# 4.6: Compact Sets

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We now pause to consider a very important kind of sets. In Chapter 3, §16, we showed that every sequence \(\left\{\overline{z}_{m}\right\}\) taken from a closed interval \([\overline{a}, \overline{b}]\) in \(E^{n}\) 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.

Definition: sequentially compact

A set \(A \subseteq(S, \rho)\) is said to be **sequentially compact** (briefly compact) iff every sequence \(\left\{x_{m}\right\} \subseteq A\) clusters at some point \(p\) in \(A .\)

If all of \(S\) is compact, we say that the metric space \((S, \rho)\) is compact.

Example \(\PageIndex{1}\)

(a) Each closed interval in \(E^{n}\) is compact (see above).

(a') However, nonclosed intervals, and \(E^{n}\) itself, are not compact.

For example, the sequence \(x_{n}=1 / n\) is in \((0,1] \subset E^{1},\) but clusters only at \(0,\) outside \((0,1] .\) As another example, the sequence \(x_{n}=n\) has no cluster points in \(E^{1} .\) Thus \((0,1]\) and \(E^{1}\) fail to be compact (even though \(E^{1}\) is complete); similarly for \(E^{n}\left(^{*} \text { and } C^{n}\right) .\)

(b) Any finite set \(A \subseteq(S, \rho)\) is compact. Indeed, an infinite sequence in such a set must have at least one infinitely repeating term \(p \in A .\) Then by definition, this \(p\) is a cluster point (see Chapter 3, §14, Note 1).

(c) The empty set is "vacuously" compact (it contains no sequences).

(d) \(E^{*}\) is compact. See Example \((\mathrm{g})\) in Chapter 3, §14.

Other examples can be derived from the theorems that follow.

Theorem \(\PageIndex{1}\)

If a set \(B \subseteq(S, \rho)\) is compact, so is any closed subset \(A \subseteq B\).

**Proof**-
We must show that each sequence \(\left\{x_{m}\right\} \subseteq A\) clusters at some \(p \in A\). However, as \(A \subseteq B,\left\{x_{m}\right\}\) is also in \(B,\) so by the compactness of \(B,\) it clusters at some \(p \in B .\) Thus it remains to show that \(p \in A\) as well.

Now by Theorem 1 of Chapter \(3, §16,\left\{x_{m}\right\}\) has a subsequence \(x_{m_{k}} \rightarrow p\). As \(\left\{x_{m_{k}}\right\} \subseteq A\) and \(A\) is closed, this implies \(p \in A\) (Theorem 4 in Chapter \(3,\) \(§16) . \quad \square\)

Theorem \(\PageIndex{2}\)

Every compact set \(A \subseteq(S, \rho)\) is closed.

**Proof**-
Given that \(A\) is compact, we must show (by Theorem 4 in Chapter 3, §16) that \(A\) contains the limit of each convergent sequence \(\left\{x_{m}\right\} \subseteq A\).

Thus let \(x_{m} \rightarrow p,\left\{x_{m}\right\} \subseteq A .\) As \(A\) is compact, the sequence \(\left\{x_{m}\right\}\) clusters at some \(q \in A,\) i.e., has a subsequence \(x_{m_{k}} \rightarrow q \in A .\) However, the limit of the subsequence must be the same as that of the entire sequence. Thus \(p=q \in A\); i.e., \(p\) is in \(A,\) as required. \(\square\)

Theorem \(\PageIndex{3}\)

Every compact set \(A \subseteq(S, \rho)\) 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 \(\varepsilon>0\) and, seeking a contradiction, assume that \(A\) cannot by any finite number of globes of that radius.

Then if \(x_{1} \in A,\) the globe \(G_{x_{1}}(\varepsilon)\) does not cover \(A,\) so there is a point \(x_{2} \in A\) such that

\[x_{2} \notin G_{x_{1}}(\varepsilon), \text{ i.e., } \rho\left(x_{1}, x_{2}\right) \geq \varepsilon\]

By our assumption, \(A\) is not even covered by \(G_{x_{1}}(\varepsilon) \cup G_{x_{2}}(\varepsilon) .\) Thus there is a point \(x_{3} \in A\) with

\[x_{3} \notin G_{x_{1}}(\varepsilon) \text{ and } x_{3} \notin G_{x_{2}}(\varepsilon), \text{ i.e., } \rho\left(x_{3}, x_{1}\right) \geq \varepsilon \text{ and } \rho\left(x_{3}, x_{2}\right) \geq \varepsilon.\]

Again, \(A\) is not covered by \(\bigcup_{i=1}^{3} G_{x_{i}}(\varepsilon),\) so there is a point \(x_{4} \in A\) not in that union; its distances from \(x_{1}, x_{2},\) and \(x_{3}\) must therefore be \(\geq \varepsilon\).

Since \(A\) is never covered by any finite number of \(\varepsilon\) -globes, we can continue this process indefinitely (by induction) and thus select an infinite sequence \(\left\{x_{m}\right\} \subseteq A,\) with all its terms at least \(\varepsilon\) -apart from each other.

Now as \(A\) is compact, this sequence must have a convergent subsequence \(\left\{x_{m_{k}}\right\},\) which is then certainly Cauchy (by Theorem 1 of Chapter 3, §17). This is impossible, however, since its terms are at distances \(\geq \varepsilon\) from each other, contrary to Definition 1 in Chapter 3, §17. This contradiction completes the proof. \(\square\)

**Note 1.** We have actually proved more than was required, namely, that no matter how small \(\varepsilon>0\) is, \(A\) can be covered by finitely many globes of radius \(\varepsilon\) 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 \(E^{n}\left(^{*} \text { and } C^{n}\right),\) however, the converse is likewise true, as we show next.

Theorem \(\PageIndex{4}\)

In \(E^{n}\left(^{*} \text { and } C^{n}\right)\) a set is compact iff it is closed and bounded.

**Proof**-
In fact, if a set \(A \subseteq E^{n}\left(^{*} C^{n}\right)\) is bounded, then by the Bolzano-Weierstrass theorem, each sequence \(\left\{x_{m}\right\} \subseteq A\) has a convergent subsequence \(x_{m_{k}} \rightarrow p .\) If \(A\) is also closed, the limit point \(p\) must belong to \(A\) itself.

Thus each sequence \(\left\{x_{m}\right\} \subseteq A\) clusters at some \(p\) in \(A,\) so \(A\) is compact.

The converse is obvious. \(\square\)

**Note 3. **In particular, every closed globe in \(E^{n}\left(^{*} \text { or } C^{n}\right)\) is compact since it is bounded and closed (Chapter 3, §12, Example \((6) ),\) so theorem 4 applies.

The converse is obvious. \(\square\)

Theorem \(\PageIndex{5}\)

(Cantor's principle of nested closed sets). Every contracting sequence of nonvoid compact sets

\[F_{1} \supseteq F_{2} \supseteq \cdots \supseteq F_{m} \supseteq \cdots\]

in a metric space \((S, \rho)\) has a nonvoid intersection; i.e., some \(p\) belongs to all \(F_{m} .\)

For complete sets \(F_{m},\) this holds as well, provided the diameters of the sets \(F_{m}\) tend to \(0 : d F_{m} \rightarrow 0 .\)

**Proof**-
We prove the theorem for complete sets first.

As \(F_{m} \neq \emptyset,\) we can pick a point \(x_{m}\) from each \(F_{m}\) to obtain a sequence \(\left\{x_{m}\right\}, x_{m} \in F_{m} .\) since \(d F_{m} \rightarrow 0,\) it is easy to see that \(\left\{x_{m}\right\}\) is a Cauchy sequence. (The details are left to the reader.) Moreover,

\[(\forall m) \quad x_{m} \in F_{m} \subseteq F_{1}.\]

Thus \(\left\{x_{m}\right\}\) is a Cauchy sequence in \(F_{1},\) a complete set (by assumption).

Therefore, by the definition of completeness (Chapter 3, §17), \(\left\{x_{m}\right\}\) has a limit \(p \in F_{1} .\) 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 \(x_{m}, x_{m+1}, \ldots,\) which, by construction, is entirely contained in \(F_{m}\) (why?), with the same limit P. Then, however, the completeness of \(F_{m}\) implies that \(p \in F_{m}\) as well. As \(m\) is arbitrary here, it follows that \((\forall m) p \in F_{m},\) i.e.,

\[p \in \bigcap_{m=1}^{\infty} F_{m}, \text{ as claimed.}\]

The proof for compact sets is analogous and even simpler. Here \(\left\{x_{m}\right\}\) need not be a Cauchy sequence. Instead, using the compactness of \(F_{1},\) we select from \(\left\{x_{m}\right\}\) a subsequence \(x_{m_{k}} \rightarrow p \in F_{1}\) and then proceed as above. \(\square\)

**Note 4.**In particular, in \(E^{n}\) we may let the sets \(F_{m}\) be closed intervals (since they are compact). Then Theorem 5 yields the principle of nested intervals: Every contracting sequence of closed intervals in \(E^{n}\) has a nonempty intersection. (For an independent proof, see Problem 8 below.)