nLab countably compact topological space

Definition

A topological space is called countably compact if every open cover consisting of a countable set of open subsets (every countable cover) admits a finite subcover, hence if there is a finite subset of the open in the original cover which still cover the space.

Examples

• A compact space is a fortiori countably compact.

• A sequentially compact space is countably compact.

• The long line is an example of a countably compact space that is not compact.

Properties

Proposition

A countably compact space is a limit point compact space.

Proof

Recall that a space is limit point compact if every closed discrete subspace is finite; equivalently, if every countable closed discrete subspace is finite.

Suppose $X$ is countably compact and $A$ is a countable closed discrete subspace. For each $a \in A$, choose an open neighborhood $U_a$ such that $U_a \cap A = \{a\}$, and let $V_a$ be the open subset $U_a \cup \neg A$. Clearly we still have $V_a \cap A = \{a\}$. Also, $\{V_a: a \in A\}$ is a countable cover of $X$, hence admits a finite subcover $V_{a_1}, \ldots, V_{a_n}$. But then

$$A = \left(\bigcup_{i=1}^n V_{a_i} \right) \cap A = \bigcup_{i=1}^n (V_{a_i} \cap A) = \{a_1, \ldots, a_n\}$$

as was to be shown.

Proposition

A space that is $T_1$ and limit point compact is countably compact.

Proof

Consider any countable open cover $U_1, U_2, \ldots$ of $X$. Put $V_n = U_1 \cup \ldots \cup U_n$ so that $V_1 \subseteq V_2 \subseteq \ldots$, and discard repetitions, i.e., consider the maximal chain of strict inclusions

$$\emptyset = V_0 \subset V_{n_1} \subset V_{n_2} \subset \ldots$$

It suffices to show this maximal chain is finite. Rename it as $\emptyset = W_0 \subset W_1 \subset W_2 \subset \ldots$.

For each $W_n$ occurring in this chain ($n \geq 1$), pick a point $x_n \in W_n \setminus W_{n-1}$. Observe that if $m \lt n$, then $x_n \notin W_m$.

Since $X$ is $T_1$ (points are closed), the set $W_m \cap \neg \{x_1, \ldots, x_{m-1}\}$ is an open neighborhood of $x_m$ that does not contain $x_n$ whenever $n \gt m$, and does not contain $x_n$ whenever $n \lt m$. Thus every point $x_m$ is open relative to $A = \{x_1, x_2, \ldots\}$, i.e., $A$ is a discrete subspace.

Finally, any point $x \notin A$ belongs to some $W_n$, and then $W_n \cap \neg \{x_1, \ldots x_n\}$ is an open neighborhood of $x$ that doesn’t intersect $A$. Thus $A$ is a closed discrete subspace, and is finite by limit point compactness. Therefore the maximal chain consisting of the sets $W_n$ is finite, as was to be shown.

Proposition

If $X$ is countably compact and $f \colon X \to Y$ is a map in Top, then the direct image $f(X)$ is a countably compact subset of $Y$.

Proof

The proof is just like the proof of the corresponding statement where the word “countably” is dropped. Suppose $\{U_n: n \in \mathbb{N}\}$ is a countable cover of $f(X)$, so that $f(X) \subseteq \bigcup_{n \in \mathbb{N}} U_n$. This is equivalent to

$$X \subseteq f^{-1}\left(\bigcup_{n \in \mathbb{N}} U_n\right) = \bigcup_{n \in \mathbb{N}} f^{-1}(U_n).$$

Thus $\{f^{-1}(U_n): n \in \mathbb{N}\}$ is a countable open cover of $X$, so by countable compactness there is a finite subcover $f^{-1}(U_{n_1}), \ldots, f^{-1}(U_{n_k})$ of $X$:

$$X \subseteq \bigcup_{i=1}^k f^{-1}(U_{n_i}) = f^{-1}\left(\bigcup_{i=1}^k U_{n_i}\right).$$

This is equivalent to $f(X) \subseteq \bigcup_{i=1}^k U_{n_i}$, as required.

• In the case of metric spaces, countably compact metric spaces are equivalently compact metric spaces.

Related concepts

• countably paracompact topological space

• compact topological space

• countably compact topological space

• sequentially compact topological space

• paracompact topological space

• strongly compact topological space