Re: Topology with first countable.

On 09-07-2008 13:20, David C. Ullrich wrote:

I know that
Let S have the cofinite or the cocountable topology. Show
S is countable iff S is 2nd countable iff S is 1st countable.
- William Elliot -

I have a question.
If X is a countable set, then (X, T) is first countable.

is this true ?
No. Just take the one-point compactification of Q, that is, Q U {oo}
with the topology T for which a set A is open when:

1) if oo does not belong to A, then A is an open subset of Q;

2) otherwise, then the complement of Q is compact.

Oops! What I meant here was that "the complement of A is compact".

I can't quite parse (2) as a definition of which sets are
open. If you mean the one-point compactification of
Q (with the discrete topology) then

(1) Any subset of Q is open
(2) Any subset of Q U {oo} with a finite complement is open.

(Ie, the complement (in Q U {oo}) of any compact subset of Q
is open.)

Except that seems second-countable to me: If F_n is a sequence of finite subsets of Q with union equal to Q
then don't the complements of the F_n give a local base
at 00?
But what I meant was the one-point compactification of Q with respect to
the usual topology.

Ok. The reason I didn't realize that's what you meant is that if that's what you meant then your version of (2) is simply wrong,
not just confusing. Because not every compact subset of Q
is finite.

Another reason I didn't guess that is that it's locally compact
spaces that _have_ a one-point compactification, and Q
is not locally compact. (If K is compact (hausorff) and oo
is a point of K then K \ {oo} is locally compact. Hence
Q has no one-point compactification.)

I should have written: apply to Q the method which, when applied to
locally compact spaces, gives a one-point compactification. The
_definition_ of the topology makes sense, although, yes, Q U {oo} is not
compact.

So right now I'm really not certain exactly what topological
space you were referring to. Whatever it is, I tend to doubt
that it can be an example of a non-first-countable countable
space, just because I doubt it's possible to give a description
of one that's _quite_ that simple.

Q U {oo} is not first countable. More precisely, there is no sequence
(A_n)_n of open neighborhoods of oo such that each neighborhood of oo
contains some A_n. An equivalent way of saying this is: there is no
sequence (K_n)_n of compact subsets of Q such that each compact subset
of Q is contained in some K_n. For each _n_ let q_n be a rational number
from ]-1/n,1/n[ which does not belong to K_n. Then, if you define

K = {0, q_1, q_2, q_3, q_4, ...}

the set K will be a compact of Q which will contain all the q_n's and
therefore cannot be a subset of some K_n.

Best regards,

Jose Carlos Santos
.

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