Certain infinite subsets of $k_\omega$-spaces have an infinite closed discrete subspace

Question

A Hausdorff space $( X , \mathcal{T} )$ is said to be a $k_\omega$-space if there is a countable collection $X_n$, $n \in \mathbb{N}$ of compact subsets of $X$ such that

1. $X_n \subseteq X_{n+1}$, for all $n$,
2. $X = \bigcup_{n=1}^\infty X_n$,
3. any subset $A$ of $X$ is closed if and only if $A \cap X_n$ is compact for each $n \in \mathbb{N}$.

If $S$ is an infinite subset of the $k_\omega$-space $( X , \mathcal{T} )$, such that $S$ is not contained in any $X_n$, $n \in \mathbb{N}$, then $S$ has an infinite discrete closed subspace.

I notice that in order to prove this, it suffices to show that there exists collection of open sets $U_n \subset X_n$ but $U_n \not\subset X_{n-1}$. However, I can not construct such collection of open sets by just the properties stated above like Hausdorff and $k_\omega$-space. Can someone give me a hint?

The above information is from the Ebook, Topology Without Tears by Sidney A. Morris around Page 290.

Answer 1

If $S$ is infinite and not contained in any single $X_{n}$, then the set $\{ n \in \mathbb{N} : S \cap (X_n \setminus X_{n-1} ) \neq \emptyset \}$ is infinite, and we may recursively pick $x_1 , x_2 , \ldots \in S$ such that for each $k \in \mathbb{N}$ there is an $n \in \mathbb{N}$ such that $S \cap X_n = \{ x_1 , \ldots , x_k \}$. (Let $n_1 = \min \{ n \in \mathbb{N} : S \cap X_n \neq \emptyset \}$ and pick $x_1 \in S \cap X_{n_1}$; given appropriate $n_1 < \cdots < n_{k-1}$ let $n_k = \min \{ n \in \mathbb{N} : S \cap ( X_n \setminus X_{n_{k-1}} ) \neq \emptyset \}$ and pick $x_k \in S \cap (X_{n_k} \setminus X_{n_{k-1}} )$.)

I claim that $A = \{ x_1 , x_2 , \ldots \}$ is closed discrete.

• Note that as the $X_n$ are increasing, by our choice of the $x_k$ it follows that $A \cap X_n$ is finite, and hence compact, for each $n$, meaning that $A$ is closed in $X$.
• In fact, using the same reasoning as above, all subsets of $A$ are closed in $X$. In particular, for each $k \in \mathbb{N}$ as $A \setminus \{ x_k \}$ is closed in $X$, then $X \setminus ( A \setminus \{ x_k \} ) = \{ x_k \} \cup ( X \setminus A )$ is an open neighborhood of $x_k$ whose intersection with $A$ is $\{ x_k \}$. Thus $A$ is a discrete subset of $X$.