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Is Q Homeomorphic to N?

  1. Oct 9, 2007 #1
    Is Q homeomorphic to N?

    I understand that there exists a bijection from Q to N but I cannot figure out how this function is continuous and it's inverse is also continuous.
  2. jcsd
  3. Oct 10, 2007 #2

    matt grime

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    With what topology? The subspace topology from being a subspace of the reals with their normal topology, or the discrete topology. They're not homeomorphic in the former, but are in the latter.
  4. Oct 10, 2007 #3
    The normal topology...this is what i came up with...not sure if its right:

    Claim: they are not homeomorphic
    Proof: Assume they are. Then There exists a continuous function f from Q to N. Therefore all of the sequences in Q are mapped to a sequence in N preserving limits. But since sequences in N converge constantly, this cannot be a bijection. therefore they are not homeomorphic.

    Does this make sense?
  5. Oct 10, 2007 #4


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    Why do sequences in N converge constantly? The sequence (1, 2, 3, ...) certainly doesn't converge. Did you mean to say that the only sequences in N that converge are the ones that are eventually constant? But then why is this a contradiction? Maybe your homeomorphism f maps convergent sequences to sequences that are eventually constant. Can this happen?

    But honestly I wouldn't bother thinking about it this way. Just think of what the topologies are. N will get the trivial topology, i.e. every set is going to be open (why?). Will Q get the trivial topology? No, it won't. Try to find a set that's not open in Q. This will be enough to show that they're not homeomorphic.
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