# How to show that if a topological space or a set in it is compact then every sequence in it contains a convergent subsequence?

Definition 1
A topological space $$(S, {\cal{T}})$$ is compact if any of its open covers contains a finite subcover.

Definition 2
In a topological space $$(S, {\cal{T}})$$, a set $$A\subset S$$ is compact if every open cover of $$A$$ contains a finite subcover.

Question:
Is it correct that if a topological space $$(S, {\cal{T}})$$ or a set $$A\subset S$$ is compact then every sequence in it contains a convergent subsequence?

• Are you referring to convergence as in nets? Or are you working with a metric space? Sep 18 at 6:53
• There is no reason why an arbitrary topological space has to be sequential. Sep 18 at 7:01
• "Every sequence in $X$ contains a convergent subsequence" is usually referred to as "$X$ is sequentially compact". Not every compact set is sequentially compact (hence different words): math.stackexchange.com/questions/220422/… Sep 18 at 7:22
• @cgb5436 By convergence I imply that a sequence $\{x_i\}$ has a limit, i.e. a point $x$ every neighbourhood of which contains all the sequence points, except at most finitely many of them. Sep 18 at 7:40
• It is generally not true that in a compact topological space every sequence has a convergent subsequence. Not without some additional assumptions, e.g. first countable. It is true that every sequence has a convergent subnet, but it is a different statement. The devil is in the details. Sep 18 at 7:58

## 1 Answer

We can show that every compact space is countably compact (meaning that every countable open cover has a finite subcover). We then show that every countably compact space that is additionally first-countable (meaning that each $$x \in X$$ has a countable neighborhood basis) is sequential compact (meaning that every sequence $$(a_n)_{n \in \mathbb{Z}^+}$$ has a convergent subsequence $$(a_{n_m})_{m \in \mathbb{Z}^+}$$).

The first part is easy. Letting $$X$$ be a compact space, since all open covers of $$X$$ contain a finite subcover, it is obviously true that all countable open covers of $$X$$ contain a finite subcover.

Now suppose $$X$$ is first-countable. Suppose further that $$X$$ is not sequentially compact: that is, there exists a sequence of points $$( a_n )$$ in $$X$$ that does not have a convergent subsequence. We use this sequence to show that $$X$$ is not countably compact. First, let $$A = \{ a_n \}$$ be the collection of points in the sequence $$( a_n )$$. If $$A$$ is finite, then it is trivial that $$A$$ has a convergent subsequence, so suppose that $$A$$ is infinite. For each $$n \in \mathbb{Z}^+$$, consider the neighborhood $$U_n = X - {\rm Cl} \{ a_k \}_{k \ge n}$$ of $$a_n$$. We claim that the countable collection $$\mathcal{U} = \{ U_n \}$$ is an open cover of $$X$$, but that it has no finite subcover.

($$\mathcal{U}$$ covers $$X$$) Suppose that $$\mathcal{U}$$ does not cover $$X$$; let $$x_0 \in X - \bigcup \mathcal{U}$$ be an arbitrary element that is not covered. By definition of $$\mathcal{U}$$ and applying DeMorgan's Laws, this means that $$x_0 \in \bigcap_{n \in \mathbb{Z}^+} {\rm Cl} \{ a_k \}_{k \ge n}$$. We claim that there is a subsequence $$( a_{n_m} )$$ of $$( a_n )$$ that converges to $$x_0$$. Since $$X$$ is first-countable, there exists a nested neighborhood basis $$B_1 \supseteq B_2 \supseteq \dots$$ about $$x_0$$. For $$m = 1$$, take an arbitrary element $$a_{n_{m = 1}} \in A \cap B_{m = 1}$$. Such an element exists because $$x_0 \in X - \bigcup \mathcal{U} \subseteq X - U_1 = {\rm Cl} \{ a_k \}_{k \ge 1} = {\rm Cl} A$$ and $$B_1$$ is a neighborhood of $$x_0$$. For $$m > 1$$, consider the set $$A_m = \{ a_n \}_{n > n_{m - 1}}$$, which is non-empty by hypothesis that $$A$$ is infinite. Take an arbitrary element $$a_{n_m} \in A_m \cap B_m$$. Such an element exists because $$x_0 \in X - \bigcup \mathcal{U} \subseteq X - \bigcup_{p = 1}^{n_{m - 1} + 1} U_p = \bigcap_{p = 1}^{n_{m - 1} + 1} {\rm Cl} \{ a_k \}_{k \ge p} \subseteq {\rm Cl} \{ a_k \}_{k > n_{m - 1}} = {\rm Cl} A_m$$ and $$B_m$$ is a neighborhood of $$x_0$$. The subsequence $$( a_{n_m} )$$ converges to $$x_0$$, since any neighborhood $$V$$ of $$x_0$$ contains some basis element $$B_{m_V}$$ of $$x_0$$, and $$a_{n_m} \in B_{m_V} \subseteq V$$ for each $$n_m > n_{m_V}$$. This contradicts the hypothesis that $$( a_n )$$ does not have a convergent subsequence, so it must be that such an $$x_0 \in X - \bigcup \mathcal{U}$$ does not exist.

(No finite subset of $$\mathcal{U}$$ covers $$X$$) Suppose there is a finite subset $$\{ U_{n_1}, \dots, U_{n_N} \} \subseteq \mathcal{U}$$ that covers $$X$$. Take $$n_{N'} > \max \{ n_1, \dots, n_N \}$$; clearly $$\bigcup_{p = 1}^N U_{n_p}$$ does not contain $$a_{n_{N'}} \in X$$, a contradiction.

The above shows that if a first-countable space $$X$$ is not sequentially compact, it is not countably compact. Equivalently, if a first-countable space $$X$$ is countably compact, it is sequentially compact, as desired.

The answer to your follow-up question on metric spaces is in the positive, since all metric spaces are first-countable: for any $$x \in X$$, the collection $$\left\{ B_d \left( x, \frac{1}{n} \right) \right\}_{n \in \mathbb{Z}^+}$$ is a countable neighborhood basis about $$x$$.