Hi, I would appreciate some guiding with this question. Thanks.
Let X be some topological space and Y a compact, Hausdorff topological space. Let and define .
Show that if is a closed subset of with the product topology, then f is continuous.
Theorem: Let be any space and be compact. Then, if is closed we have that is continuous.
Proof:
Lemma: Let be compact, then is a closed.
Proof: Let be closed. We prove that is closed by proving that it's compliment is open. So, let . Then, it follows that , and for each has a basic open neighborhood . Since the tube is compact (since it's homeomorphic to ) there is a finite subcovering . The neighborhood is clearly disjoint from . The conclusion follows.
Now, it is easily proved that since is closed so is . But, it is clearly a bijective continuous mapping and since it's closed it's also a homeomorphism. The problem follows after considering the following lemma
Lemma: A mapping is continuous if is a homeomorphism.
Proof: We must merely note that .
I think I came up with a solution. Tell me if you see anything wrong please.
I want to show that f is continuous, ie. that if is open then is open as well.
Let V be an open subset of Y. Let , then U is closed, and so is . Since is closed as well, is also closed.
From a previous result, we know that the projection is a closed mapping when Y is compact. Therefore, is closed.
However, note that is exactly the set of points in X which f does not map into V, and it is closed, therefore is open and we are done.
I know it is a bit informal, but that is simply the outline.
Thanks again for helping.