Lemma 15.19.20022849340F28

Proof. Choose $\varepsilon >0$ with $2\varepsilon \geq \operatorname {dis}R$. Give the disjoint union $E\sqcup F$ its original distances within each space, and use the symmetric cross distance

It is finite and at least $\varepsilon $. The triangle inequalities involving one cross distance follow from the triangle inequality within the corresponding space and the infimum. For two cross distances, take $(x_0,y_0),(x_1,y_1)\in R$. The distortion bound gives

Taking the two infima gives $d_ E(x,x')\leq D(x,y)+D(y,x')$. The same argument with $E,F$ exchanged treats the other case. Hence $D$ is a metric extending both metrics. Each point has a correspondent at cross distance $\varepsilon $, so the Hausdorff distance between $E$ and $F$ in $(E\sqcup F,D)$ is at most $\varepsilon $. Letting $\varepsilon $ decrease to half the distortion proves the upper bound, including zero distortion.

For the lower bound, suppose two isometric images have Hausdorff distance $h$. For every $\delta >0$, two points in the first image have points in the second within $h+\delta $. Their distance is at most $\operatorname {diam}F+2h+2\delta $. Taking the supremum and then $\delta \downarrow 0$ gives $\operatorname {diam}E\leq \operatorname {diam}F+2h$. Exchanging the images and taking the infimum over embeddings proves the lower bound. $\square$