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[SOLVED] Version of Baire Category Theorem

joypav

Active member
Mar 21, 2017
151
Problem:
If X is a locally compact Hausdorff space, then X is not the union of countably many nowhere dense sets.


I've tried working on this problem a couple times and I always seem to get nowhere or go in a circle.
In class we have not yet mentioned or learned anything about Baire spaces. So I am not sure if I am supposed to use the property of a Baire space (countable union of open dense sets their intersection is dense).

Our professor encouraged us to use the following...
$\cdot$ If X is compact and G is a monotonic collection of closed subsets of X, then $\cap_{g \in G} g \neq \emptyset$.
$\cdot$ M is nowhere dense in X iff $X-\overline{M}$ is a dense open subset of X.

I would really appreciate a push in the right direction or idea for the proof.
 

Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,890
Hi joypav ,

Suppose $\{E_n\}_{n = 1}^\infty$ is a sequence of nowhere dense sets in $X$. For each $n$, $X - \overline{E_n}$ is a dense open set. If you can show that the intersection of all $X - \overline{E_n}$ is dense, it follows that this intersection is nonempty, and consequently $X$ is not the union of the $E_n$.

Set $S = \bigcap\limits_{n = 1}^\infty (X - \overline{E_n})$, and fix a nonempty open set $U\subset X$. Since $X - \overline{E_1}$ is dense, there exists $x_1\in U \cap (X - \overline{E_1})$. As $X$ is locally compact Hausdorff, there is an open neighborhood $O_1\ni x_1$ such that $\overline{O_1}$ is compact and $\overline{O_1}\subset U\cap (X - \overline{E_1})$. Since $O_1$ is open and $X - \overline{E_2}$ is dense, there exists $x_2\in O_1 \cap (X - \overline{E_2})$, and by the locally compact Hausdorff condition, there corresponds an open neighborhood $O_2 \ni x_2$ with $\overline{O_2}$ compact and $\overline{O_2}\subset O_1 \cap (X - \overline{E_2})$. Continue the process to extract an infinite sequence of open sets $\{O_n\}_{n = 1}^\infty$ having the property $$\overline{O_{n+1}} \subset O_n \cap (X - \overline{E_{n+1}})\quad (n = 1,2,3,\ldots)$$
See what you can do with this.
 

joypav

Active member
Mar 21, 2017
151
Hi joypav ,

Suppose $\{E_n\}_{n = 1}^\infty$ is a sequence of nowhere dense sets in $X$. For each $n$, $X - \overline{E_n}$ is a dense open set. If you can show that the intersection of all $X - \overline{E_n}$ is dense, it follows that this intersection is nonempty, and consequently $X$ is not the union of the $E_n$.

Set $S = \bigcap\limits_{n = 1}^\infty (X - \overline{E_n})$, and fix a nonempty open set $U\subset X$. Since $X - \overline{E_1}$ is dense, there exists $x_1\in U \cap (X - \overline{E_1})$. As $X$ is locally compact Hausdorff, there is an open neighborhood $O_1\ni x_1$ such that $\overline{O_1}$ is compact and $\overline{O_1}\subset U\cap (X - \overline{E_1})$. Since $O_1$ is open and $X - \overline{E_2}$ is dense, there exists $x_2\in O_1 \cap (X - \overline{E_2})$, and by the locally compact Hausdorff condition, there corresponds an open neighborhood $O_2 \ni x_2$ with $\overline{O_2}$ compact and $\overline{O_2}\subset O_1 \cap (X - \overline{E_2})$. Continue the process to extract an infinite sequence of open sets $\{O_n\}_{n = 1}^\infty$ having the property $$\overline{O_{n+1}} \subset O_n \cap (X - \overline{E_{n+1}})\quad (n = 1,2,3,\ldots)$$
See what you can do with this.
Sorry for the delay. I've been very busy with midterms and grading! A deadly combo...

From what you've given...
We have an infinite sequence of open sets $\{O_n\}_{n = 1}^\infty$ having the property $\overline{O_{n+1}} \subset O_n \cap (X - \overline{E_{n+1}})\quad (n = 1,2,3,\ldots)$.

Then, we have $\overline{O_{n+1}} \subset O_n \cap (X - \overline{E_{n+1}}) \subset O_n \subset \overline{O_n}$

$\implies$ we obtain a nested sequence of nonempty closed set
$\overline{O_1} \supset \overline{O_2} \supset ...$

(By our info given) $\implies \exists x \in \cap_{n\ge1}\overline{O_n}$

$\implies x \in O_n \cap (X-\overline{E_{n+1}})$ for every n
$\implies x \in \cap_{n\ge1}(X-\overline{E_n})$
$\implies x \in \cap_{n\ge1}(\overline{E}^c)$
$\implies x \in \overline{E_n}^c$ for all n, $\implies x \notin \overline{E_n}$ for all n
$\implies x \notin \cup_{n\ge1}\overline{E_n} \implies \cup_{n\ge1}\overline{E_n} \ne X$

Then, $\cup_{n\ge1}E_n \subset \cup_{n\ge1}\overline{E_n} \ne X$.
 

Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,890
Yes, that's correct.