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Math Help - Topological counterexample

  1. #1
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    Topological counterexample

    How do I show, I am assuming by example, that X can be homeomorphic to a subspace of Y and Y is homeomorphic to a subspace of X but X and Y are not homeomorphic?

    I have tried things involving the indiscrete and the discrete topologies as these seemed simple to work with, however I have been unable to come up with anything.

    Edit: I have added an extra question later on in the thread that I am stuck on.
    Last edited by ihateyouall; September 2nd 2010 at 03:25 AM.
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  2. #2
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    You're making it too complicated I'm afraid. Can you think of two subsets of \mathbb{R}, with the usual topology, which would work in this case?
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  3. #3
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    I thought about this before. I was finding difficulty because I was considering using something like rationals as one subspace however this does not seem so easy to get to be homeomorphic to a subset of something. If one set is countable and the other is uncountable then it would seem that I will have problems getting the larger to be homeomorphic to a subset of the smaller; so I think I might need to have two uncountable sets, or maybe some countable ones. I don't think that I can have two intervals as these would be homeomorphic. Any ideas?
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  4. #4
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    What about [0,1] and (0,1)?
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  5. #5
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    Of course.
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  6. #6
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    I am sorry to ask again but I have a related problem. I have to find X, Y which are not homeomorphic but for whom X*I and Y*I are, where I is the unit interval with the normal topology. I am thinking that I must have to make some move in the added 'dimension' to the space X*I that gets to something that can be called Y*I and the move does not correspond to something valid for homeomorphism in the non-extended state. I have the hint to consider a disk with two arcs attached. Any ideas?
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