Countably Infinite Set in Countably Compact Space has Omega-Accumulation Point

Theorem
Let $\left({X, \tau}\right)$ be a countably compact topological space.

Let $A \subseteq X$ be countably infinite.

Then $A$ has an $\omega$-accumulation point in $X$.

Proof
Proof by Contradiction.

that $A$ does not have an $\omega$-accumulation point in $X$.

Let $\mathcal S \subseteq \mathcal P \left({A}\right)$ be the set of all finite subsets of $A$.

By Set of Finite Subsets of Countable Set is Countable, we have that $\mathcal S$ is countable.

For all (finite) $F \in \mathcal S$, define:
 * $U_F = \left({F \cup \left({X \setminus A}\right)}\right)^{\circ}$

where $^{\circ}$ denotes the interior.

By Image of Countable Set under Mapping is Countable, we have that $\mathcal C = \left\{{U_F: F \in \mathcal S}\right\}$ is countable.

By the definition of an $\omega$-accumulation point, it follows that:
 * $\forall x \in X: \exists U \in \tau: x \in U: \exists F \in \mathcal S: U \cap A = F$

By Set Difference Union Intersection, we have:
 * $U = \left({U \cap A}\right) \cup \left({U \setminus A}\right) \subseteq F \cup \left({X \setminus A}\right)$

By Set Interior is Largest Open Set, it follows that $\mathcal C$ is a countable open cover for $X$.

Hence, by the definition of a countably compact space, there exists a finite subcover $\mathcal C'$ of $\mathcal C$ for $X$.

By the Principle of Finite Choice, there exists an indexed family $\left\langle{G_V}\right\rangle_{V \in \mathcal C'}$ of elements of $\mathcal S$ such that:
 * $\forall V \in \mathcal C': V = U_{G_V} = \left({G_V \cup \left({X \setminus A}\right)}\right)^{\circ}$

Define:
 * $\displaystyle B = \bigcup_{V \mathop \in \mathcal C'} G_V \subseteq A$

By Union of Finite Sets is Finite, it follows that $B$ is finite.

We can use the definition of a cover to conclude that:
 * $X \subseteq B \cup \left({X \setminus A}\right)$

By Relative Complement inverts Subsets, it follows that:
 * $\left({X \setminus B}\right) \cap A = \varnothing$

Hence, $A \subseteq B$.

But then by Subset of Finite Set is Finite, it follows that $A$ is finite, a contradiction.

Also see

 * Infinite Set in Compact Space has $\omega$-Accumulation Point