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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:

  1. 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 $x\notin\overline V$.
  2. 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).

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    $\begingroup$ To prove the exercise, you wish to show that given any non-empty open set $U$, $U$ contains a point outside $\bigcup A_n$. In the same manner as the proof of theorem 27.7, you wish to construct a decreasing sequence of open sets which avoid $A_n$. $\endgroup$ – lc2r43 Nov 17 '19 at 5:44
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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.

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  • $\begingroup$ Thanks, but where exactly did you use Hausdorffness? $\endgroup$ – Math1000 Nov 17 '19 at 23:52
  • $\begingroup$ I think I see it: Since $\bigcap_{n=1}^\infty \overline O_n$ and $\bigcup_{n=1}^\infty A_n$ are disjoint, there exist points $x\in\bigcap_{n=1}^\infty \overline O_n$ and $y\in\bigcup_{n=1}^\infty A_n$ with disjoint neighborhoods $U$ and $V$. But this means that $V\subset \bigcup_{n=1}^\infty A_n$, contradicting the assumption that $\bigcup_{n=1}^\infty A_n$ has empty interior. Is that correct? $\endgroup$ – Math1000 Nov 18 '19 at 0:13
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    $\begingroup$ @Math1000 I use it when I apply regularity in the shrinking (closures) of open sets. The contradiction lies in the fact that $x\in U$ but $x\notin \bigcup_n A_n$ despite the initial assumptions of non-empty interior (which gave us $O$ in the first place). $\endgroup$ – Henno Brandsma Nov 18 '19 at 5:11
  • $\begingroup$ I didn't see a $U$ in your post. Are you referring to the $U$ in my comment? $\endgroup$ – Math1000 Nov 18 '19 at 18:10
  • $\begingroup$ And as for "regularity," $X$ is regular because it is compact Hausdorff, correct? Sorry, I have not yet studied regular spaces. $\endgroup$ – Math1000 Nov 18 '19 at 18:13

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