# Is X is a nonempty and Hausdorff space a sufficient condition to prove that if X has no isolated points, X is uncountable?

I'm studying Munkres' Topology now and came across this theorem:

Let X be a nonempty compact Hausdorff space. If X has no isolated points, then X is uncountable.

I was trying to prove it for myself and the proof I came up with only required the Hausdorff condition, so I was hoping if someone could point out where I went wrong.

Proof: Suppose X is nonempty and Hausdorff. Suppose X is countable, we can index the points in X. Let $$x_0 \in U_1$$, $$x_1 \in V_1$$, where $$U_1$$ and $$V_1$$ are disjoint open sets such that $$U_1 \cap V_1 = \emptyset$$, which exist as X is Hausdorff.

For all points in X, let $$x_n \in U_n$$, $$x_1 \in V_n$$ such that $$U_n \cap V_n = \emptyset$$. Then, let V = $$\bigcap_{i \in \mathbb{N}} V_i$$. As $$x_1 \in V_n$$ and $$x_n \notin V_n$$ for all n, V = {$$x_1$$}, implying that $$x_1$$ is an isolated point as V is an open set.

Thus, by contrapositive, if X has no isolated points, X is uncountable.

Thank you!

• Why should $V$ be open?
– Ulli
Jan 19 at 14:31
• Infinite intersections of open sets need not be open. Jan 19 at 14:31
• Interestingly, you seem to have been under the impression that countable intersections of open sets are open but uncountable ones need not be (as otherwise this proof would have proved not that $X$ is uncountable but that in any Hausdorff space whatsoever all points are isolated). Jan 19 at 15:01

As pointed out in the comment, the problem is $$V$$ as an infinite intersection of open sets is not necessarily open.
If compactness is dropped, there is a clear counter-example: $$\mathbb Q$$ with the usual topology induced by the metric $$d(a,b)=|a-b|$$.
To solve this problem, it's very similar to how one can show the Cantor set is uncountable. That is to pick two points and separate them by two non-intersecting compact neighborhoods, and for each of them, we can further divide. That is, for each finite $$0$$-$$1$$ sequence $$a_1a_2\cdots a_n$$, we have a corresonding non-empty subset $$V_{a_1a_2\cdots a_n}$$, where e.g. $$V_0$$ is the first compact neighborhood, and $$V_{00}$$ and $$V_{01}$$ are the further divisions of $$V_0$$, etc. Now we can show that for each infinite sequence $$a_1a_2\cdots$$, we have $$\cap_{n=1}^\infty V_{a_1\cdots a_n}$$ is non-empty, and $$\cap_{n}V_{b_1\cdots b_n}\cap_n V_{a_1\cdots a_n}$$ is empty unless $$a=b$$, since if $$a_n\not=b_n$$, then $$V_{a_1\cdots a_n}\cap V_{b_1\cdots b_n}=\emptyset$$.
Another way is to recognize locally compact Hausdorff spaces are all Baire spaces, i.e. countable intersection of open dense sets is still dense. If $$X=\{x_1, x_2, \cdots\}$$ is countable, since it doesn't contain any isolated point, $$X\setminus\{x_1, \cdots, x_n\}$$ is still dense (and open by Hausdorff-ness), and eventually we get $$\emptyset$$ is dense by Baire-ness.