I was playing with Baire's Theorem, and seemed to deduce the following:
In a complete metric space $X$ that has no isolated points, any countable intersection of open dense sets is uncountable.
Proof:
Let $S$ be the countable intersection of open dense sets $\{U_n\}$. By Baire's Theorem $S$ is dense thus $U_n$ are open and dense. This implies that $U_n^c$ are nowhere dense.
$$X = S\cup S^c = S\cup (\bigcap_n U_n)^c = S\cup \bigcup_n U_n^c.$$ If $S$ were countable, then $X = \bigcup_{s\in S}\{s\} \cup \bigcup_n U_n^c$. Since $X$ has no isolated points, every singleton set is closed and has empty interior, thus nowhere dense. We have just written $X$ has a countable union of nowhere dense sets. A contradiction to Baire's corollary.
As a corollary, any connected complete metric space, such as $\mathbb{R}$, must be uncountable (this gives a topological proof that $\mathbb{R}$ is uncountable). Also $\mathbb{Q}$ is not the intersection of open dense sets of $\mathbb{R}$.
I am a little bit suspicious about my proof, is it correct?
Best Answer
It seems like a more obvious proof is to avoid the corollary and use the theorem directly:
If $S=\{s_1,s_2,\dots\}$ is countable, define $V_n=X\setminus \{s_n\}$, which are dense and open since $X$ has no isolated points. Then $\emptyset = \bigcap V_n\cap\bigcap U_n$ is the the countable intersection of dense open sets.
This is essentially the dual of your proof.