# Math Help - [SOLVED] Help! Topology problem :=\

1. ## [SOLVED] Help! Topology problem :=\

I need to prove the following...

Let X be a topological space with a countable basis, that satisfies the "Uniqueness of limit" axiom, meaning, that every sequence converges only into one limit.
Then X is hausdorff.

In class, we have studied of spereration axioms, marked as T0, T1, T2, etc...
A T1 space is one that for every x,y, x<>y (x isn't y), you can find an open set of x, not intersecting y, and vice versa.
A T2 space is an hausdorff space - for every x,y, x<>y, you can find two DIJOINT neighborhoods of x, y.
I have shown that the uniqueness of limit means that X is T1. I cannot, however, find a way to use the countability of the basis to form two disjoint open sets around x and y....

Tomer.

2. Originally Posted by aurora
Let X be a topological space with a countable basis, that satisfies the "Uniqueness of limit" axiom, meaning, that every sequence converges only into one limit. Then X is hausdorff.
I think that you mean a “countable local basis”. Then this is a standard theorem.
Without the LaTeX server it is difficult to present a good answer.
But here is an outline. I will skip any finite cases.
Suppose that there are two points a & b such that there does not exist two disjoint neighborhoods one containing a the other b. Now there are countable infinite local bases, one at a U_n and one at b V_n. By the hypothesis, V_n intersect U_n is not empty. So using the axiom of choice, you can chose a sequence that must converge to both a & b.
If you have access to a good mathematics library then look for the topology text by Helen F Cullen. She has given a beautiful proof of this theorem.

3. erm, actually, I mean what's written in the exercise: X is a second countable space => it has a countable base. I don't even know what a local basis is :-\

Thank you very much for the response, anyhow - though I still need help

4. Originally Posted by aurora
X is a second countable space => it has a countable base. - though I still need help
Well being second countable does not change the proof I outlined.