A countable basis vs. countably locally finite problem

[radou]

A "countable basis" vs. "countably locally finite" problem

Homework Statement

Sometimes it's fairly difficult to name a thread for a specific problem.

So, one needs to show that, if X has a countable basis, a collection A of subsets of X is countably locally finite of and only if it is countable.

(A collection A is countably locally finite if it can be expressed as A = U Ai, where is goes over the positive integers, and where for each i, the collection Ai is locally finite.)

The Attempt at a Solution

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Assume A is a countable collection of subsets of X, so it can be indexed with the positive integers, so we have A1, A2, ... Ai, ... . Now, let pick some x in X. Trivially, X is a neighborhood of x which intersects Ai, for any i, and hence only finitely many members of the collection Ai (Ai is the only member of Ai). So, A is countably locally finite.

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Let A be a countably locally finite collection of subsets of X. Pick x in X, and some element Ai from the collection A (Ai is a collection of sets itself). Then there exists some neighborhood U of x which intersects Ai in only finitely many elements. Let B be a countable basis for X. Pick an element Bi containing x and contained in U. Then Bi intersects Ai in only finitely many elements too (possibly even in none of them, but that meany finitely, too). Do this for every x in X. Now here's probably a wrong conclusion: if I do this for every x, I can't arrive at a countable collection of basis elements {Bi} which cover X, right? Since X may be uncountable. Since if this was true, then {Bi} would be an open cover for X, and every element of {Bi} would intersect Ai in a finite number of elements. Hence, Ai would be countable, which would make A countable.

But I fear this won't work. Since if X is uncountable, I can't choose, for every x in X, the a basis element containing x and contained in the neighborhood Ux of x which intersects Ai in finitely many elements. Or?

 

[micromass]

Well, I think you've proven it, but you don't realize it yet...
For every x, you can choose a set Bi. Now, the resulting collection of Bi is countable (even if X is uncountable). This is since the entire basis B is countable. So the collection of all the Bi must be countable...

 

[radou]

Well, I think you've proven it, but you don't realize it yet...
For every x, you can choose a set Bi. Now, the resulting collection of Bi is countable (even if X is uncountable). This is since the entire basis B is countable. So the collection of all the Bi must be countable...

Well, then this works after all! Great!

For some reason, I thought something wasn't right here... But indeed, since B is countable, we arrive at a countable collection at the end.