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Lindelof + Metrizable ==> Second Countable
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James
science forum Guru Wannabe


Joined: 02 May 2005
Posts: 195

PostPosted: Tue Jul 05, 2005 2:03 pm    Post subject: Lindelof + Metrizable ==> Second Countable Reply with quote
Hello,

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).

Let x in X. Let U be open in X and x in U. Then my problem is that in my
countable subcollection, it may be the case that x_i is outside of U for
every i. Then there is no way I can get a B(x_n ; r_n) to contain x and at
the same time be contained in U, no matter how small the radius of this ball
may be.

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)

James
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José Carlos Santos
science forum Guru


Joined: 25 Mar 2005
Posts: 1111

PostPosted: Tue Jul 05, 2005 2:28 pm    Post subject: Re: Lindelof + Metrizable ==> Second Countable Reply with quote
On 05-07-2005 17:03, James wrote:

Quote:
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!).

Quote:
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
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James
science forum Guru Wannabe


Joined: 02 May 2005
Posts: 195

PostPosted: Tue Jul 05, 2005 3:00 pm    Post subject: Re: Lindelof + Metrizable ==> Second Countable Reply with quote
"José Carlos Santos" <jcsantos@fc.up.pt> wrote in message
news:3ivqouFnjnjqU1@individual.net...
Quote:
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

Got it thanks!
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