Show that $\phi$ is an isometry

Question

The following Lemma can be obtained in 'Geometric Nonlinear Functional Analysis' by Benyamini and Lindenstrauss, page $2.$

Lemma: Every metric space $X$ is isometric to a subset of $\ell_{\infty}(\Gamma)$ for some set $\Gamma.$

Proof: Fix any $x_0 \in X,$ and take $\Gamma$ to be the set $X$ itself. The embedding $\phi:X \rightarrow \ell_{\infty}(\Gamma)$ is defined by $$\phi(x)(y) = d(x,y) - d(x_0,y).$$

Question: How to show that $\phi$ is an isometry between $X$ and $\ell_{\infty}(\Gamma)?$

We want to show that $ d(\phi(x), \phi(z)) = d(x,z)$ for every $x,z \in X.$

Observe that for each $x \in X,$ $$\sup_{y \in X}|\phi(x)(y)| = \sup_{y \in X}|d(x,y)-d(x_0,y)| \leq \sup_{y \in X}|d(x,x_0)| = |d(x,x_0)|.$$ I do not know how to continue.

Any hint is appreciated.

Answer 1

Note that: $$\sup_{y\in X}| d(x,y) - d(x_0,y) - d(z,y)+d(x_0,y)|=\sup_{y\in X}|d(x,y)-d(z,y)|≥|d(x,z)-d(z,z)|=d(x,z)$$ and via the lower triangle inequality you have $|d(x,y)-d(z,y)|≤d(x,z)$ for any $y$. So $$d(\phi(x),\phi(z))=\|\phi(x)-\phi(z)\|_\infty = d(x,z)$$