# Does Every infinite Hausdorff space contains a countably infinite discrete subspace?

I found a answer here Every infinite Hausdorff space has an infinite discrete subspace but I don't understand why $$X \setminus \bigcup_{i \in \mathbb{N}} \overline{U_i}$$ open or not doesn't matter in that answer, any help thanks!

• Well, why do you think it does matter? – Eric Wofsey Oct 21 '19 at 1:57

It doesn't matter. The property that is used in the answer is that $$X \setminus \bigcup_{i=0}^n \overline{U_i}$$ is open for each $$n$$. Actually, as shown in the comments to the answer, the set $$X \setminus \bigcup_{i \in \mathbb{N}} \overline{U_i}$$ is not necessarily open.

A different proof: Let $$X$$ be $$T_2$$ and infinite. Let $$X^i$$ be the set of isolated points of $$X$$. That is, $$p\in X^i$$ iff $$\{p\}$$ is open.

(i). Now $$X^i$$ is a discrete subspace of $$X$$. So if $$X^i$$ is infinite then we're done.

(ii). So suppose $$X^i$$ is finite.Then $$X\setminus X^i\ne\emptyset.$$ And because $$X$$ is $$T_2$$ and $$X^i$$ is finite, any nbhd of any $$p\in X\setminus X^i$$ contains some (infinitely many, in fact) $$q$$ with $$p\ne q\in X\setminus X^i.$$

So let $$p_1\in X\setminus X^i$$ and $$U_1=X.$$ Recursively, for $$n\in \Bbb N,$$ suppose $$p_n\in X\setminus X^i,$$ and $$p_n\in U_n$$ where $$U_n$$ is open and $$U_n\cap \{p_j:j

Take $$p_{n+1}\in U_n\setminus X^i$$ with $$p_{n+1}\ne p_n$$ Take disjoint open subsets $$V_n, U_{n+1}$$ of $$U_n$$ with $$p_n\in V_n$$ and $$p_{n+1}\in U_{n+1}.$$

Then $$p_n\not \in \overline {\{p_j:j\ne n\}}$$ because the open set $$V_n$$ contains $$p_n$$ and we have $$V_n\cap \{p_j:j while $$V_n\cap \{p_j:j>n\}\subset V_n\cap U_{n+1}=\emptyset.$$

So $$\{p_n:n\in \Bbb N\}$$ is an infinite discrete subspace of $$X.$$