Why the space of Katětov functions with finite support is separable?

92 Views Asked by At

Let $(X, \rho)$ be a metric space. A function $f: X \rightarrow \mathbb R$ is called Katětov map if $$|f(x) - f(y)| \le \rho(x, y) \le f(x) + f(y) \quad \forall x, y \in X.$$ Denote the space of all Katětov maps on $X$ by $E(X)$. This is a metric space with sup-metric.

For $Y \subset X, f \in E(Y)$ let $\hat f(x) := \sup\limits_{y \in Y} \left(f(y)+\rho(x, y)\right)$; the map $\hat f$ is called the Katětov extension of $f$. We say that $S \subset X$ is a support of $f \in E(X)$ if $f$ is a Katětov extension of $f|_S$. Consider $E(X, \omega) = \{f \in E(X): f \text{ has a finite support}\}$. A Kuratowski map $x \mapsto \rho(\cdot, x)$ embeds $X$ isometrically into $E(X, \omega)$.

I want to know, why if $X$ is separable then $E(X, \omega)$ is separable too. This fact is used in a construction of Urysohn space. I would be very much appreciated if someone could provide me with any hints, proofs or references.