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Homework Help: Show that a distance preserving map T:X->X is onto

  1. May 16, 2012 #1
    1. The problem statement, all variables and given/known data

    I'm trying to show that a distance preserving map is 1:1 and onto. The 1:1 part was easy, but I'm stuck on proving it's onto...

    2. Relevant equations

    X is compact
    T(X)[itex]\subseteq[/itex]X
    THere's a hint saying to consider a point y in X\T(X) and consider the minimum distance
    between y and x[itex]\in[/itex] T(X) (the infinum of d(x,y) for all x[itex]\in[/itex] T(X) where d is an undefined metric, and call it deltA)
    Then it says to consider the sequence
    yn=Tn(y)

    3. The attempt at a solution

    Because X is compact, and yn is a subset of X, it must be bounded (I think?) and have a convergent subsequence, and inf(d(T^n(y),T^n(x))=delta for all n (and x_n=T^n(x) will also have a convergent subsequence). Show that the limits of y_n and x_n provide a contradiction?
     
  2. jcsd
  3. May 16, 2012 #2

    Dick

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    I'm really not sure where that hint is supposed to lead. Here is an alternative hint. You know X can be covered by a finite number of open balls of radius delta/2. Let N be the minimum number of open balls of radius delta/2 you need to cover X. Now throw away a ball containing x. It's doesn't intersect f(X). That means f(X) can be covered by N-1 balls. Do you see a problem looming here? Think about inverse images.
     
  4. May 17, 2012 #3
    Is the next step something like:
    Take a=sup(T(x'),T(x)) and because X is compact, it is closed, therefore using distance preservation of T there exists at least on x in X such that
    d(x',x)=a
    then lead this to contradiction?
     
  5. May 17, 2012 #4

    Dick

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    I don't see what contradiction that leads to.
     
  6. May 18, 2012 #5
    Sorry - that one was completely off track, but I think I got it out now, thanks
     
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