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Let $X$ be a complete metric space. Let $E_n$ be a nowhere dense subset of $X$ for every $n$. Let $M_n$ be a dense open subset of $X$ for every $n$.

Show that $\bigcap_n M_n$ is not contained in $\bigcup_n E_n$.

My attempt:

Since $X$ is a complete metric space, then $X$ is a Baire space. Hence $\bigcap_n M_n$ is dense in $X$ and $\bigcup_n E_n$ has empty interior.

However, this is not enough to show the statement. There must be something missing but I can't find out.

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user430647
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1 Answers1

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Suppose $\bigcap_nM_n\subset \bigcup_nE_n$. Let $U_n$ be the complementary subset of $E_n$, it is a dense open subset. The family of open subsets $U_m,M_n$ is numerable. Baire theorem implies that $\bigcap_{n,m}M_n\cap U_m$ is dense, but $\bigcap_{n,m}M_n\cap U_m\subset (\bigcup_nE_n)\cap_mU_m=\bigcup_n(E_n\cap_mU_m) $ is empty. Contradiction.

Tsemo Aristide
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  • If $E_n=\mathbb{Z}/n$, then $E_n$ is nowhere dense. $\bigcup_n E_n$ has empty interior. But that doesn't mean the interior of its complementary subset is not empty. Actually the complementary of $\bigcup_n E_n$ is $\mathbb{R}\backslash \mathbb{Q}$ and has empty interior. – user430647 May 14 '17 at 21:42