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Absolute value topology?

  1. Nov 18, 2005 #1
    Hi all, I have the following question.
    Are the following spaces homeomorphic in the real number space with absolute value topology?
    1) [a,b) and (a,b]
    2) (a,b) and (r,s)U(u,v) where r < s < u < v.

    For 1), I got that they are not homeomorphic because it fails the topological property that sequences in a set converges to a point in the set. As seen, a sequence could converge to a point a in [a,b), which is in the set, but is not contained in the set (a,b].
    -Is this correct?

    For 2), I'm not really sure at all. I was wondering if I should try to create a function that maps (a,b) to (r,s)U(u,v) and show the properties for homeomorphisms under topological spaces but I can't find out a function.
     
  2. jcsd
  3. Nov 18, 2005 #2

    StatusX

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    What exactly is the absolute value topology? It sounds like it is is symmetric with respect to the two kinds of half open intervals, in which case the first two spaces are homeomorphic, but it's hard to answer this without knowing exactly what the toopology is.
     
  4. Nov 18, 2005 #3
    The absolute value topology is exactly what it should be on the real number line. Open sets are open intervals, closed sets are closed intervals.
     
  5. Nov 18, 2005 #4

    matt grime

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    that is the standard, or metric topology.

    Of course the first two sets are homeomorphic, I don't get your objection.

    And, if you believe tow spaces are not homeomorphic, it suffices to pretend there is a homeomorphism, you don't actually have to construct it or write it down.
     
  6. Nov 18, 2005 #5

    Hurkyl

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    I think he's been confused by the fact the identity map is not a homeomorphism of [a, b) and (a, b].
     
  7. Nov 18, 2005 #6
    For 2), think of connectedness.
     
  8. Nov 18, 2005 #7

    StatusX

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    The second part involves connectedness, which apparently you aren't familiar with. A space is connected if it cannot be decomposed into the union of two disjoint, non-empty open sets. Clearly two disjoint open intervals are not connected, and it can be shown that an interval is. This proof is fairly straightforward but a little long, so if you can't just use the fact that an interval is connected, you might want to look for a different approach. But it can be easily shown that any space homeomorphic to a connected space is connected from the definition I gave above, so this method will get you there, it just might not be the easiest way.
     
  9. Nov 18, 2005 #8
    Unfortunately, you are correct, my course has not covered connectedness yet. So far, this is what I've covered in the course regarding topological spaces: derived sets, continuity, homeomorphisms, and the separation axioms.

    Also, for [a,b) and (a,b], is it neccessary that I find a function that maps to one another in order to prove it's a homeomorphism?

    Thanks for the help, by the way.
     
  10. Nov 18, 2005 #9

    matt grime

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    It is necessary to prove such a map exists, whether or not you have to exhibit it explicitly is another matter. certainly writing it out is sufficient to demonstrate it exists, and i'm sure if you think geometrically you can describe the map easily enough. it is after all clear how to map [-1,1) to (-1,1] I hope, if you think geometrically.
     
  11. Nov 18, 2005 #10

    NateTG

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    For 2:
    What are the sets that are both open and closed in (a,b)?
    What are the sets that are both open and closed in (r,s)U(u,v)?
     
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