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 be a topological space, a compact topological space and such that is continuous, onto and closed (maps any closed set in X to a closed set in Y), and such that is compact.
Prove that is compact.
Thanks for any help.
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 is compact.
You are given that is a closed mapping.
Can you use the FIP to prove that is compact?
I did try approaching this way, seemed rational as the mapping is closed, but could not really finish it:
take some collection of closed sets in that has the FIP. Then maps it to a collection of closed sets in , and we know that a closed subset of a compact space is compact itself, so for every , is compact...
Does being compact also imply that also has FIP (this seems to be correct intuitively...)? And also, if that is true, then isn't which would mean we are done? :\
I know this is not much, but I'm just having a brainfart, so excuse me.
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 where is the collection of images of p. But we may have that 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 is empty.
For each point in is not in one of those sets.
So we cover with finite collect of open sets that contains no point is a finite sub-collection of .
That is not possible.