Show that in a compact Hausdorff space, any countable collection of dense open sets has dense intersection.

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Here is this week's POTW:

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Show that in a compact Hausdorff space, any countable collection of dense open sets has dense intersection.

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Congratulations to Opalg for his correct solution, which you can read below:

Spoiler

This is one form of the Baire category theorem. The proof relies on the fact that a compact Hausdorff space $X$ is regular, meaning that if $U$ is an open subset of $X$ and $x\in U$ then there is an open subset $V$ such that $x\in V\subseteq \overline{V} \subseteq U$, where $ \overline{V}$ is the closure of $V$.

Suppose that $\{U_n\}_{n\in\Bbb N}$ is a countable collection of dense open subsets of $X$. Let $W$ be an open subset of $X$. Then we want to show that \(\displaystyle W\cap \bigcap_{n\in\Bbb N}U_n \ne\emptyset\).

Since $U_1$ is dense in $X$, $W\cap U_1\ne\emptyset$. So there exists a nonempty open set $V_1$ such that $\overline V_1 \subseteq W\cap U_1$.

Since $U_2$ is dense in $X$, $U_2\cap V_1\ne\emptyset$. So there exists a nonempty open set $V_2$ such that $\overline V_2 \subseteq V_1\cap U_2$.

Continuing in this way, we get a decreasing sequence of nonempty open sets $\{V_n\}_{n\in\Bbb N}$ such that \(\displaystyle \overline V_n \subseteq W\cap \bigcap_{k=1}^n U_k\). The sets $\overline V_n$ are compact and satisfy the finite intersection property. So the intersection \(\displaystyle \bigcap_{n\in\Bbb N}\overline V_n\) is nonempty and contained in \(\displaystyle W\cap \bigcap_{n\in\Bbb N}U_n\). That completes the proof.