹ƒi∈IOi.

  2. X=Aβˆͺ(Xβˆ–A).

  3. Then, XβŠ†(Xβˆ–A)βˆͺ⋃i∈IOi.

  4. Since all Oi are subsets of X, hence we can write it as: X=(Xβˆ–A)βˆͺ⋃i∈IOi.

  5. Since A is closed, hence Xβˆ–A is open.

  6. Thus, (Xβˆ–A)βˆͺ⋃i∈IOi is an open cover of X.

  7. But X is compact. Hence, there exist finite indices i1,…,in such that X=(Xβˆ–A)βˆͺOi1βˆͺβ‹―βˆͺOin.

  8. But then AβŠ†X and A∩(Xβˆ–A)=βˆ… imply that: AβŠ†Oi1βˆͺβ‹―βˆͺOin.

  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)β†’(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, fβˆ’1 exists. Let g=fβˆ’1.

  2. Then, due to bijection property gβˆ’1(A)=f(A) holds for every subset A of X.

  3. Thus, gβˆ’1(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 fβˆ’1=g are continuous.

  6. Thus, f is a homeomorphism.

Theorem 3.83 (Compact domain + Continuity = Uniform continuity)

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

Proof. We proceed as follows.

  1. Let Ο΅>0.

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

  3. The collection of open balls B(x,rx) covers X; i.e., X=⋃x∈XB(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 Ξ΄=min{rx1,…,rxn}.

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

  7. Now, pick any x,y∈X 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))<ϡ.

  10. Now, by triangle inequality:

    d(y,xi)≀d(y,x)+d(x,xi)<Ξ΄+rxi≀2rxi

    holds true.

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

  12. Thus,

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

3.8.10. Euclidean SpacesΒΆ

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

d(x,y)β‰œ(βˆ‘i=1m|xiβˆ’yi|2)12.

By 0∈Rm 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,y∈A.

  2. Fix an element y∈A.

  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 1≀i≀m.

  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 1≀i≀m, 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)β†’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}βŠ‚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 SpacesΒΆ

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,…,xn∈X 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)}x∈X.

  3. Then X=⋃x∈XB(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,…,nβˆ’1n.

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

  5. knβˆ’r<knβˆ’1n=kβˆ’1n.

  6. Thus,

    (kβˆ’1n,k+1n)βŠ†B(kn,r)
  7. Thus,

    (0,1)=⋃k=1nβˆ’1B(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 x∈X 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 F1βŠ‚X such that

    X=⋃x∈F1B(x,1).
  3. We can extend the open balls with closed balls without problem:

    X=⋃x∈F1B[x,1].
  4. Thus:

    A=A∩X=⋃x∈F1(A∩B[x,1]).
  5. Since A is infinite, there is some x1∈F1 such that A1=A∩B[x1,1] is an infinite set.

    1. If A∩B[x,1] were finite for each x∈F, 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 F2βŠ‚X such that

    X=⋃x∈F2B[x,12].
  7. Since A1 is infinite, there is some x2∈F2 such that A2=A1∩B[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

    A∩B[x1,1]∩B[x2,12]βˆ©β‹―βˆ©B[xn,1n]∩B[xn+1,1n+1]

    is infinite.

  9. Define

    En=B[x1,1]∩B[x2,12]βˆ©β‹―βˆ©B[xn,1n]

    for each n.

  10. Then for each n:

    1. En is nonempty and closed.

    2. A∩En is infinite.

    3. En+1βŠ†En.

    4. diamEn≀2n. limdiamEn=0.

  11. Due to Theorem 3.59, there exists a∈X, such that a∈En for each n.

  12. Now, if y∈A∩En, 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 y∈A∩En, y∈B(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 MetricsΒΆ

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 AβŠ†X 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 AβŠ†β‹ƒi∈IOi 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, ⋃i∈IOi 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 AβŠ†Oi1βˆͺβ‹―βˆͺOin.

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

  6. Thus, AβŠ†Oi1βˆͺβ‹―βˆͺOin 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.