12
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Stefan Hamcke May 24 '15 at 16:50
  • $\begingroup$ @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. $\endgroup$ – Saaqib Mahmood Dec 20 '18 at 21:13
  • $\begingroup$ @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. $\endgroup$ – Saaqib Mahmood Dec 20 '18 at 21:14
  • 1
    $\begingroup$ 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. $\endgroup$ – Nowras May 29 '19 at 14:47
2
$\begingroup$

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.

$\endgroup$

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.