# Special Case of the Baire Category Theorem

1. Dec 7, 2012

### sammycaps

1. So the solutions claim a different and better proof of this, but I just wanted to see if mine made sense.

Question - Let X be a compact Hausdorff space; let {An} be a countable collection of closed sets of X. Show that if each set An has empty interior in X, then the union $\bigcup$An has empty interior in X.

2. Relevant equations

If X is compact Hausdorff with no isolated points, for any nonempty open set U of X and any point x of X, there exists a nonempty open set V contained in U such that x$\notin$$\overline{V}$.

3. The attempt at a solution

Now that I've seen the solution it seems obvious, but here is what I did.

If we let Int($\bigcup$An)=U and take a point xi of Ai, then there exists an open set Vi $\subset$ Int($\bigcup$An) s.t. x$\notin$$\overline{V}$i. We can form this construction because we can choose a point in Int($\bigcup$An) different from x, because if not, then we would either have an empty interior (in the case where x$\notin$ Int($\bigcup$An) ) or we would have a one-point open set (in the case where x$\in$ Int($\bigcup$An) , which would mean that Ai for some i has non-empty interior. Then by the Hausdorff property, there is a W1 and W2 about x and y respectively, that are disjoint. Then W2$\bigcap$Int($\bigcup$An) is an open set contained in Int($\bigcup$An), not containing x. It follow that the closure of this set, which I'll call V, does not contain x.

Then we can choose $\overline{V}$1 $\supset$ $\overline{V}$2 $\supset$ .....

And since X is compact, there exists an x in this infinite intersection. Then, since U is the union of {An}, x must be contained in Ai for some i. But then $\overline{V}$i $\cap$ Ai is a closed set containing x but not xi. Then the complement of $\overline{V}$i $\cap$ Ai in Ai is open and non empty in Ai, implying that Ai has non-empty interior, a contradiction.

Last edited: Dec 7, 2012
2. Dec 7, 2012

### micromass

Staff Emeritus
It doesn't really seem obvious to me that the x that you constructed actually lies in U. Sure, every $V_i$ lies in U. But you're taking the intersections of the $\overline{V_i}$. Do those necessarily lie in U?

also, you seem to make LaTeX more difficult than it needs to be. If you want to type $\overline{V_i}$, then you can do this by

Code (Text):

[itеx]\overline{V_i}[/itex]

This seems a lot easier than using the [NOPARSE]...[/NOPARSE] tags.

Here is a LaTeX FAQ: https://www.physicsforums.com/showpost.php?p=3977517&postcount=3

Last edited: Dec 7, 2012
3. Dec 7, 2012

### sammycaps

O, I did not see that. Its not obvious to me that x lies in U, and I don't know whether or not its true. Thanks for pointing that out.

Maybe I don't actually need x to be in U, only for x to be in $\bigcup${An}. Still though, its not clear to me that is true.

Anyway, maybe this proof just doesn't work.

4. Dec 7, 2012

### micromass

Staff Emeritus
I'm thinking that you might be able to fix it by taking an open set $U^\prime$ such that $\overline{U^\prime}\subseteq U$ that is nonempty. Then you might take care that all the $V_i\subseteq U^\prime$. That would for the intersection of the $\overline{V_i}$ to be in U.

I didn't really work it out, but you can try some trickery like this.