# Confused by proof in Rudin Functional Analysis, metrization of topological vector space with countable local base

I'm working through Rudin's Functional Analysis, and I am confused by a step in his proof for Theorem 1.24, which states that if X is a topological vector space with a countable local base, then there is a metric d on X such that a) d is compatible with the topology of X b) the open balls are balanced c) d is invariant and, if X is locally convex, then all open balls are convex.

Here is a slimmed down version of his proof:

By his theorem 1.14, X has a balanced local base $\{V_n\}$ such that $V_{n+1} + V_{n+1} + V_{n+1} + V_{n+1} \subset V_n$.

D is the set of all rational numbers r of the form $\Sigma_{n=1}^{\infty} c_n(r) 2^{-n}$, where each of the digits $c_i(r)$ is 0 or 1 and only finitely many are 1. Define A(r) = X if $r \geq 1$. For any r in D, define A(r) = $\Sigma c_n(r)V_n$. (Note each is a finite sum). Define $f(x) = \inf \{r : x \in A(r)\}$ for $x \in X$ and d(x,y) = f(x - y). Rudin proves an inclusion $A(r) + A(s) \subset A(r + s)$ and uses it to show that $\{A(r)\}$ is totally ordered by set inclusion and that $f(x + y) \leq f(x) + f(y)$.

As each A(r) is balanced, $f(x) = f(-x)$. $f(0) = 0$. If x$\not = 0$, then $x \not \in V_n = A(2^{-n})$ for some n, so $f(x) \geq 2^{-n} > 0$.