Special Case of the Baire Category Theorem

[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 {A_n} be a countable collection of closed sets of X. Show that if each set A_n has empty interior in X, then the union $\bigcup$A_n has empty interior in X.

Homework 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}$.

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($\bigcup$A_n)=U and take a point x_i of A_i, then there exists an open set V_i $\subset$ Int($\bigcup$A_n) s.t. x$\notin$$\overline{V}$_i. We can form this construction because we can choose a point in Int($\bigcup$A_n) different from x, because if not, then we would either have an empty interior (in the case where x$\notin$ Int($\bigcup$A_n) ) or we would have a one-point open set (in the case where x$\in$ Int($\bigcup$A_n) , which would mean that A_i 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($\bigcup$A_n) is an open set contained in Int($\bigcup$A_n), 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 {A_n}, x must be contained in A_i for some i. But then $\overline{V}$_i $\cap$ A_i is a closed set containing x but not x_i. Then the complement of $\overline{V}$_i $\cap$ A_i in A_i is open and non empty in A_i, implying that A_i has non-empty interior, a contradiction.

 

[micromass]

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 {A_n} be a countable collection of closed sets of X. Show that if each set A_n has empty interior in X, then the union $\bigcup$A_n has empty interior in X.

Homework 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}$.

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($\bigcup$A_n)=U and take a point x_i of A_i, then there exists an open set V_i $\subset$ Int($\bigcup$A_n) s.t. x$\notin$$\overline{V}$_i. We can form this construction because we can choose a point in Int($\bigcup$A_n) different from x, because if not, then we would either have an empty interior (in the case where x$\notin$ Int($\bigcup$A_n) ) or we would have a one-point open set (in the case where x$\in$ Int($\bigcup$A_n) , which would mean that A_i 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($\bigcup$A_n) is an open set contained in Int($\bigcup$A_n), 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 {A_n}, x must be contained in A_i for some i. But then $\overline{V}$_i $\cap$ A_i is a closed set containing x but not x_i. Then the complement of $\overline{V}$_i $\cap$ A_i in A_i is open and non empty in A_i, implying that A_i has non-empty interior, a contradiction.

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

[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

 

[sammycaps]

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?

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 $\bigcup${A_n}. Still though, its not clear to me that is true.

Anyway, maybe this proof just doesn't work.

 

[micromass]

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 $\bigcup${A_n}. Still though, its not clear to me that is true.

Anyway, maybe this proof just doesn't work.

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.