Every isometry is Lipchitz-continuous

Question

Let $(X, d_X)$ and $(Y, d_Y)$ be metric spaces, $f:(X,d_X) \rightarrow (Y,d_Y)$ be an isometry. Then $f$ is Lipchitz-continuous.

Attempt:

Suppose that $f$ is an isometry. Then for all $x_1,x_2$ in $X$, we have $d_Y(f(x_1),f(x_2)) = d(x_1,x_2)$...

From here, how do I prove that there exists $k>0$ such that for all $x_1,x_2 \in X,$ $d_Y(f(x_1),f(x_2))\leq kd_X(x_1,x_2)$

Guacho Perez · Accepted Answer · 2019-05-21 22:00:05Z

3

The Lipschitz constant is $K=1$.

answered May 21 '19 at 22:00

Guacho Perez

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$\begingroup$ How did you get that? $\endgroup$ – RUNN May 21 '19 at 22:04
2

$\begingroup$ From $d_Y(f(x),f(y))=d_X(x,y)\le 1\cdot d_X(x,y)$ $\endgroup$ – Guacho Perez May 21 '19 at 22:06

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