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.

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?

  • $\begingroup$ @Mathmo123 No, he's talking about complements, not closure. $\endgroup$ – Thomas Andrews Aug 4 '14 at 22:11
  • $\begingroup$ It's correct, but the argument that $U_n^c$ is closed and has empty interior is quite roundabout. That is just the statement that $U_n$ is dense and open. $\endgroup$ – Daniel Fischer Aug 4 '14 at 22:18

It seems like a more obvious proof is to avoid the corollary and use the theorem directly:

If $x$ is not an isolated point, then $X\setminus\{x\}$ is dense and open in $X$.

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.

  • $\begingroup$ Under the assumption that $X$ has no isolated points. $\endgroup$ – mez Aug 4 '14 at 22:36
  • $\begingroup$ Which was in the question, yes. @mez $\endgroup$ – Thomas Andrews Aug 4 '14 at 22:36
  • 1
    $\begingroup$ Yea. Just to warn the reader that the lemma in your box is not true in general. $\endgroup$ – mez Aug 4 '14 at 22:37
  • $\begingroup$ Changed it to use the second argument, which is easier anyway. @mez $\endgroup$ – Thomas Andrews Aug 4 '14 at 22:40
  • $\begingroup$ The argument as given by Thomas is a nice way for showing that $\mathbb{Q}$ cannot be a $G_{\delta}$ set. $\endgroup$ – akech Oct 20 '14 at 2:36

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