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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?

Saaqib Mahmood
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    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
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    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

1 Answers1

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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.

Stefan Hamcke
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