Every infinite Hausdorff space has an infinite discrete subspace

I want to show that any infinite Hausdorff space contains an infinite discrete subspace.

I am motivated by the role of $\mathbb N$ in $\mathbb R$. We know that if a Hausdorff space is finite, then it is a discrete space, but an infinite subspace of a Hausdorff space is obviously not necessarily discrete.

Solution 1:

This seems to be an old question, so I'll give a pretty full answer.

Suppose that $X$ is an infinite Hausdorff space. We inductively pick nonempty open sets $\{ U_i : i \in \mathbb{N} \}$ such that $X \setminus \bigcup_{i=0}^n \overline{U_i}$ is infinite, and $U_{n+1} \subseteq X \setminus \bigcup_{i=0}^n \overline{U}_i$. If we can do this, then choosing $x_i \in U_i$ will give us an infinite discrete subset of $X$.

Note that we only need to show that given any infinite Hausdorff space $X$ there is a nonempty open $U \subseteq X$ such that $X \setminus \overline{U}$ is infinite, since in the next step we consider the open subspace $Y = X \setminus \overline{U}$ of $X$. (If $V \subseteq Y$ is open in $Y$,then by openness of $Y$ in $X$ it follows that $V$ is open in $X$. If $Y \setminus \mathrm{cl}_Y (V)$ is infinite, then as $\mathrm{cl}_Y ( V ) = \overline{V} \cap Y$ it follows that $X \setminus ( \overline{U} \cup \overline{V} )$ is infinite.)

Well, suppose not. Then take any nonempty open $U \subseteq X$ such that $\overline{U} \neq X$ (such must exist by Hausdorffness). It follows that the subspace $X \setminus \overline{U}$ is finite and T$_1$ and so it is discrete. But then for any $x \in X \setminus \overline{U}$ we have that $U_0 = \{ x \}$ is open in $X \setminus \overline{U}$, and since this is an open subspace of $X$ it is also open in $X$. By T$_1$-ness $U_0$ is also closed in $X$, and so $X \setminus \overline{U_0} = X \setminus \{ x \}$ is infinite!