# Prob. 5, Sec. 27 in Munkres' TOPOLOGY, 2nd ed: Every compact Hausdorff space is a Baire space

Here is Prob. 5, Sec. 27, in the book Topology by James R. Munkres, 2nd edition:

Let $$X$$ be a compact Hausdorff space; let $$\left\{ A_n \right\}$$ be a countable collection of closed sets of $$X$$. If each set $$A_n$$ has empty interior in $$X$$, then the union $$\bigcup A_n$$ has empty interior in $$X$$.

How to show this fact? What if we have an uncountable collection of closed sets, each set having empty interior? Does the conclusion still hold?

My effort:

First a preliminary result:

Let $$X$$ be a compact Hausdorff space, and let $$A$$ be a closed subset of $$X$$. If $$U$$ is a non-empty open set in $$X$$ such that $$U \not\subset A$$, then there is a non-empty open set $$V$$ in $$X$$ such that $$\overline{V} \subset U-A$$, that is, $$\overline{V} \subset U$$ and $$\overline{V} \cap A = \emptyset$$.

Am I right?

Proof:

Since $$U \not\subset A$$, the set $$U - A$$ is non-empty. Let $$x \in U-A$$.

Let us put $$B \colon= A \cup (X-U). \tag{Definition 0}$$ Then $$B$$ is closed in $$X$$, and also $$x \not\in B$$.

Now since $$X$$ is compact and since $$B$$ is closed in $$X$$, therefore $$B$$ is also compact (as a subspace of $$X$$), by virtue of Theorem 26.2 in Munkres.

Since $$X$$ is a Hausdorff space, since $$x \in X$$, and since $$B$$ is a compact subspace of $$X$$ such that $$x \not\in B$$, therefore by Lemma 26.4 in Munkres there are open sets $$V$$ and $$W$$ in $$X$$ such that $$x \in V, \qquad B \subset W, \qquad \mbox{ and } \qquad V \cap W = \emptyset. \tag{1}$$ Therefore we have \begin{align} X- W &\subset X-B \qquad \mbox{ [because B \subset W \subset X] } \\ &= X - \big( A \cup (X-U) \big) \qquad \mbox{ [by (Definition 0) above ] } \\ &= (X-A)\cap \big( X-(X-U) \big) \qquad \mbox{ [a DeMorgan's law] } \\ &= (X-A) \cap U \\ & \qquad \mbox{ [the compelement of the complement set U\subset X equals U itself] }\\ &= U-A \qquad \mbox{ [a set-theoretic identity] }, \end{align} that is, $$X-W \subset U-A; \tag{2}$$ and also from (1) we have $$V \subset X-W.$$

Moreover, since $$X-W$$ is closed in $$X$$ and since $$V \subset X - W$$, therefore we can also conclude that $$\overline{V} \subset X-W. \tag{3}$$

Thus from (1), (2), and (3) above we obtain $$x \in V \subset \overline{V} \subset X-W \subset U-A.$$

That is, $$V$$ is a non-empty open set in $$X$$ and $$\overline{V} \subset U-A$$, as required.

Is this proof correct?

Now for the main proof:

Let us put $$A \colon= \bigcup A_n . \tag{A}$$

We show that $$A$$ has empty interior. For this, we show that there is no non-empty open set $$U$$ in $$X$$ such that $$U \subset A$$.

Let $$U$$ be any non-empty open set in $$X$$.

Let us put $$V_0 \colon= U. \tag{0}$$ This is just for notational convenience.

Then since $$A_1$$ has empty interior in $$X$$ and since $$V_0$$ is a non-empty open set in $$X$$, therefore the set $$V_0$$ is not contained in $$A_1$$. So there exists a non-empty open set $$V_1$$ in $$X$$ such that $$\overline{V_1} \subset V_0 -A_1. \tag{1}$$

Again as the set $$A_2$$ has empty interior in $$X$$ and as $$V_1$$ is a non-empty open set in $$X$$, so the set $$V_1 \not\subset A_2$$, and thus there exists a non-empty open set $$V_2$$ in $$X$$ such that $$\overline{V_2} \subset V_1 - A_2. \tag{2}$$

Now suppose that the non-empty open sets $$V_1, \ldots, V_{n-1}$$ (for $$n= 3, 4, 5, \ldots$$) have been chosen such that $$\overline{V_k} \subset V_{k-1} - A_k \ \mbox{ for each } \ k = 1, 2, \ldots, n-1. \tag{3}$$

Now as the set $$A_n$$ has empty interior in $$X$$ and as the set $$V_{n-1}$$ is a non-empty open set in $$X$$, so $$V_{n-1} \not\subset A_n$$, which implies that there exists a non-empty open set $$V_n$$ in $$X$$ such that $$\overline{V_n} \subset V_{n-1} - A_n. \tag{4}$$

From (1) we note that $$\overline{V_1} \subset V_0 \subset \overline{V_0} \qquad \mbox{ and also } \qquad \overline{V_1} \cap A_1 = \emptyset.$$

From (2) we find that $$\overline{V_2} \subset V_1 \subset \overline{V_1} \qquad \mbox{ and also } \qquad \overline{V_2} \cap A_2 = \emptyset.$$

And so on, from (4) we find that $$\overline{V_n} \subset V_{n-1} \subset \overline{V_{n-1}} \qquad \mbox{ and also } \qquad \overline{V_n} \cap A_n = \emptyset$$ for all $$n \in \mathbb{N}$$ such that $$n > 1$$.

Thus we have a sequence
$$\overline{V_0}, \overline{V_1}, \overline{V_2}, \overline{V_3}, \ldots$$ of non-empty closed sets in $$X$$ such that, for each $$n = 1, 2, 3, \ldots$$, we have $$\overline{V_n} \subset \overline{V_{n-1}} \qquad \mbox{ and } \qquad \overline{V_n} \cap A_n = \emptyset. \tag{5}$$ In particular, $$\overline{V_1} \subset U$$, by virtue of (1) and (0) above.

And, if $$\overline{V_{n_1}}, \ldots, \overline{V_{n_k}}$$ are any finitely many of these sets, then we have $$\bigcap_{i=1}^k \overline{V_{n_i}} = \overline{V_{n_0}} ,$$ where $$n_0 \colon= \max\left\{ n_1, \ldots, n_k \right\},$$ and so $$\bigcap_{i=1}^k \overline{V_{n_i}}$$ is non-empty.

Thus the nested sequence $$\left( \overline{V_n} \right)_{n \in \mathbb{N}}$$ of non-empty closed sets of $$X$$ has the finite intersection property, and as $$X$$ is compact, so by Theorem 26.9 in Munkres these sets have a non-empty intersection. That is, $$\bigcap \overline{V_n} \neq \emptyset.$$

Suppose $$x \in \bigcap \overline{V_n}$$. Then $$x$$ is in each set $$\overline{V_n}$$, which implies that $$x \in U$$ and also that $$x$$ is not in any set $$A_n$$, and so $$x \not\in A$$, refer to Def. (A) above. Thus $$x \in U - A$$ and therefore $$U \not\subset A$$.

But $$U$$ was any arbitrarily chosen non-empty open set in $$X$$. Hence $$A = \bigcup A_n$$ has empty interior.

Is my proof correct? If so, is it clear enough in each and every detail? If not, then where is it in need of improvement / correction?

• A space in which any countable union of nowhere dense sets has empty interior is called a Baire space. Any locally compact Hausdorff space is a Baire space. – Stefan Hamcke May 24 '15 at 16:50
• @StefanHamcke I'm really sorry but I'm not yet very comfortable with the notion of local compactness, which is the topic of Sec. 29 in Munkres. – Saaqib Mahmood Dec 20 '18 at 21:13
• @StefanHamcke can you please have a look at my post once again? I've tried to improve readability of the main proof in my original post. – Saaqib Mahmood Dec 20 '18 at 21:14
• There is a problem in the proof of the preliminary result. Lemma 26.4 only applies when $B$ is compact. The correction is simple: Since $X$ is compact and $A, X-U \in X$ are closed subspaces of $X$, then they are both compact and so is their union. The rest of the proof seems correct. – Nowras May 29 '19 at 14:47

Your proof looks fine. You are showing that a compact Hausdorff space is a Baire space, that is a space in which any countable union of nowhere dense sets has empty interior. However, the preceding lemma also holds for the broader class of locally compact Hausdorff space: If $A$ is closed and $U$ is an open set not contained in $A$, then any point $x\in U-A$ has a compact neighborhood $K$ in $U-A$. Since $X$ is Hausdorff, $K$ is closed. Now if $V$ is the interior of $K$, then $\overline V$ is a closed subset of $K$, thus compact. Hence $U-A$ contains a non-empty open set with a compact closure.
In the main proof, you can now alter the sets $V_i$ to be non-empty open sets with compact closure. Then the collection $\dots\subseteq\overline{V_{i+1}}\subseteq\overline{V_i}\subseteq\dots$ has the FIP within $\overline{V_1}$, and thus has a non-empty intersection.