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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.

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  • $\begingroup$ Using images instead of MathJax makes it much more difficult for others to search for this question. I have transcribed the images, but please check to make sure I have not made any errors. Also including information about the source of those images (book, website, author(s), etc.) would aid others who might have the same question from the same source. $\endgroup$ – бір-төрт-төрт-үш-жеті-бес Jul 24 '18 at 9:46
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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 $X$.
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