Birkhoff-Kakutani Theorem/Topological Vector Space/Corollary

Corollary
Let $\GF \in \set {\R, \C}$.

Let $\struct {X, \tau}$ be a topological vector space over $\GF$.

Then $\struct {X, \tau}$ is metrizable $\struct {X, \tau}$ is first-countable and Hausdorff.

Further, if $\struct {X, \tau}$ is metrizable then there exists an invariant metric $d$ on $X$ such that:
 * $(1) \quad$ $d$ induces $\tau$
 * $(2) \quad$ the open balls in $\struct {X, d}$ are balanced.

Sufficient Condition
Suppose that $\struct {X, \tau}$ is first-countable and Hausdorff.

We replicate the construction in Birkhoff-Kakutani Theorem: Topological Vector Space.

Construct the local basis $\sequence {V_n}_{n \mathop \in \N}$ to have:
 * $V_{n + 1} + V_{n + 1} \subseteq V_n$ for each $n \in \N$.

Let $D$ be the set of real numbers with a terminating binary notation.

For $r \ge 1$, set $\map A r = X$.

For $r \in D$, set:
 * $\ds \map A r = \sum_{j \mathop = 1}^\infty \map {c_j} r V_j$

where $\ds \sum_{j \mathop = 1}^\infty$ denotes linear combination.

Define $f : X \to \hointr 0 \infty$ by:
 * $\map f x = \inf \set {r \in D \cup \hointr 1 \infty : x \in \map A r}$

for each $x \in X$.

Define:
 * $\map d {x, y} = \map f {x - y}$

In Birkhoff-Kakutani Theorem: Topological Vector Space, it is shown that $d$ is an invariant pseudometric and induces $\tau$.

It remains to show that is satisfied.

We have that ${\mathbf 0}_X \in \map A 0 = \set { {\mathbf 0}_X}$.

Hence, we have $\map f { {\mathbf 0}_X} = 0$.

So we have $\map d {x, x} = 0$ for each $x \in X$.

In view of proving that if $\map d {x, y} = 0$, then $x = y$, we prove that if $\map f x = 0$ then $x = {\mathbf 0}_X$.

We have already established that $\map f { {\mathbf 0}_X} = 0$.

Conversely, suppose that $x \ne {\mathbf 0}_X$.

Since $X$ is Hausdorff, there exists an open neighborhood $U$ of ${\mathbf 0}_X$ such that $x \not \in U$.

Since $\sequence {V_n}_{n \mathop \in \N}$ is a local basis for ${\mathbf 0}_X$, there exists $n \in \N$ such that $V_n \subseteq U$.

So, we have $x \not \in V_n = \map A {2^{-n} }$.

Since $\map A {r_1} \subseteq \map A {r_2}$ for $r_1 < r_2$, we have $x \not \in \map A r$ for $r < 2^{-n}$.

So, we have $\map f x \ge 2^{-n}$.

Hence $\map f x \ne 0$.

So we have $\map f x = 0$ $x = {\mathbf 0}_X$.

So we have $\map d {x, y} = \map f {x - y} = 0$ $x = y$.

So $d$ is a metric, and our proof is complete.