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

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

    Hello everyone! thanks for reading!
    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....

    Any hints/help/answers greatly appreciated
    Tomer.
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  2. #2
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    Quote Originally Posted by aurora View Post
    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.
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  3. #3
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    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
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  4. #4
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    Quote Originally Posted by aurora View Post
    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.
    Did you read it and try to follow the logic?
    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.
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  5. #5
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    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
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  6. #6
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    k, I've checked it up on Wiki and found what you meant. I got it now, thank you very much!!!
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