Compact spaces

• Jan 22nd 2010, 11:27 AM
Defunkt
Compact spaces
Hi, I would like some help with this problem if possible. Been trying to solve it for a while but haven't really got anything meaningful...

Let $\displaystyle X$ be a topological space, $\displaystyle Y$ a compact topological space and $\displaystyle p:X\to Y$ such that $\displaystyle p$ is continuous, onto and closed (maps any closed set in X to a closed set in Y), and such that $\displaystyle \forall y \in Y, ~ p^{-1}(\{y\})$ is compact.

Prove that $\displaystyle X$ is compact.

Thanks for any help.
• Jan 22nd 2010, 12:04 PM
Drexel28
Quote:

Originally Posted by Defunkt
Hi, I would like some help with this problem if possible. Been trying to solve it for a while but haven't really got anything meaningful...

Let $\displaystyle X$ be a topological space, $\displaystyle Y$ a compact topological space and $\displaystyle p:X\to Y$ such that $\displaystyle p$ is continuous, onto and closed (maps any closed set in X to a closed set in Y), and such that $\displaystyle \forall y \in Y, ~ p^{-1}(\{y\})$ is compact.

Prove that $\displaystyle X$ is compact.

Thanks for any help.

Have you considered that $\displaystyle X=\bigcup_{y\in Y}p^{-1}(\{y\})$?
• Jan 22nd 2010, 12:30 PM
Defunkt
Quote:

Originally Posted by Drexel28
Have you considered that $\displaystyle X=\bigcup_{y\in Y}p^{-1}(\{y\})$?

Yes I have, but is it true that *any* union of compact sets is compact?

It is trivially correct for a finite union, but I don't know if it is for inifinte.
• Jan 22nd 2010, 12:41 PM
Plato
Have you considered the Finite Intersection Property?
A topological space is compact if and only if every collection of subsets having the FIP has itself a nonempty intersection.
You are given that $\displaystyle Y$ is compact.
You are given that $\displaystyle p$ is a closed mapping.

Can you use the FIP to prove that $\displaystyle X$ is compact?
• Jan 22nd 2010, 02:38 PM
Defunkt
Quote:

Originally Posted by Plato
Have you considered the Finite Intersection Property?
A topological space is compact if and only if every collection of subsets having the FIP has itself a nonempty intersection.
You are given that $\displaystyle Y$ is compact.
You are given that $\displaystyle p$ is a closed mapping.

Can you use the FIP to prove that $\displaystyle X$ is compact?

I did try approaching this way, seemed rational as the mapping is closed, but could not really finish it:

take some collection $\displaystyle \mathcal{S}$ of closed sets in $\displaystyle X$ that has the FIP. Then $\displaystyle p$ maps it to a collection of closed sets $\displaystyle \mathcal{T} = \{p(S) : S \in \mathcal{S}\}$ in $\displaystyle Y$, and we know that a closed subset of a compact space is compact itself, so for every $\displaystyle T \in \mathcal{T}$, $\displaystyle T$ is compact...

Does $\displaystyle Y$ being compact also imply that $\displaystyle \mathcal{T}$ also has FIP (this seems to be correct intuitively...)? And also, if that is true, then isn't $\displaystyle p^{-1}(\bigcap_{T \in \mathcal{T}}T) \subset \bigcap_{S \in \mathcal{S}}S$ which would mean we are done? :\

I know this is not much, but I'm just having a brainfart, so excuse me.
• Jan 22nd 2010, 03:12 PM
Plato
It means that $\displaystyle \mathcal{T}$ has the FIP. You may need to prove that.
Because $\displaystyle Y$ is compact then $\displaystyle {\exists t \in \bigcap\limits_{v \in \mathcal{T}} v }$.
You know that $\displaystyle p$ is onto.
Does that mean that the intersection of the sets in $\displaystyle \mathcal{S}$ is not empty?
• Jan 22nd 2010, 03:33 PM
Defunkt
Quote:

Originally Posted by Plato
It means that $\displaystyle \mathcal{T}$ has the FIP. You may need to prove that.
Because $\displaystyle Y$ is compact then $\displaystyle {\exists t \in \bigcap\limits_{v \in \mathcal{T}} v }$.
You know that $\displaystyle p$ is onto.
Does that mean that the intersection of the sets in $\displaystyle \mathcal{S}$ is not empty?

Well, since we know $\displaystyle \exists t \in \bigcap\limits_{v \in \mathcal{T}} v$ we get that $\displaystyle \forall S \in \mathcal{S}, ~t \in p(S)$ and $\displaystyle p$ being onto means $\displaystyle \forall S \in \mathcal{S} ~ \exists s \in S : p(s) = t$

Now, if $\displaystyle |p^{-1}(t)| = 1$ we are done, otherwise, we (I) seem to be in a problem. I guess we need to use that $\displaystyle p^{-1}(\{t\})$ is compact at this point, but I don't see how...
• Jan 22nd 2010, 04:07 PM
Plato
Frankly I have no idea what all of that means.
Here is what we must do.
Given any collection of closed subset of $\displaystyle X$ having the FIP then that collection must have a non-empty interection.
• Jan 24th 2010, 11:58 AM
Defunkt
Quote:

Originally Posted by Plato
Frankly I have no idea what all of that means.
Here is what we must do.
Given any collection of closed subset of $\displaystyle X$ having the FIP then that collection must have a non-empty interection.

I realize that, but I still can't seem to solve it.

Given a collection of closed subsets in X that has the FIP, its image (of p) also has the FIP since p is closed. Now, since Y is compact, there exists some $\displaystyle t \in \bigcap_{T \in \mathcal{T}}T$ where $\displaystyle \mathcal{T}$ is the collection of images of p. But we may have that $\displaystyle p^{-1}(\{t\})$ contains more than one element "spread" over the collection of closed subsets in X, and then I don't see how we are sure that there is an element in the intersection of that collection.

Sorry for being a bother, I just can't seem to see how to get the desired result. I, of course, appreciate your help.
• Jan 24th 2010, 03:06 PM
Plato
Quote:

Originally Posted by Defunkt
I realize that, but I still can't seem to solve it.

Given a collection of closed subsets in X that has the FIP, its image (of p) also has the FIP since p is closed. Now, since Y is compact, there exists some $\displaystyle t \in \bigcap_{T \in \mathcal{T}}T$ where $\displaystyle \mathcal{T}$ is the collection of images of p. But we may have that $\displaystyle p^{-1}(\{t\})$ contains more than one element "spread" over the collection of closed subsets in X, and then I don't see how we are sure that there is an element in the intersection of that collection.

Sorry for being a bother, I just can't seem to see how to get the desired result. I, of course, appreciate your help.

This is the last response I will give you.
The remainder of the proof is extremely messy.
Suppose that $\displaystyle \bigcap_{T \in \mathcal{T}}T$ is empty.
For each point in $\displaystyle p^{-1}(\{t\})$ is not in one of those sets.
So we cover $\displaystyle p^{-1}(\{t\})$ with finite collect of open sets that contains no point is a finite sub-collection of $\displaystyle \mathcal{T}$.
That is not possible.
• Jan 24th 2010, 06:21 PM
Drexel28
I'm sorry. I misread the question.

Quote:

Originally Posted by Defunkt
Yes I have, but is it true that *any* union of compact sets is compact?

It is trivially correct for a finite union, but I don't know if it is for inifinte.

No. $\displaystyle \mathbb{R}=\bigcup_{n\in\mathbb{N}}\left[-n,n\right]$
• Jan 25th 2010, 02:44 AM
Defunkt
Quote:

Originally Posted by Plato
This is the last response I will give you.
The remainder of the proof is extremely messy.
Suppose that $\displaystyle \bigcap_{T \in \mathcal{T}}T$ is empty.
For each point in $\displaystyle p^{-1}(\{t\})$ is not in one of those sets.
So we cover $\displaystyle p^{-1}(\{t\})$ with finite collect of open sets that contains no point is a finite sub-collection of $\displaystyle \mathcal{T}$.
That is not possible.

Thanks, I understand now.