Equivalence of Definitions of Countably Compact Space

Theorem
Let $T = \left({S, \tau}\right)$ be a topological space.

Then the following statements are equivalent:
 * $(1): \quad$ $T$ is countably compact, i.e. every countable open cover for $S$ 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, $T$ satisfies the countable finite intersection axiom.


 * $(3): \quad$ Every infinite sequence in $S$ has an accumulation point in $S$.


 * $(4): \quad$ Every countably infinite subset of $S$ has an $\omega$-accumulation point in $S$.


 * $(5): \quad$ Every infinite subset of $S$ has an $\omega$-accumulation point in $S$.

$(1) \iff (2)$
This is proven in Countably Compact Space satisfies Countable Finite Intersection Axiom.

$(1) \implies (4)$
This is proven in Countably Infinite Set in Countably Compact Space has $\omega$-Accumulation Point.

$(4) \implies (3)$
Let $\left\langle{x_n}\right\rangle_{n \in \N}$ be an infinite sequence in $S$.

Let $A \subseteq S$ be the range of $\left\langle{x_n}\right\rangle$.

If $A$ is finite, then consider the equality:
 * $\displaystyle \N = \bigcup_{y \mathop \in A} \left\{{n \in \N: x_n = y}\right\}$

Therefore, there exists a $y \in A$ such that $\left\{{n \in \N: x_n = y}\right\}$ is an infinite set.

Hence, $y$ is an accumulation point of $\left\langle{x_n}\right\rangle$.

Otherwise, $A$ is countably infinite.

By assumption, $A$ has an $\omega$-accumulation point $x \in S$.

It follows that $x$ is an accumulation point of $\left\langle{x_n}\right\rangle$.

$(3) \implies (1)$
We use a Proof by Contradiction.

Suppose $S$ has a countably infinite open cover $\left\{{U_n: n \in \N}\right\}$ which does not have a finite subcover.

Then, using the axiom of countable choice, we can obtain a sequence $\left \langle {x_n} \right \rangle_{n \in \N}$ in $S$ such that:
 * $\displaystyle \forall n \in \N: x_n \notin \bigcup_{k \mathop = 0}^n U_k$.

Let $x \in S$.

Then, by the definition of a cover, there exists an $n \in \N$ such that $x \in U_n$.

By construction, $U_n \subseteq \left\{{x_0, x_1, \ldots, x_{n-1}}\right\}$.

Hence, $x$ is not an accumulation point of $\left \langle {x_n} \right \rangle$.

$(4) \implies (5)$
Follows directly from Infinite Set has Countably Infinite Subset and the definition of an $\omega$-accumulation point.