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Consider the following theorem known as Baire's Category Theorem (BCT).

Theorem.[BCT] A non-empty complete metric space $X$ is not a countable union of nowhere dense sets.

I am interested on how to prove the following using BCT.

The intersection of every countable collection of dense open subsets of a complete metric space is dense.

My draft: Let $(A_k)$ be a sequence of dense open sets in $X$. Then for each $k$, we get $$ A_k \mbox{ is dense }\Leftrightarrow \overline{A_k}=X \Leftrightarrow \mbox{int }A_{k}^c=\varnothing. $$ Since $A_k$ is also open, we get $$ \mbox{int }\overline{A_{k}^c}=\mbox{int }A_{k}^c=\varnothing $$ and so $A_{k}^c$ is nowhere dense. We now consider the set $$\bigcup_{k=1}^{\infty}A_{k}^c.\qquad \qquad \qquad(*)$$ I am convinced that upon taking the complement of (*), the resulting set must be dense. But I don't know how to verify that step. Any tips?

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2 Answers 2

up vote 3 down vote accepted

Let me give you some extra tools to use, here. Some details are left to you to verify.


Definitions: A subset $A$ of a topological space $X$ is said to be nowhere-dense (in $X$) if for every non-empty open subset $U$ of $X$ there is a non-empty open subset $V$ of $X$ such that $V\subseteq U$ and $V\cap A=\emptyset$. (You should be able to prove that this is equivalent to your definition of nowhere-dense.) A topological space $X$ is said to be a Baire space if every countable union of nowhere-dense subsets of $X$ has empty interior.

Lemma 1: $X$ is a Baire space if and only if every countable intersection of open dense subsets of $X$ is dense in $X$.

Proof: Suppose $X$ is a Baire space and $\{G_n\}_{n=1}^\infty$ is a countable collection of open dense sets, so each $X\setminus G_n$ is nowhere dense. (Why?) Thus, for any non-empty open set $U$, we have that $$U\nsubseteq\bigcup_{n=1}^\infty (X\setminus G_n)=X\setminus\left(\bigcap_{n=1}^\infty G_n\right),$$ so there must be some point of $\bigcap_{n=1}^\infty G_n$ in $U$. Thus, $\bigcap_{n=1}^\infty G_n$ is dense.

Suppose countable unions of open dense sets are dense, and let $A_n$ be nowhere dense for each $n$, so each $X\setminus\overline{A_n}$ is open and dense. (Why?) Take any non-empty open $U$, so that $U$ intersects $$\bigcap_{n=1}^\infty\left(X\setminus\overline{A_n}\right)=X\setminus\left(\bigcup_{n=1}^\infty\overline{A_n}\right)\subseteq X\setminus\left(\bigcup_{n=1}^\infty A_n\right),$$ so cannot be a subset of $\bigcup_{n=1}^\infty A_n$. Thus, $X$ is a Baire space. $\Box$

Lemma 2: $X$ is a Baire space if and only if every countable union of closed nowhere-dense subsets of $X$ has empty interior. (I leave the proof to you. Lemma 1 and DeMorgan's Laws should help.)

Lemma 3: If $X$ is a non-empty complete metric space with metric $d:X\times X\to\Bbb R$, and $F$ is a non-empty closed subset of $X$, then $F$ is a complete metric space with metric $d_F$ defined by $d_F(x,y)=d(x,y)$ for all $x,y\in F$. (I leave the proof of this to you.)


Proposition: Assuming BCT holds, every non-empty complete metric space is a Baire space.

Proof: Suppose by way of contradiction that $X$ is a non-empty complete metric space, but isn't a Baire space. Then by Lemma 2, there is some countable collection $\{A_n\}_{n=1}^\infty$ of closed nowhere-dense subsets of $X$ whose union doesn't have empty interior--meaning there is some non-empty open $U$ such that $$U\subseteq\bigcup_{n=1}^\infty A_n.$$ Now, in particular, we can take a non-empty open set $V$ such that $\overline V\subseteq U.$ (Why?) Then $$\overline V\subseteq\bigcup_{n=1}^\infty A_n,$$ so $$\overline V=\overline V\cap\left(\bigcup_{n=1}^\infty A_n\right)=\bigcup_{n=1}^\infty(\overline V\cap A_n).\tag{#}$$ But $\overline V$ is a non-empty complete metric space by Lemma 3, so by BCT, $\overline V$ is not a countable union of closed nowhere-dense subsets of $\overline V$. But each $\overline V\cap A_n$ is closed and nowhere-dense in $\overline V$ (why?), so $(\#)$ gives us the desired contradiction. $\Box$

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I prefer the statement of BCT as "every completely metrisable space and every locally compact Hausdorff space is a Baire space". Of course that forces a rather large change in how the results are presented. –  kahen Mar 7 '13 at 21:23
    
@kahen: As it turns out, "every completely metrizable space is a Baire space" is equivalent (in ZF, at least) to the Principle of Dependent Choices. This, in turn implies the Principle of Dependent Multiple Choices, which is equivalent (in ZF) to "every locally compact Hausdorff space is Baire." In ZF, then, the bit about locally compact Hausdorff spaces is redundant to the statement of BCT. Still, those equivalence results take a good bit of finagling to prove, so it's nicer to just include the locally compact Hausdorff bit for good measure. –  Cameron Buie Mar 7 '13 at 21:53
    
@CameronBuie, thank you very much.:)Your answer is more than I expected. –  juniven Mar 7 '13 at 22:57

Hint #1. The following three properties of a set $E\subset X$ are equivalent:

  • $E^c$ is not dense
  • $E$ contains an open ball (of positive radius)
  • $E$ contains a closed ball (of positive radius)

Hint #2. A closed subset of a complete metric space is also a complete metric space.

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1  
+1: Nice summary. –  copper.hat Mar 7 '13 at 19:44
    
@5pm, thanks for pointing out the summary. Its nice.:) –  juniven Mar 7 '13 at 23:02

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