This is exercise 27.5 in Munkres:
Let $X$ be a compact Hausdorff space; Let $\{A_n\}$ be a countable collection of closed sets of $X$. Show that if each set $A_n$ has empty interior in $X$, then the union $\bigcup A_n$ has empty interior in $X$. [Hint: Imitate the proof of Theorem 27.7.]
Here is Theorem 27.7:
Let $X$ be a nonempty compact Hausdorff space. If $X$ has no isolated points, then $X$ is uncountable.
The proof of Theorem 27.7 boils down to this:
- Show that given any nonempty open set $U$ of $X$ and any $x\in X$, there exists a nonempty open set $V$ contained in $U$ such that $x\notin\overline V$.
- Given $f:\mathbb Z_+\to X$, use 1. to construct a sequence of points $x_n$ all distinct from $x$. It follows that $f$ is not surjective, and hence $X$ is uncountable.
I don't see any relation between the exercise and the proof of the theorem. The theorem constructs a sequence of points, while the exercise requires me to show that a set has empty interior. Any hints on how to relate these two would be greatly appreciated (I'd like to complete the proof myself).
Best Answer
Suppose that $O \subseteq \bigcup_n A_n$ is non-empty open.
$A_1$ has empty interior and is closed, so there is some $O_1$ non-empty open with $\overline{O_1} \subseteq O$ such that $O_1 \cap A_1 = \emptyset$.
$A_1 \cup A_2$ also has empty interior and is still closed, so there is some non-empty open $O_2$ such that $O_2 \subseteq \overline{O_2} \subseteq O_1$ and $O_2 \cap (A_1 \cup A_2)=\emptyset$.
Continue finding $O_n$ this way and note that $\bigcap_n \overline{O_n}$ is non-empty by compactness and see what contradiction you find.
With a minor adaptation at the start, this also works for locally compact Hausdorff spaces.
The common theme with the older proof is the decreasing open sets intersecting to achieve a "countable goal" when finite goals are achievable.