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Is the metric space Q of rationals homeomorphic to N, the natural numbers?

  1. Oct 1, 2008 #1

    gmn

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    I don't know if this is more appropriate for the topology forum, but I am learning this in analysis. I am asked to say whether or not Q and N are homeomorphic to eachother and to justify why. I am confused as to how to prove precisely that two spaces are homeomorphic, for there are no formal proofs of this in my text, only that (0,1] is not homeomorphic to the unit circle. everything else is just a visualisation of a donut and a coffee cup or something like that, but that doesn't really help me when thinking about N and Q. I know there exists a bijection, but I feel like it might not qualify for a homeomorphism because it doesn't necessarily send points that are close together or far away from eachother in N to points that are close together or far away in Q. I'm confused. Please help!

    thanks
     
  2. jcsd
  3. Oct 1, 2008 #2

    morphism

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    Try to find a topological property that N possesses but Q does not. Hint: what can you say (topologically) about an arbitrary subset of N?
     
  4. Oct 2, 2008 #3

    mathman

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    It depends on the topology you put on the sets. If you use a discrete topology for the rationals, then they will be homeomorphic. If you use the "natural" topology for the rationals, they will not be.
     
  5. Oct 2, 2008 #4

    Hurkyl

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    And he specified he's using the metric topology.
     
  6. Oct 2, 2008 #5
    doesn't the metric function of the space have to continuous?
     
  7. Oct 2, 2008 #6
    I don't see this being made clear anywhere.

    edit: Whoops. It was in the title. Well, that is a hidden place...

    Yes. One never has to worry about it, though, because the metric function is always continuous in the topology it itself defines.
     
  8. Oct 3, 2008 #7

    mathman

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    The metric function does not have to be continuous!
    Examples: Integers d(x,y) = |x-y|.
    Rational numbers: Express all rationals in lowest terms, d(a/b,c/f)=|a-c|+|b-f|.

    These will both lead to discrete topologies.
     
  9. Oct 3, 2008 #8

    Hurkyl

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    And in both cases, that makes the metric a function from a discrete space into the reals.

    Exercise: Prove that any function with a discrete domain is continuous.
     
  10. Oct 3, 2008 #9
    the topology on Q is probably the subspace topology inherited from R, a metric space.
     
  11. Oct 4, 2008 #10

    mathman

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    The distances in the topology I described are all integers (which is of course a subset of the reals). The point I was trying to make is that with the topologies I defined, the integers and the rationals are homeomorphic.
     
  12. Oct 8, 2008 #11
    c'mon that's easy
     
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