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Math Help - Isometry between metric spaces

  1. #1
    Senior Member
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    Isometry between metric spaces

    Let f:X->Y be isometry between metric spaces (X, d) and (Y, d'). Show that for each a \in X and r > 0,

    f(B_{d}(a,r)) = B_{d'}(f(a),r).


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    Definition of "isometry".
    Let (X,d) and (Y,d') be metric spaces. Bijective function f is called an isometry if
    d(a, b) = d'(f(a), f(b)) for all a, b in X.
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hello,
    Quote Originally Posted by aliceinwonderland View Post
    Let f:X->Y be isometry between metric spaces (X, d) and (Y, d'). Show that for each a \in X and r > 0,

    f(B_{d}(a,r)) = B_{d'}(f(a),r).
    You can start by showing that f(B_{d}(a,r)) \subseteq B_{d'}(f(a),r), that is to say that for all x\in B_{d}(a,r) we have f(x)\in B_{d'}(f(a),r).
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