Density of the function class Let $X$ be any set and let $[0,1]^X$ (the class of all functions $X\to[0,1]$) be endowed with the metric given by $\rho(f,g):=\sup_{x\in X}|f(x) - g(x)|$. Consider any class of functions $\mathscr F\subseteq [0,1]^X$. Define:
$$
  d_\mathscr F(x,y):=\sup_{f\in \mathscr F}|f(x) - f(y)|.
$$
Clearly, $d_{\mathscr F}$ is a pseudo-metric on $X$, that is it is symmetric and satisfies the triangular inequality, but $d_{\mathscr F}(x,y) = 0$ does not necessarily imply $x = y$. Define the corresponding Lipshitz functions:
$$
  \mathrm{Lip}(d_{\mathscr F}):=\left\{f\in [0,1]^X:\sup_{d_\mathscr F(x,y)\neq 0}\frac{|f(x) - f(y)|}{d_{\mathscr F}(x,y)}\leq 1\right\}.
$$
Is it always true that $\mathscr F$ is dense in $(\mathrm{Lip}(d_\mathscr F),\rho)$?

In general, $\mathscr{F}$ is not dense in $\operatorname{Lip}(d_\mathscr{F})$. As an extreme case, consider $X = [0,1]$ and $\mathscr{F} = \{\operatorname{id}_{[0,1]}\}$. Then $d_\mathscr{F}$ is the usual distance, and $\mathscr{F}$ is closed in $\operatorname{Lip}(d_\mathscr{F})$. Since $\operatorname{Lip}(d_\mathscr{F})$ always contains the constants, no finite family can be dense.
