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.


1 Answer 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!

  • 1
    $\begingroup$ How do you know that doing this same process infinitely many times will give you open sets. I mean $X- \bigcup\overline{U}_i$ is not necessarily open, or is it? $\endgroup$
    – user419934
    May 17, 2017 at 21:39
  • $\begingroup$ @3927 It doesn't matter. In fact, if $X$ happens to be limit-point compact, the set difference $X \setminus \bigcup_{i \in \mathbb{N}} \overline{U_i}$ will not be open. But we're only looking to find an infinite subset of $X$ which as a subspace is discrete. (I know that some authors use "discrete" to mean something stronger, something which is called "closed discrete" elsewhere. But an infinite Hausdorff space may not have an infinite closed discrete subset. For example the unit interval $[0,1]$.) $\endgroup$
    – user642796
    May 18, 2017 at 3:34
  • $\begingroup$ Can this argument be tweaked for finite cases (the checked answer)? $\endgroup$
    – user837396
    Feb 7, 2021 at 2:06

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