1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Show that a distance preserving map T:X->X is onto
Loading...