Thread: [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.
Suppose that there are not two disjoint neighborhoods one containing and the other containing b. Now there is a countable base, U_n, at a, also one at b, V_n. By the assumption U_n intersect V_n is not empty for every n. Using countably and choice we can with care find a sequence that must converge to both a & b.

5. Thank you, great one

Still, something is not clear to me: you say "there is a countable base at a". What does that mean? That you can take a countable number of sets which contain a and form a basis to the topology?
If I'll understand that, I'll probably just see the rest of it come to light

6. k, I've checked it up on Wiki and found what you meant. I got it now, thank you very much!!!