# Thread: Isometry between metric spaces

1. ## Isometry between metric spaces

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

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

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

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