# 4.4: Compact Sets

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Suppose \(T \subset \mathbb{R} .\) If \(A\) is a set, \(U_{\alpha}\) is an open set for every \(\alpha \in A,\) and

\[T \subset \bigcup_{\alpha \in A} U_{\alpha},\]

then we call \(\left\{U_{\alpha}: \alpha \in A\right\}\) an open cover of \(T\).

For \(n=3,4,5, \dots,\) let

\[U_{n}=\left(\frac{1}{n}, \frac{n-1}{n}\right).\]

Then \(\left\{U_{n}: n=3,4,5, \ldots\right\}\) is an open cover of the open interval \((0,1)\).

Suppose \(\left\{U_{\alpha}: \alpha \in A\right\}\) is an open cover of \(T \subset \mathbb{R} .\) If \(B \subset A\) and

\[T \subset \bigcup_{\beta \in B} U_{\beta},\]

then we call \(\left\{U_{\beta}: \beta \in B\right\}\) a subcover of \(\left\{U_{\alpha}: \alpha \in A\right\} .\) If \(B\) is finite, we call \(\left\{U_{\beta}: \beta \in B\right\}\) a finite subcover of \(\left\{U_{\alpha}: \alpha \in A\right\}\).

Show that the open cover of \((0,1)\) given in the previous example does not have a finite subcover.

We say a set \(K \subset \mathbb{R}\) is compact if every open cover of \(K\) has a finite sub cover.

As a consequence of the previous exercise, the open interval \((0,1)\) is not compact.

Show that every finite subset of \(\mathbb{R}\) is compact.

Suppose \(n \in \mathbb{Z}^{+}\) and \(K_{1}, K_{2}, \ldots, K_{n}\) are compact sets. Show that \(\bigcup_{i=1}^{n} K_{i}\) is compact.

If \(I\) is a closed, bounded interval, then \(I\) is compact.

**Proof**-
Let \(a \leq b\) be finite real numbers and \(I=[a, b] .\) Suppose \(\left\{U_{\alpha}: \alpha \in A\right\}\) is an open cover of \(I .\) Let \(\mathcal{O}\) be the set of sets \(\left\{U_{\beta}: \beta \in B\right\}\) with the properties that \(B\) is a finite subset of \(A\) and \(a \in \bigcup_{\beta \in B} U_{\beta} .\) Let

\[(s-\epsilon, s+\epsilon) \subset U_{\alpha}.\]

Moreover, there exists a \(\left\{U_{\beta}: \beta \in B\right\} \in \mathcal{O}\) for which

\[\left[a, s-\frac{\epsilon}{2}\right] \subset \bigcup_{\beta \in B} U_{\beta}.\]

But then

\[\left\{U_{\beta}: \beta \in B\right\} \cup\left\{U_{\alpha}\right\} \in \mathcal{O}\]

and

\[\left[a, s+\frac{\epsilon}{2}\right] \subset\left(\bigcup_{\beta \in B} U_{\beta}\right) \cup U_{\alpha},\]

contradicting the definition of \(s .\) Hence we must have \(s=b .\) Now choose \(U_{\alpha}\) such that \(b \in U_{\alpha} .\) Then, for some \(\epsilon>0\),

\[(b-\epsilon, b+\epsilon) \subset U_{\alpha}.\]

Moreover, there exists \(\left\{U_{\beta}: \beta \in B\right\} \in \mathcal{O}\) such that

\[\left[a, b-\frac{\epsilon}{2}\right] \subset \bigcup_{\beta \in B} U_{\beta}.\]

Then

\[\left\{U_{\beta}: \beta \in B\right\} \cup\left\{U_{\alpha}\right\} \in \mathcal{O}\]

is a finite subcover of \(I .\) Thus \(I\) is compact. \(\quad\) Q.E.D.

If \(K\) is a closed, bounded subset of \(\mathbb{R},\) then \(K\) is compact.

**Proof**-
Since \(K\) is bounded, there exist finite real numbers \(a\) and \(b\) such that \(K \subset[a, b] .\) Let \(\left\{U_{\alpha}: \alpha \in A\right\}\) be an open cover of \(K .\) Let \(V=\mathbb{R} \backslash K .\) Then

\[\left\{U_{\alpha}: \alpha \in A\right\} \cup\{V\}\]

in the latter case, we have

\[K \subset[a, b] \backslash V \subset \bigcup_{\beta \in B} U_{\beta}.\]

In either case, we have found a finite subcover of \(\left\{U_{\alpha}: \alpha \in A\right\}\). \(\quad\) Q.E.D.

Show that if \(K\) is compact and \(C \subset K\) is closed, then \(C\) is compact.

If \(K \subset \mathbb{R}\) is compact, then \(K\) is closed.

**Proof**-
Suppose \(x\) is a limit point of \(K\) and \(x \notin K .\) For \(n=1,2,3, \ldots,\) let

\[U_{n}=\left(-\infty, x-\frac{1}{n}\right) \cup\left(x+\frac{1}{n},+\infty\right).\]

Then

\[\bigcup_{n=1}^{\infty} U_{n}=(-\infty, x) \cup(x,+\infty) \supset K .\]

However, for any \(N \in \mathbb{Z}^{+},\) there exists \(a \in K\) with

\[a \in\left(x-\frac{1}{N}, x+\frac{1}{N}\right),\]

and hence

\[a \notin \bigcup_{n=1}^{N} U_{n}=\left(-\infty, x-\frac{1}{N}\right) \cup\left(x+\frac{1}{N},+\infty\right).\]

Thus the open cover \(\left\{U_{n}: n \in \mathbb{Z}^{+}\right\}\) does not have a finite subcover, contradicting the assumption that \(K\) is compact. \(\quad\) Q.E.D.

Suppose that for each \(\alpha\) in some set \(A, K_{\alpha}\) is compact. Show that \(\bigcap_{\alpha \in A} K_{\alpha}\) is compact.

If \(K \subset \mathbb{R}\) is compact, then \(K\) is bounded.

**Proof**-
Suppose \(K\) is not bounded. For \(n=1,2,3, \ldots,\) let \(U_{n}=(-n, n) .\) Then

\[\bigcup_{n=1}^{\infty} U_{n}=(-\infty, \infty) \supset K .\]

But, for any integer \(N,\) there exists \(a \in K\) such that \(|a|>N,\) from which it follows that

\[a \notin \bigcup_{n=1}^{N} U_{n}=(-N, N).\]

Thus the open cover \(\left\{U_{n}: n \in \mathbb{Z}^{+}\right\}\) does not have a finite subcover, contradicting the assumption that \(K\) is compact. \(\quad\) Q.E.D.

Taken together, the previous three propositions yield the following fundamental result:

A set \(K \subset \mathbb{R}\) is compact if and only if \(K\) is closed and bounded.

If \(K \subset \mathbb{R}\) is compact and \(\left\{x_{n}\right\}_{n \in I}\) is a sequence with \(x_{n} \in K\) for every \(n \in I,\) then \(\left\{x_{n}\right\}_{n \in I}\) has a convergent subsequence \(\left\{x_{n_{k}}\right\}_{k=1}^{\infty}\) with

\[\lim _{k \rightarrow \infty} x_{n_{k}} \in K .\]

**Proof**-
Since \(K\) is bounded, \(\left\{x_{n}\right\}_{n \in I}\) has a convergent subsequence \(\left\{x_{n_{k}}\right\}_{k=1}^{\infty}\). Since \(K\) is closed, we must have \(\lim _{k \rightarrow \infty} x_{n_{k}} \in K\).

Suppose \(K \subset \mathbb{R}\) is such that whenever \(\left\{x_{n}\right\}_{n \in I}\) is a sequence with \(x_{n} \in K\) for every \(n \in I,\) then \(\left\{x_{n}\right\}_{n \in I}\) has a subsequence \(\left\{x_{n_{k}}\right\}_{k=1}^{\infty}\) with \(\lim _{k \rightarrow \infty} x_{n_{k}} \in K .\) Then \(K\) is compact.

**Proof**-
Suppose \(K\) is unbounded. Then we may construct a sequence \(\left\{x_{n}\right\}_{n=1}^{\infty}\) such that \(x_{n} \in K\) and \(\left|x_{n}\right|>n\) for \(n=1,2,3, \ldots\) Hence the only possible subsequential limits of \(\left\{x_{n}\right\}_{n=1}^{\infty}\) would be \(-\infty\) and \(+\infty,\) contradicting our assumptions. Thus \(K\) must be bounded.

Now suppose \(\left\{x_{n}\right\}_{n \in I}\) is a convergent sequence with \(x_{n} \in K\) for all \(n \in I .\) If \(L=\lim _{n \rightarrow \infty} x_{n},\) then \(L\) is the only subsequential limit of \(\left\{x_{n}\right\}_{n \in I} .\) Hence, by the assumptions of the proposition, \(L \in K .\) Hence \(K\) is closed.

Since \(K\) is both closed and bounded, it is compact. \(\quad\) Q.E.D.

Show that a set \(K \subset \mathbb{R}\) is compact if and only if every infinite subset of \(K\) has a limit point in \(K\).

Show that if \(K\) is compact, then \(\sup K \in K\) and \(\inf K \in K\).

Given a set \(K \subset \mathbb{R},\) the following are equivalent:

1. Every open cover of \(K\) has a finite subcover.

2. Every sequence in \(K\) has a subsequential limit in \(K\).

3. Every infinite subset of \(K\) has a limit point in \(K\).

Suppose \(K_{1}, K_{2}, K_{3}, \ldots\) are nonempty compact sets with

\[K_{n+1} \subset K_{n}\]

for \(n=1,2,3, \ldots\) Show that

\[\bigcap_{n=1}^{\infty} K_{n}\]

is nonempty.

We say a collection of sets \(\left\{D_{\alpha}: \alpha \in A\right\}\) has the finite intersection property if for every finite set \(B \subset A\),

\[\bigcap_{\alpha \in B} D_{\alpha} \neq \emptyset .\]

Show that a set \(K \subset \mathbb{R}\) is compact if and only for any collection

\[\left\{E_{\alpha}: \alpha \in A, E_{\alpha}=C_{\alpha} \cap K \text { where } C_{\alpha} \subset \mathbb{R} \text { is closed }\right\}\]

which has the finite intersection property we have

\[\bigcap_{\alpha \in A} E_{\alpha} \neq \emptyset .\]