4.6: Compact Sets
( \newcommand{\kernel}{\mathrm{null}\,}\)
We now pause to consider a very important kind of sets. In Chapter 3, §16, we showed that every sequence \boldsymbol{¯zm} taken from a closed interval \boldsymbol[¯a,¯b] in \boldsymbolEn must cluster in it (Note 1 of Chapter 3, §16). There are other sets with the same remarkable property. This leads us to the following definition.
A set \boldsymbolA⊆(S,ρ) is said to be sequentially compact (briefly compact) iff every sequence \boldsymbol{xm}⊆A clusters at some point \boldsymbolp in \boldsymbolA.
If all of \boldsymbolS is compact, we say that the metric space \boldsymbol(S,ρ) is compact.
(a) Each closed interval in \boldsymbolEn is compact (see above).
(a') However, nonclosed intervals, and \boldsymbolEn itself, are not compact.
For example, the sequence \boldsymbolxn=1/n is in \boldsymbol(0,1]⊂E1, but clusters only at \boldsymbol0, outside \boldsymbol(0,1]. As another example, the sequence \boldsymbolxn=n has no cluster points in \boldsymbolE1. Thus \boldsymbol(0,1] and \boldsymbolE1 fail to be compact (even though \boldsymbolE1 is complete); similarly for \boldsymbolEn(∗ and Cn).
(b) Any finite set \boldsymbolA⊆(S,ρ) is compact. Indeed, an infinite sequence in such a set must have at least one infinitely repeating term \boldsymbolp∈A. Then by definition, this \boldsymbolp is a cluster point (see Chapter 3, §14, Note 1).
(c) The empty set is "vacuously" compact (it contains no sequences).
(d) \boldsymbolE∗ is compact. See Example \boldsymbol(g) in Chapter 3, §14.
Other examples can be derived from the theorems that follow.
If a set \boldsymbolB⊆(S,ρ) is compact, so is any closed subset \boldsymbolA⊆B.
- Proof
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We must show that each sequence {xm}⊆A clusters at some p∈A. However, as A⊆B,{xm} is also in B, so by the compactness of B, it clusters at some p∈B. Thus it remains to show that p∈A as well.
Now by Theorem 1 of Chapter 3,§16,{xm} has a subsequence xmk→p. As {xmk}⊆A and A is closed, this implies p∈A (Theorem 4 in Chapter 3, §16).◻
Every compact set A⊆(S,ρ) is closed.
- Proof
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Given that A is compact, we must show (by Theorem 4 in Chapter 3, §16) that A contains the limit of each convergent sequence {xm}⊆A.
Thus let xm→p,{xm}⊆A. As A is compact, the sequence {xm} clusters at some q∈A, i.e., has a subsequence xmk→q∈A. However, the limit of the subsequence must be the same as that of the entire sequence. Thus p=q∈A; i.e., p is in A, as required. ◻
Every compact set A⊆(S,ρ) is bounded.
- Proof
-
By Problem 3 in Chapter 3, §13, it suffices to show that A is contained in some finite union of globes. Thus we fix some arbitrary radius ε>0 and, seeking a contradiction, assume that A cannot by any finite number of globes of that radius.
Then if x1∈A, the globe Gx1(ε) does not cover A, so there is a point x2∈A such that
x2∉Gx1(ε), i.e., ρ(x1,x2)≥ε
By our assumption, A is not even covered by Gx1(ε)∪Gx2(ε). Thus there is a point x3∈A with
x3∉Gx1(ε) and x3∉Gx2(ε), i.e., ρ(x3,x1)≥ε and ρ(x3,x2)≥ε.
Again, A is not covered by ⋃3i=1Gxi(ε), so there is a point x4∈A not in that union; its distances from x1,x2, and x3 must therefore be ≥ε.
Since A is never covered by any finite number of ε -globes, we can continue this process indefinitely (by induction) and thus select an infinite sequence {xm}⊆A, with all its terms at least ε -apart from each other.
Now as A is compact, this sequence must have a convergent subsequence {xmk}, which is then certainly Cauchy (by Theorem 1 of Chapter 3, §17). This is impossible, however, since its terms are at distances ≥ε from each other, contrary to Definition 1 in Chapter 3, §17. This contradiction completes the proof. ◻
Note 1. We have actually proved more than was required, namely, that no matter how small ε>0 is, A can be covered by finitely many globes of radius ε with centers in A. This property is called total boundedness (Chapter 3, §13, Problem 4).
Note 2. Thus all compact sets are closed and bounded. The converse fails in metric spaces in general (see Problem 2 below). In En(∗ and Cn), however, the converse is likewise true, as we show next.
In En(∗ and Cn) a set is compact iff it is closed and bounded.
- Proof
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In fact, if a set A⊆En(∗Cn) is bounded, then by the Bolzano-Weierstrass theorem, each sequence {xm}⊆A has a convergent subsequence xmk→p. If A is also closed, the limit point p must belong to A itself.
Thus each sequence {xm}⊆A clusters at some p in A, so A is compact.
The converse is obvious. ◻
Note 3. In particular, every closed globe in En(∗ or Cn) is compact since it is bounded and closed (Chapter 3, §12, Example (6)), so theorem 4 applies.
The converse is obvious. ◻
(Cantor's principle of nested closed sets). Every contracting sequence of nonvoid compact sets
F1⊇F2⊇⋯⊇Fm⊇⋯
in a metric space (S,ρ) has a nonvoid intersection; i.e., some p belongs to all Fm.
For complete sets Fm, this holds as well, provided the diameters of the sets Fm tend to 0:dFm→0.
- Proof
-
We prove the theorem for complete sets first.
As Fm≠∅, we can pick a point xm from each Fm to obtain a sequence {xm},xm∈Fm. since dFm→0, it is easy to see that {xm} is a Cauchy sequence. (The details are left to the reader.) Moreover,
(∀m)xm∈Fm⊆F1.
Thus {xm} is a Cauchy sequence in F1, a complete set (by assumption).
Therefore, by the definition of completeness (Chapter 3, §17), {xm} has a limit p∈F1. This limit remains the same if we drop a finite number of terms, say, the first m−1 of them. Then we are left with the sequence xm,xm+1,…, which, by construction, is entirely contained in Fm (why?), with the same limit P. Then, however, the completeness of Fm implies that p∈Fm as well. As m is arbitrary here, it follows that (∀m)p∈Fm, i.e.,
p∈∞⋂m=1Fm, as claimed.
The proof for compact sets is analogous and even simpler. Here {xm} need not be a Cauchy sequence. Instead, using the compactness of F1, we select from {xm} a subsequence xmk→p∈F1 and then proceed as above. ◻
Note 4. In particular, in En we may let the sets Fm be closed intervals (since they are compact). Then Theorem 5 yields the principle of nested intervals: Every contracting sequence of closed intervals in En has a nonempty intersection. (For an independent proof, see Problem 8 below.)