Is X metrizable if two of its closed subspaces are metrizable?

  • Thread starter radou
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In summary: Thanks for the feedback!In summary, the author was unable to solve the problem he was given. He was told to use a countable basis, but didn't realize that this was necessary. After following the hint from the book, the author was able to find the error in his proof.
  • #1
radou
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Homework Statement



Let X be a compact Hausdorff space that is the union of the closed subspaces X1 and X2. If X1 and X2 are metrizable, show that X is metrizable.

The Attempt at a Solution



The problem wit this is that the proof which I found seems pretty simple, and the hint in the book points to something a bit different. But still, I can't find the (eventual) error in my proof.

So.

Since X1 and X2 are both closed subspaces of a compact space, they are compact. Since they are metrizable too, they are second countable. Let {B1} be a countable basis for X1, and {B2} for X2. I claim that the collection B = {B1 U B2} is a countable basis for X.

Let U be an open set in X and x an element in U. U is either in X1, or in X2, or intersects both of these sets. In any of these cases, the intersection of U with any of these spaces is open in the respective subspace topology. Now, if x is in U, it is in some open set of one of the subspace topologies on X1 or X2. Then there exists some element of {B1} or {B2} such that x is containe din that element, and that that element is contained in U.

Since X is compact Hausdorss and has a countable basis, it follows that X is metrizable.

Any comments are welcome, thanks in advance.
 
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  • #2
..any thoughts?
 
  • #3
My best guess: For the collection {B} to be a countable basis for X, you need that each element of {B} is an open subset of X. While you do note that each element of {B} is open in its respective subspace topology, I don't think that this implies it's open in X.

I really know very little topology though, so I'm probably completely off the mark. I figured I'd give you some feedback though :)
 
  • #4
jgens said:
My best guess: For the collection {B} to be a countable basis for X, you need that each element of {B} is an open subset of X. While you do note that each element of {B} is open in its respective subspace topology, I don't think that this implies it's open in X.

I really know very little topology though, so I'm probably completely off the mark. I figured I'd give you some feedback though :)

Actually, that's a very good point, thanks! :) What a shame I didn't see that! My basis would work for U contained in X1 or in X2, since then some basis element would be open in U, and U is open in X, so the element would be open in X. But if U is not contained in X1 or X2, then the basis element is open in the relative topology only. But the open set which contains this basis element needn't be open in X.

I guess I should follow the hint from the book.
 

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