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Math Help - Non-Hausdorff Quotient Space

  1. #1
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    Non-Hausdorff Quotient Space

    This is a homework problem I've been struggling with, so I'd prefer a gentle nudge rather than the absolute solution.

    Consider a topological space X that is the disjoint union of two real lines.
    ie. X = R + R.

    Define the equivalence relation ~ :
    (a,0) ~ (b,1) iff a = b, but neither equal to zero.

    The quotient space looks like a real line with two separate 0's. Obviously this isn't Hausdorff.

    How can I prove this?

    My progress is claiming that if it is Hausdorff, then there exist disjoint open sets U0 and U1 in the quotient space containing [(0,0)] and [(0,1)] respectively. Then since the pi(x)=[x] function is continuous, there exist open preimages of U0 and U1 in X, and they must also be disjoint.

    However, it should be easy to prove that the preimages of U1 and U0 can't be disjoint (a contradiction) and therefore the quotient space isn't Hausdorff.
    But I can't really get that argument going.

    Any advice?
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  2. #2
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    All right I finally solved it. What I had forgotten to use was the fact that X = R + R, so it wasn't just any topology but a metric space (a nice Euclidean one even). As soon as I knew that, I could construct epsilon neighborhoods around any point I wanted, and showing what I wanted became very simple.

    Thanks anyway.
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  3. #3
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    Quote Originally Posted by Soltras View Post
    All right I finally solved it. What I had forgotten to use was the fact that X = R + R, so it wasn't just any topology but a metric space (a nice Euclidean one even). As soon as I knew that, I could construct epsilon neighborhoods around any point I wanted, and showing what I wanted became very simple.

    Thanks anyway.
    Ummm...Yeah. What he said. I understood the problem, not the solution. Ah well. Back to my Topology book...

    -Dan
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