Q:

If $K$ is compact and $(f_k)\subseteq C(K)$ is dense, then $x_n\to x$ in $K$ iff $f_k(x_n)\to f_k(x)$ for each $k$

Let $(K,d)$ be a compact metric space and $(f_k)_{k\in\mathbb N}\subseteq C(K)$ be dense (wrt the supremum norm).

Let $(x_n)_{n\in\mathbb N}\subseteq E$ and $x\in E$. How can we show that $d(x_n,x)\xrightarrow{n\to\infty}0$ iff $f_k(x_n)\xrightarrow{n\to\infty}f_k(x)$ for each $k\in\mathbb N$?

The "only if" part is trivial, but how can we show the converse? Moreover, I would like to conclude that if $(a_n)_{n\in\mathbb N}\subseteq(0,\infty)$ with $\sum_{n\in\mathbb N}a_n<\infty$, then $$\rho(x,y):=\sum_{k\in\mathbb N}a_n(|f_k(x)-f_k(y)|\wedge1)\;\;\;\text{for }x,y\in E$$ is a metric on $K$ equivalent to $d$? (And I read that since the identity mapping is uniformly continuous with respect to $d$ and $\rho$, we may assume that $d=\rho$ ... I don't get that.)

A:

Assume that $(x_n)$ does not converg to $x$. Since $K$ is compact then $(x_n)$ has a subsequence say $(a_n)\subseteq(x_n)$ convergent to $a\neq x$. By continuity of any $f_k$ we have $f_k(a_n)\to f_k(a)$. So all we need now is to find $f_k$ such that $f_k(a)\neq f_k(x)$ to obtain a contradiction with uniqueness of the limit (among subsequences).
So assume that $f_k(a)=f_k(x)$ for all $k$. By the Tietze extension theorem there is continuous $g:K\to\mathbb{R}$ such that $g(a)=0$ and $g(x)=1$. The point is that $g(a)\neq g(x)$. And therefore it cannot be a limit of $f_k$ contradicting them being dense.

