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).

  • 1
    $\begingroup$ 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

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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.