Let $X$ be a compact Hausdorff space and let $\{U_n\}$ be a countable collection of subsets that are open and dense in $X$. Show that the intersection $$\bigcap\limits_{n=1}^\infty U_n$$ is dense.
I tried to show that the closure of this intersection is equal to the intersection of the closures of each set, but I'm not getting anywhere.
Any help would be greatly appreciated.
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$\begingroup$Let $X$ be a compact and Hausdorff space.
Let $\{U_n\}$ be a countable collection of subsets that are open and dense in X.
Let $O$ be an open set. Since $U_1$ is dense, it must intersect $U_1$ in a non-empty open set $O_1$. Let $x_1 \in O_1$ , and since the space is Hausdorff and compact exists $B_1(x_1)$ a open set that contains $x_1$ and $\overline{B_{1}(x_1)}$ is contained in $O_1$. Now, because $U_2$ is dense, intersects $B_{1}(x_1)$ in a non-empty open set $O_2$. In the same way, let $x_2 \in O_2$ , and take $B_2(x_2)$ such that the closure $\overline{B_{2}(x_2)}$ is contained in $O_2$.
With this process we obtained a nested sequence of non-empty closed sets $\overline{B_1} \supseteq \overline{B_2} \supseteq \overline{B_3} \supseteq ... \supseteq \overline{B_n} \supseteq ... $ . As $X$ is compact, then exists $ x \in \bigcap\limits_{n=1}^{\infty} \overline{B_n} $. In this way, $x \in O_n$, for each $n$ and hence $x \in O \cap \bigcap\limits_{n=1}^{\infty} {U_n}$. Thus $\bigcap\limits_{n=1}^{\infty} {U_n}$ intersects each non-empty set $O$ in a least one point, that is precisely, that $\bigcap\limits_{n=1}^{\infty} {U_n}$ is dense in $X$.
$\endgroup$ 1 $\begingroup$What you are looking for is Baire's theorem. (The proof on Wikipedia is for a complete metric space, but the proof is similar for a compact Hausdorff space.)
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