Another paracompactness problem

[radou]

Homework Statement

Let X be regular. If X is a countable union of compact subspaces of X, then X is paracompact.

The Attempt at a Solution

Let X = U Ci. Let an open cover for Ci be given. Denote its finite subcover with {Ci1', ... , Cin'}. Clearly, X can now be expressed as the countable union of the elements Cij'. Denote this collection with C. Now, let U be an open cover for X. Then the collection {A$$\cap$$Cij' : A is in U, Cij' is in C} covers X and is an open refinement of U. But nothing more.

The idea is to

a) either conclude that X is Lindeloff, since paracompactness would then follow
b) find a refinement of U which is an open cover for X and locally finite, paracompactness would be satisfied by definition
c) either find a refinement of U which is open, covers X and countably locally fininte, or simply covers X and is locally finite, or is closed, covers X and is locally finite, since then paracompactness would follow again from a lemma

Am I on the right track here? Any discrete hints are welcome...

 

[micromass]

Let's not make it harder then it is: go for (a)

 

[radou]

Let's not make it harder then it is: go for (a)

Ha, OK, I'll do that. It turns out to he an elegant set-up. OK, I'll post a bit later...

 

[radou]

Ahh, this turns out to be very easy, actually.

Let A be an open cover for X. The A is an open cover for every compact subspace of X, Ci. For every Ci, let {Ai} be the finite subcover of A covering Ci. Obviously the collection of all the fininte subsovers Ai is countable and a subcollection of A covering X.

 

[micromass]

That is a perfect solution!

 

[radou]

That is a perfect solution!

Thanks, btw I don't see any other one.