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Non homeomorphic spaces

  1. May 25, 2013 #1
    Why are the irrationals R-Q and the product space (R-Q)XQ not homeomorphic?
    The first space i Baire space.may be the second space is not?
     
  2. jcsd
  3. May 25, 2013 #2

    Office_Shredder

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    Their completions aren't homeomorphic, I think that does the trick but maybe there's some weird counterexample
     
  4. May 25, 2013 #3

    micromass

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    The completions of ##\mathbb{Q}## and ##\{x\in \mathbb{Q}~\vert~0<x<1\}## aren't homeomorphic either, even though the two spaces are homeomorphic. The problem is that completion is a metric concept and not a topological concept.

    I think looking at Baire spaces is the way to go
     
  5. May 26, 2013 #4

    Bacle2

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    Well, how about from the perspective that R-Q can be embedded in R, but , at least that I can

    tell, (R-Q)xQ cannot?
     
  6. May 26, 2013 #5
    why not?
    it seems that a homeomorphism actually can be defind by using continous fractions.Isn't it?
     
    Last edited: May 26, 2013
  7. May 26, 2013 #6
    (R-Q)XQ isn't Baire. To prove that look at "slices".
     
  8. May 26, 2013 #7
    what is "slices" ? can you give me a link for the proof?
    Thank's a lot.
     
    Last edited: May 26, 2013
  9. May 26, 2013 #8
    Take the sets (R-Q)x{q} for rational q. This is a countable set of closed sets with nonempty interior but their union is the entire space. Hence, it isn't Baire.
     
  10. May 26, 2013 #9
    You mean with EMPTY interior right?
     
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