Compactness

• topological space and its subspace
• T1 topological space and T2 topological space
• Lindelöf topological space
• sequence convergence in a topological space
• continuity of functions between topological spaces
• first countability of topological spaces
• pseudometric space and metric space
• separable metric space

Two equivalent definition of a compact topological space \((X,\mathcal{T})\):

1. Any open cover of \(X\) has a finite open subcover¹.
2. Any family of closed sets \(\mathcal{F}\) with Finite Intersection Property has \(\bigcap\mathcal{F}\neq\emptyset\)².

A subset \(K\) in \((X,\mathcal{T})\) is called a compact subset if the subspace \((K, \mathcal{T}_K)\) is compact³.

Three more different notions of compactness.

sequentially compact

every sequence has a convergent subsequence

countably compact

every countable open cover of \(X\) has a finite subcover

pseudocompact

every continuous function \(f:X\to\mathbb{R}\) is bounded.

In addition, if \(X\) is a pseudometric space, there is another important property.

totally bounded

for each \(\epsilon > 0\), \(X\) can be covered by a finite number of \(\epsilon\)-balls.

Let \((X, \mathcal{T})\) be a topological space.

Theorem 8.5. Let \(K \subset X\).

1. \(K\) is closed \(\Longrightarrow\) \(K\) is compact, if \(X\) is compact.

2. \(K\) is compact \(\Longrightarrow\) \(K\) is closed, if \(X\) is Hausdorff.

Theorem 8.11. The following implications hold generally. \[ X\text{ is (sequentially) compact }\Rightarrow X\text{ is countably compact } \Rightarrow X\text{ is pseudocompact}.\]

Lemma 8.12. \(X\) is countably compact \(\Longrightarrow\) \(X\) is sequentially compact, if \(X\) is first countable.

Lemma 8.16. A totally bounded pseudometric space is separable, and thus Lindelöf.

Theorem 8.17. In a pseudometric space \((X,d)\), the property of compactness, sequentially compactness, countably compactness and pseudocompactness are all equivalent.

Proof to Theorem 8.5. Assume \(K\) is closed and \(X\) is compact, then any open cover of \(K\) would form an open cover of \(X\) if augmented by \(K^\mathsf{c}\). This implies that a finite subcover of \(K\) exists.

Assume \(K\) is compact and \(X\) is Hausdorff. Pick any \(p\in K^\mathsf{c}\). For any \(q \in K\), there exists a neighborhood \(V_q\) of \(q\) and a neighborhood \(W_q\) of \(p\) such that \(V_q \cap W_q = \emptyset\). By the compactness of \(K\), there exists finite many \(q_i\) such that \(\bigcup_{i=1}^n V_{q_i}\supset K\). This implies that \(\bigcap_{i=1}^n W_{q_i}\subset K^\mathsf{c}\) is a neighborhood of \(p\). Hence, \(p\) is an interior point of \(K^\mathsf{c}\).

Q.E.D.

Proof to Theorem 8.11.

Compact \(\Rightarrow\) countably compact. Obviously.

Sequentially compact \(\Rightarrow\) countably compact. If not, there exists a countable open cover \((V_i)_{i=0}^\infty\) of \(X\) which has no finite subcover. Let \(x_k\in\bigl(\bigcup_{i=0}^k V_i\bigr)^\mathsf{c}\). Clearly, the sequence \((x_k)\) has no convergent subsequence. This contradicts with the hypothesis.

Countably compact \(\Rightarrow\) pseudocompact. For any real-valued continuous function \(f:X\to\mathbb{R}\), \((f^{-1}(-n,n))_{n=1}^{\infty}\) is a countable open cover of \(X\), and thus has a finite subcover. WLOG, assume the finite subcover is \((f^{-1}(-n,n))_{n=1}^{M}\). Clearly, \(f\) is bounded by \(M\).

Q.E.D.

Proof to Lemma 8.12. First, we prove that \(X\) is countably compact if and only if every sequence has a cluster point. The if part has been proved in Theorem 8.11. To prove the only if part, assume \(X\) be countably compact and \((x_n)\) is a sequence with no cluster point. Let \(T_n\) be the tail set of \((x_n)\): \[T_n:=\{x_k\mid k\geq n\}.\] Let \(\overline{T}_n\) be the closure of \(T_n\). Since \(X\) is countably compact, the countable family of closed sets \(\{\overline{T}_n \mid n\in\mathbb{N}\}\) must have \[\bigcap_{n=0}^\infty\overline{T}_n\neq\emptyset.\] Pick \(x\in \bigcap_{n=0}^\infty\overline{T}_n\). For any \(n\), there is \(x\in \overline{T}_n\). Hence, for any neighborhood \(N_x\) of \(x\), there is \(N_x\cap T_n\neq\emptyset\). Recalling the definition of \(T_n\), we conclude that \(x\) is a cluster point of \((x_n)\). However, this contradicts with the hypothesis that \((x_n)\) has no cluster point. We finish the only if part.

Then we prove this lemma. Let \(X\) be countably compact and first countable. For any sequence \((x_n)\), it has a cluster point \(x\). There must exist a subsequence of \((x_n)\) which converges to \(x\).

• Let \((B_k)\) be a countable shrinking neighborhood base at \(x\). Since \((x_n)\) is frequently in \(B_1\), we can pick \(n_1\) so that \(x_{n_1}\in B_1\). Since \((x_n)\) is frequently in \(B_2\), we can pick \(n_2 > n_1\) so that \(x_{n_2}\in B_2\). Continue inductively: having chosen \(n_1 < n_2 < \cdots < n_k\) so that \(x_{n_k}\in U_k \subset U_{k-1} \subset \cdots U_1\), we can then choose \(n_{k+1} > n_k\) so that \(x_{n_{k+1}}\in U_{k+1}\subset U_k\). Clearly, \((x_{n_k})_{k=1}^\infty\) converges to \(x\).

In conclusion, if \(X\) is countably compact, then every sequence has a cluster point. Since \(X\) is first countable, we conclude that every sequence has a convergent subsequence.

Q.E.D.

Proof to Lemma 8.16. Let \((X,d)\) be a totally bounded pseudometric space.

First, we prove that a totally bounded pseudometric space is separable. For each \(n\in\mathbb{N}\), there exists finite many points \(x^{(n)}_1,x^{(n)}_2,\ldots x^{(n)}_{k_n}\) such that \(X\) can be covered by \(\frac{1}{n}\)-balls centered at these points. We claim that \[ E:=\bigcup_{n=1}^\infty \{x^{(n)}_i\mid 1\leq i\leq k_n\} \] is a dense subset of \(X\).

• For any \(x\in X\) and arbitrary small \(\epsilon > 0\), we can find \(x^{(n)}_i\in E\) such that \(d( x^{(n)}_i, x ) < \epsilon\). This is done by choosing \(n > \frac{1}{\epsilon}\) and \(i=1\).

Then, we prove that a separable pseudometric space is second countable. Let \(D=\{x_k\mid x\in\mathbb{N}\}\) be a dense subset of \(X\). We claim that \[ \mathcal{O} := \bigcup_{k=1}^\infty \{B_{\frac{1}{n}}(x_k)\mid n\in\mathbb{N}\} \] is a countable topological base.

• For any \(x\in V\in\mathcal{T}_d\), there exists some \(\epsilon > 0\) such that \(B_\epsilon(x)\subset V\). As \(D\) is dense in \(X\), there is \(x_k\in D\) such that \(d(x_k, x) < \frac{\epsilon}{2}\). Choose \(n\) such that \(\frac{1}{n} < \frac{\epsilon}{2}\), we have \(B_{\frac{1}{n}}(x_k)\subset B_\epsilon(x)\subset V\).

Finally, we prove that a second countable pseudometric space is Lindelöf. Let \(\mathcal{B}\) be a countable base of \(\mathcal{T}\), and let \(\mathcal{U}\) be an arbitrary open cover of \(X\). For any \(x\in X\), there exists \(U_x\in\mathcal{U}\) such that \(x\in U_x\). Since \(\mathcal{B}\) is a base, for each \(x\), there exists a \(B_x\in\mathcal{B}\) such that \(x\in B_x\subset U_x\). Therefore, \[ \mathcal{V}:=\bigcup_{x\in X}B_x \] forms an open cover of \(X\). However, \(\mathcal{V}\subset\mathcal{B}\) must be countable. Hence, \(\mathcal{V}\) can be represented as \[ \mathcal{V}:=\bigcup_{i=1}^\infty B_{x_i}. \] We conclude that \(\bigcup_{i=1}^\infty U_{x_i}\) is a countable subcover.

Q.E.D.

Proof to Theorem 8.17. Based on Theorem 8.11, we need only to prove the following implications.

Countably compactness implies sequentially compactness

As any pseudometric space is first countable, then countably compactness implies sequentially compactness by Lemma 8.12.

Countably compactness implies compactness

By Lemma 8.16, any totally bounded pseudometric space is Lindelöf. Hence, it is sufficient to prove that a countably pseudometric space is totally bounded.

If a countably compact pseudometric space is not totally bounded, then there exists \(\epsilon > 0\) such that \(X\) cannot be covered by finite many \(\epsilon\)-balls. Obviously, \(X\) is nonempty. Pick \(x_1 \in X\). As \(\{B_\epsilon(x_1)\}\) cannot cover \(X\), we can pick \(x_2\in X\) such that \(d(x_2,x_1) \geq \epsilon\). Again, as \(\{B_\epsilon(x_1), B_\epsilon(x_2)\}\) cannot cover \(X\), we can pick \(x_3\in X\) such that \(d(x_3,x_1) \geq \epsilon\) and \(d(x_3,x_2)\geq \epsilon\). Continue inductively, we may construct a sequence \((x_n)\) such that \(d(x_i,x_j) \geq \epsilon\) for each \(i\neq j\). Clearly, this sequence has no convergent subsequence. However, since we have proved that countably compactness implies sequentially compactness, \((x_n)\) must have a convergent subsequence, leading to a contradiction.

Pseudocompactness implies countably compactness

Assume \(X\) is pseudocompact but is not countably compact. As \(X\) is not countably compact, there exists a sequence \((x_n)\) with no cluster point.

• STEP I. Ensure \(d(x_n,x_m) > 0\) for all \(n\neq m\). If not, we may pick a subsequence \((x_{a_k})\) such that \(d(x_{a_n},x_{a_m}) > 0\) for all \(n\neq m\).
  • The subsequence is constructed by observing the following fact: /for any \(n\), the set \[ E_n:= \{m > n \mid d(x_m,x_n) =0\} \] must be finite. Otherwise, there would be convergent subsequence, and contradicts with the assumption that \((x_n)\) has no cluster point.
• STEP II. Construct a sequence of open sets \((U_n)_{n=1}^\infty\) such that 1) \(x_n\in U_n\); 2) \(U_i\cap U_j=\emptyset\) if \(i\neq j\); 3) \(\mathrm{diam}\,U_n\to 0\).
  • For any \(x_m\), there exists an open ball \(B_{\delta_m}(x_m)\) containing only finte many \(x_n\)'s (since \(x_m\) is not a cluster point of \((x_n)\)). Because \(d(x_n,x_m) > 0\) for all \(n\neq m\), the \(\delta_m\) may be shrunk such that \(B_{\delta_m}(x_m)\) contains no other \(x_n\)'s except \(x_m\). In other words, \[ d(x_m, x_n) \geq \delta_m,\qquad\forall n\neq m. \] Let \(\epsilon_m=\min(\delta_m/3, \frac{1}{m})\). Then we claim that \(U_n=B_{\epsilon_n}(x_n)\) forms the desired sequence of open sets. Clearly, \(x_n\in U_n\) and \(\operatorname{diam}U_n\to0\). To see that \(U_n\cap U_m=\emptyset\) if \(n\neq m\), we need only to note that \(\epsilon_n + \epsilon_m \leq \delta_n/3 + \delta_m/3 < d(x_n,x_m)\).
• STEP III. Show \(f_n(x)=n\frac{d(x_n,U_n^\mathsf{c})}{d(x_n,U_n^\mathsf{c})}\) and \(f=\sum_{n=1}^\infty f_n\) are well defined.
  • As \(x_n\not\in U_n^\mathsf{c}\) and \(U_n^\mathsf{c}\) is closed, the distance \(d(x_n,U_n^\mathsf{c}\) cannot be 0. Moreover, \(d(x,U_n^\mathsf{c}\neq 0\) if and only \(x\in U_n\). Hence, for any \(x\), there exists at most one \(f_n\) such that \(f_n(x)\neq 0\). Therefore, \(f=\sum_n f_n\) is finite at any \(x\).
• STEP IV. Prove that \(f\) is an unbounded continuous function.

  • Since \(f(x_n)=f_n(x_n)=n\), we have \(f\) is unbounded. To prove \(f\) is continuous, we first note that \(f_n\) is continuous.

    • If \(d\) is a metric on \(X\) and \(E\) is a subset of \(X\), then for any \(x,y\in X\), there is \[ d(x,E) \leq d(y,E) + d(x,y). \] Therefore, \(d(\cdot,E)\) must be continuous. (If \(d(x_n,x)\to0\), then \(|d(x_n,E)-d(x,E)|\leq d(x_n,x)\to0\).)

    Then we prove that for any \(a\in X\), there exists a open set \(V_a\) such that \(f|_{V_a}=\sum_{n=1}^Nf_n\) for some \(N\).

    • If \(d(a,x_n)=0\) for some \(n\), then \(V_a\) can be set to \(U_n\) and \(N=n\).
    • Suppose \(d(a,x_n)>0\) for all \(n\). Since \(a\) is not a cluster point of \(x_n\), there exists an open ball \(B_\delta(a)\) containing no \(x_n\). In other words, \[ d(a,x_n) \geq \delta,\qquad\forall n. \] Let \(V_a=B_{\delta/2}(a)\). Then for any \(x\in V_a\), \[ d(x,x_n) \geq d(a,x_n) - d(a,x) \geq \delta/2,\qquad\forall n. \] Recalling that \(\operatorname{diam}U_n\to0\), there must exist \(N\) such that for all \(n\geq N\), \(\operatorname{diam}U_n < \delta/2\). Therefore, for any \(x\in V_a\), we have \(x\not\in U_n\) for all \(n\geq N\). In otherwords, \(f_n(x)=0\) for all \(n\geq N\) if \(x\in V_a\).

    Therefore, \(f\) is continous at any point \(a\).