Re: Lindelof + Metrizable ==> Second Countable

On 05-07-2005 17:03, James wrote:

I am trying to prove that Lindelof + Metrizable ===> Second Countable.

Unfortunately I can't prove this. Here is my attempt. X = \/ B(x ; r) where the union runs over all x in X and over all r in Q (rational numbers).

Since X is Lindelof, there is a countable subcollection, let's call it { B(x_n ; r_n) }

Now I want to show that this is a basis (but I can't do this).

Right, you can't do this. And here's why: suppose that X is an uncountable discrete topological space. Yes, I know, it's not Lindelof. However, what you've proved is still true: there's a countable set of open balls whose union is X; in fact, you can take one single ball, namely B(x,2) for some in X. However, you can't deduce from this that your space is second countable (because it ain't!).

So I think I am starting off on the wrong foot. Maybe I am taking the wrong open cover of X to begin with?

Any hint? (Please I would prefer to not receive a solution)

The set (B(x,1))_{x in X} is a covering of X.

And so is the set (B(x,1/2))_{x in X}.

And so is the set (B(x,1/3))_{x\in X}.


Best regards,

Jose Carlos Santos