Countably compact space 📖 Wikipedia

(1) ${\displaystyle \Rightarrow }$ (2): Suppose (1) holds and A is an infinite subset of X without ${\displaystyle \omega }$-accumulation point. By taking a subset of A if necessary, we can assume that A is countable. Every ${\displaystyle x\in X}$ has an open neighbourhood ${\displaystyle O_{x}}$ such that ${\displaystyle O_{x}\cap A}$ is finite (possibly empty), since x is not an ω-accumulation point. For every finite subset F of A define ${\displaystyle O_{F}=\cup \{O_{x}:O_{x}\cap A=F\}}$. Every ${\displaystyle O_{x}}$ is a subset of one of the ${\displaystyle O_{F}}$, so the ${\displaystyle O_{F}}$ cover X. Since there are countably many of them, the ${\displaystyle O_{F}}$ form a countable open cover of X. But every ${\displaystyle O_{F}}$ intersect A in a finite subset (namely F), so finitely many of them cannot cover A, let alone X. This contradiction proves (2).

(2) ${\displaystyle \Rightarrow }$ (3): Suppose (2) holds, and let ${\displaystyle (x_{n})_{n}}$ be a sequence in X. If the sequence has a value x that occurs infinitely many times, that value is an accumulation point of the sequence. Otherwise, every value in the sequence occurs only finitely many times and the set ${\displaystyle A=\{x_{n}:n\in \mathbb {N} \}}$ is infinite and so has an ω-accumulation point x. That x is then an accumulation point of the sequence, as is easily checked.

(3) ${\displaystyle \Rightarrow }$ (1): Suppose (3) holds and ${\displaystyle \{O_{n}:n\in \mathbb {N} \}}$ is a countable open cover without a finite subcover. Then for each ${\displaystyle n}$ we can choose a point ${\displaystyle x_{n}\in X}$ that is not in ${\displaystyle \cup _{i=1}^{n}O_{i}}$. The sequence ${\displaystyle (x_{n})_{n}}$ has an accumulation point x and that x is in some ${\displaystyle O_{k}}$. But then ${\displaystyle O_{k}}$ is a neighborhood of x that does not contain any of the ${\displaystyle x_{n}}$ with ${\displaystyle n>k}$, so x is not an accumulation point of the sequence after all. This contradiction proves (1).

(4) ${\displaystyle \Leftrightarrow }$ (1): Conditions (1) and (4) are easily seen to be equivalent by taking complements.