Suppose that $X$ is a compact Hausdorff space. Let $\{X_i\}_{i \in \mathbb{N}}$ be a sequence of dense open subsets of $X$. Let $E$ be a closed subset of $X$. Can it happen that $\left(\cap_{i \in \mathbb{N}}X_i\right)$ has empty intersection with $E$?

First by Baire Category theorem, $\left(\cap_{i \in \mathbb{N}}X_i\right)$ is dense in $X$. Moreover, since $E$ is closed, it is compact. If the intersection were empty, then $E \subset \cup_{i \in \mathbb{N}}X\setminus X_i$. But, this doesn't yield any contradiction.

A hint would be greatly appreciated.

Thanks for the help!!


Yes, the intersection can be empty. Suppose that $X=[0,1]$, with its usual topology. Take $E=\{0\}$ and $X_n=(0,1]$, for each natural $n$.

  • $\begingroup$ You can also take a Cantor set $C$ and $X_n = [0,1] \setminus C$. Then $C$ is closed and uncountable! Since $C$ is nowhere dense, $X_n$ is dense. $\endgroup$ – p4sch Nov 21 '18 at 20:05

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