Re: Lindelof + Metrizable ==> Second Countable
On 05072005 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
.
Relevant Pages
 Re: Lindelof + Metrizable ==> Second Countable
... >> where the union runs over all x in X and over all r in Q (rational ... > uncountable discrete topological space. ... Yes, I know, it's not Lindelof. ... > open balls whose union is X; in fact, you can take one single ball, ... (sci.math)  Re: HeineBorel Theorem for E^n (Real Analysis)
... open balls at *0 vector*? ... complements of the closed disks centered at _p_. ... their union and therefore E is a subset of the union of a finite subset ... (sci.math)  Re: apparent error in Big Rudin
... I guess he means that the closure of \tau under finite intersection is a topology. ... As stated, \tau does not include for instance the intersection of two open balls A and B when A and B are not disjoint and neither A nor B is a subset of the other, since such an intersection is neither an open ball nor the union of open balls. ... The intersection of A and B is a union of open balls, since each point in either A or B is the center of an open ball. ... (sci.math)  Re: Topology
... > How do you show that those things form a topology? ... > intersection of two open balls is not a union of open balls? ... of T is an union of "open balls" then the "open balls" must be a basis ... (sci.math) 
