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Why is an isometry bijective?

  1. Oct 31, 2009 #1
    A function from the plane to itself which preserves the distance between any two points is called an isometry. Prove that an isometry must be a bijection.

    To prove that an isometry is injective is easy:
    For an isometry: [tex]||f(x)-f(y)||=||x-y||[/tex]
    If [tex]x\neq y[/tex] then [tex]||x-y||>0[/tex] and therefore [tex]||f(x)-f(y)||>0[/tex] and [tex]f(x)\neq f(y)[/tex].
    But to prove that an isometry is surjective... how should I do that?
    Last edited: Oct 31, 2009
  2. jcsd
  3. Oct 31, 2009 #2


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    Have you thought about what all this means in terms of high-school Euclidean geometry?
  4. Oct 31, 2009 #3
    Well, I tried to make some drawings in the plane, but they do not seem to lead to a solution...
  5. Oct 31, 2009 #4


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    Well, at least have you found other things that must be preserved by an isometry?
  6. Oct 31, 2009 #5
    Well, if you have three points with [tex]||f(x)-f(y)||=||f(y)-f(z)||=||f(z)-f(x)||[/tex] this is also true for [tex]||x-y||=||y-z||=||z-x||[/tex]. This means that a triangle remains a triangle. But I do not know what first step I have to see to do the proof...
  7. Nov 1, 2009 #6
    Is there anyone who can help me on this?
  8. Nov 1, 2009 #7
    Consider the tiling of the plane by regular triangles with all edges 1. What does your isometry do to its vertices? Once you know that, consider a general point (it is in one of those triangles) and see what you can say about what the isometry does to it.
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