The question is pretty much in the title, I need to show that given $X$ is a compact Hausdorff space and $\left\{ A_n\right\}_{n=1}^\infty$ is a collection of closed subsets of $X$ each with empty interior, then $\bigcup_{n\in\mathbb{N}}A_n$ has empty interior as well.
Out of curiosity only I was wondering whether the fact the collection is countable is necessary for the claim or if it's true for any such collection of subsets.
Help would be appreciated 🙂
Best Answer
A space with the property that the union of countably many nowhere dense closed sets has empty interior is called a Baire space.
It's easier if you dualize and show that the intersection of countably many open dense sets $D_n$ is dense. For if $A$ is closed and has empty interior, its complement $D$ is open and dense. Start with an open set $U$ and intersect it with $D_1$. Take a point $x_1$ in the intersection. There is an open neighborhood $U_1$ with compact closure $C_1\subset U\cap D_1$. Then intersect $U_1$ with $D_2$ and repeat this process for a point $x_2$ in $U_1\cap D_2$.