Equivalent Definitions of Countably Compact

Theorem

Let $X$ be a topological space.

The following statements are equivalent:

$(1): \quad$ $X$ is countably compact, i.e. every countable open cover of $X$ has a finite subcover.

$(2): \quad$ Every countable set of closed sets whose intersection is empty has a finite subset whose intersection is empty. That is, $X$ satisfies the Countable Finite Intersection Axiom.

$(3): \quad$ Every infinite subset of $X$ has an $\omega$-accumulation point in $X$.

$(4): \quad$ Every infinite sequence in $X$ has an accumulation point in $X$.

Proof

$(1) \iff (2)$

Let every countable open cover of $X$ have a finite subcover.

Let $\mathcal A$ be any set of closed subsets of $X$ satisfying $\bigcap \mathcal A = \varnothing$.

We define the set:

$\mathcal V := \left\{{X \setminus A : A \in \mathcal A}\right\}$

which is clearly an open cover of $X$.

From De Morgan's Laws:

$\displaystyle X \setminus \bigcup \mathcal V = \bigcap \left\{{X \setminus V : V \in \mathcal V}\right\} = \bigcap \left\{{A : A \in \mathcal A}\right\} = \varnothing$

and therefore $X = \bigcup \mathcal V$.

By definition, there exists a finite subcover $\tilde{\mathcal V} \subseteq \mathcal V$.

We define:

$\tilde{\mathcal A} := \left\{{X \setminus V : V \in \tilde{\mathcal V}}\right\}$

then $\tilde{\mathcal A} \subseteq \mathcal A$ by definition of $\mathcal V$.

Because $\tilde{\mathcal V}$ covers $X$, it follows directly that:

$\displaystyle \bigcap \tilde{\mathcal A} = \bigcap \left\{{X \setminus V : V \in \tilde{\mathcal V}}\right\} = X \setminus \bigcup \tilde{\mathcal V} = \varnothing$

Thus, in every countable set $\mathcal A$ of closed subsets of $X$ satisfying $\displaystyle \bigcap \mathcal A = \varnothing$ exists a finite subset $\tilde{\mathcal A}$ such that $\displaystyle \bigcap \tilde{\mathcal A} = \varnothing$.

That is, $X$ satisfies the Countable Finite Intersection Axiom.

$\Box$

The converse works exactly as the previous, but with the roles of the open cover and $\mathcal A$ reversed.

$\blacksquare$

$(3) \iff (4)$

By definition, a point is an $\omega$-accumulation point iff it is an accumulation point of that set viewed as an infinite sequence.

$\blacksquare$

$(1) \iff (3)$

Suppose $X$ has a countable open cover $\left\{{U_i}\right\}_{i \in \N}$ which does not have a finite subcover.

Thus we can find a sequence $\left \langle {x_n} \right \rangle$ in $X$ such that $\displaystyle x_n \notin \bigcup_{i=1}^n U_i$.

Now every point in $X$ has a neighborhood, i.e. one of the $U_i$ to which it belongs.

But $U_i$ intersects only finitely many points of the sequence $\left \langle {x_n} \right \rangle$.

So the sequence $\left \langle {x_n} \right \rangle$ can have no $\omega$-accumulation point in $X$.

Now suppose that $S \subseteq X$ is a countably infinite subset of $X$ which does not have an $\omega$-accumulation point.

Then each $x \in X$ would have an open neighborhood $U_x$ which intersects at most finitely many points of $S$.

For each finite subset $F$ of $S$, we define $\displaystyle U_F = \bigcup \left\{{U_x: U_x \cap S = F}\right\}$.

Then $\left\{{U_F}\right\}$ is a countable open cover of $X$ such that every finite subset of it includes at most finitely many points of $S$.

Thus no finite subcollection can cover $X$.

$\blacksquare$

Sources

• Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (1970)... (previous)... (next): $\text{I}: \ \S 3$: Global Compactness Properties