Proof that a metric space $X$ isometrically embeds to the set of bounded continuous functions from $X$ to $\mathbb R$

Question

Given any metric space $(X,d)$ there is an embedding from $X$ to $C_b(X,\mathbb R)$, which is the set of bounded continuous functions from X to $\mathbb R$, with the $\ell_{\infty}$ norm. Given an $x_0\in X$ let a function $G$ be defined where $x\in X$ goes to the function $f_x$ defined as $f_x(y)= d(y,x)-d(y,x_0)$. Showing that $G$ is a function and that for all $x$, $G(x)=f_x$ is a continuous bounded function from $X$ to $\mathbb R$ is easy. What I don't understand is how $G$ is an isometric embeding. My approach was given two $f_x$ and $f_z$ their distance is clearly less or equal to $d(x,z)$. To show they are greater or equal I note that $d(f_x, f_z)$ (defined by the $\ell_\infty$ norm) must be greater than $d(f_x(x),f_z(x))$ (this is in the metric $d$ of $X$) which in turn is equal to $d(x,z)$, Is this correct? Am I missing something?

Answer 1

As in the comment by @PaulSinclair we have $d(f_x,f_z)=\sup_{t\in X}|d(x,t)-d(z,t)|.$

The "subtraction version of the $\triangle$ inequality" is $|d(x,t)-d(y,t)|\le d(x,z)\; (See \;Footnote).$ Therefore $$d(f_x,f_z)\le d(x,z).$$

For any $t\in X$ we have $d(f_x,f_z)\ge |d(x,t)-d(z,t)|.$ In particular when $t=z$ we have $$d(f_x,f_z)\ge |d(x,z)-d(z,z)|=d(x,z).$$

Therefore $d(f_x,f_z)=d(x,z).$

$Footnote.$ $d(x,t)-d(z,t)\le d(x,z)$ because $d(x,t)\le d(x,z)+d(z,t).$ And $d(z,t)-d(x,t)\le d(x,z)$ because $d(z,t)\le d(z,x)+d(x,t)=d(x,z)+d(x,t).$ Therefore $|d(x,t)-d(z,t)|\le d(x,z).$