Rudin's functional analysis, section 3.8 (c) proof.

Question

From Rudin's functional analysis.

If $X$ is a compact topological space and if some sequence $\left\{ f_n \right\}$ of continuous real-valued functions separates points on $X$, then $X$ is metrizable.

In this proposition there's the following function

$$ d(p,q) = \sum_{n=1}^{\infty}2^{-n} \left| f_n(p) - f_n(q) \right| $$

Apparently the fact $d$ is a metric follows from the fact the family $\left\{ f_n \right\}$ separates point on $X$, how is this used?

Answer 1

One of the defining properties of a metric is that for a given pair $p\ne q$, we must have $d(p,q)>0$.

Now, since $\{f_n\}_n$ separates points of $X$, we can find $f_k$ such that $f_k(p)\ne f_k(q)$. This implies that $$ |f_k(p)- f_k(q)|>0 $$ and hence $d(p,q)>0$.

Answer 2

You need to know that the $f_n$ separate points to show that $d(p,q)>0$ if $p\ne q$.