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Simple quesion about metric on extended real line

  1. Apr 13, 2012 #1
    I was told the extended real [itex]\hat{R}=R\cup\{-\infty,\infty\}[/itex] is homeomorphic to [0,1], I was wondering if the mapping
    [tex]
    h: [0,1]\rightarrow\hat{R}, h(x)=\cot^{-1}(\pi x), 0<x<1, h(0)=\infty, h(1)=-\infty
    [/tex]
    is a valid homeomorphism, so that a metric may be defined by the metric on [0,1]? Thank you.
     
  2. jcsd
  3. Apr 15, 2012 #2

    quasar987

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    Yes, of course!

    The topology on the extended real line is the topology generated by the usual open sets of R plus the "neigborhoods of infinity"; i.e. the sets of the form [itex](a,+\infty][/itex] and [itex][-\infty,b)[/itex].

    Similarly, the topology on [0,1] is generated by the usual open sets of (0,1) plus the sets of the form [itex](a,1][/itex] and [itex][0,b)[/itex]

    It suffices to observe that your h sends a generator of the topology of [0,1] to a generator of the topology of the extended real line, and similarly for h-1.
     
    Last edited: Apr 15, 2012
  4. Apr 16, 2012 #3
    Thank you! When you say generator of the topology, is it referring to a base of the topology? So if I had a bijective (no need to be continuous, which need not be defined) mapping h from one base (collections of open sets) to another, then h would also be a homeomorphism between the two spaces generated by the two bases?
     
  5. Apr 16, 2012 #4

    quasar987

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    Yes and yes.

    More generally, to check continuity of a map f:X-->Y, it is not necessary to check that f-1(U) is open in X for all open sets in Y. It is sufficient to check it for the elements U of a basis of the topology on Y. (Easy exercice in "set theory")
     
  6. Apr 17, 2012 #5
    That makes a lot of sense. Thank you.
     
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