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

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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 $X$.


Corollary

Let $\left\langle{x_n}\right\rangle_{n \mathop \in \N}$ be an infinite sequence in $S$.


Then $\left\langle{x_n}\right\rangle$ has an accumulation point in $T$.


Proof

Proof by Contradiction.

Aiming for a contradiction, suppose 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.

$\blacksquare$


Also see


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