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!
Best Answer
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.