Birkhoff-Kakutani Theorem/Topological Vector Space

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

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

Then $\struct {X, \tau}$ is pseudometrizable $\struct {X, \tau}$ is first-countable.

Further, if $\struct {X, \tau}$ is pseudometrizable then there exists an invariant pseudometric $d$ 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.

Let $\sequence {U_n}_{n \mathop \in \N}$ be a local basis for ${\mathbf 0}_X$ in $\struct {X, \tau}$.

Let $V_1 = U_1$.

From Open Neighborhood of Point in Topological Vector Space contains Sum of Open Neighborhoods: Corollary 2:
 * for $j \ge 2$, we can inductively pick an open neighborhood $V_j$ of ${\mathbf 0}_X$ such that:
 * $V_j + V_j \subseteq V_{j - 1} \cap U_{j - 1}$
 * so that $V_j + V_j \subseteq V_{j - 1}$.

Since $V_j \subseteq U_j$ for each $j \in \N$, and ${\mathbf 0}_X \in V_j$ for each $j \in \N$, $\sequence {V_n}_{n \mathop \in \N}$ is also a local basis for ${\mathbf 0}_X$ in $\struct {X, \tau}$.

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

That is, the real numbers $r \in \R$ of the form:
 * $\ds r = \sum_{j \mathop = 1}^\infty \map {c_j} r 2^{-j}$

with $\map {c_j} r \in \set {0, 1}$ such that:
 * there exists $N \in \N$ such that $\map {c_j} r = 0$ for $j > N$.

From the Basis Representation Theorem, the coefficients $\map {c_j} r$ uniquely identify $r$.

Note that if $r, s \in D$ and $r + s < 1$, then $r + s \in D$.

Now, 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.

Note that from Linear Combination of Balanced Sets is Balanced, $\map A r$ is balanced for each $r \in D$.

Since $\map A 1 = X$, we have that:
 * $\set {r \in D \cup \hointr 1 \infty : x \in \map A r} \ne \O$

for each $x \in X$.

Hence:
 * $\inf \set {r \in D \cup \hointr 1 \infty : x \in \map A r}$ is finite.

So, we can 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:
 * $\map d {x, y} = \map f {x - y}$

We aim to show that $d$ is a pseudometric.

We require the following lemma.

Lemma
If $r, s \in D$ and $r + s \ge 1$, we have $\map A {r + s} = X$, hence we have the inclusion:
 * $\map A r + \map A s \subseteq \map A {r + s}$

for all $r, s \in D$.

If $r \ge 1$ or $s \ge 1$, then $\map A r = X$ or $\map A s = X$, while $\map A {r + s} = X$ from $r + s \ge 1$.

So we obtain this inclusion for all $r, s \in D \cup \hointr 1 \infty$.

Proof of
We first prove:
 * $\map f {x + y} \le \map f x + \map f y$ for all $x, y \in X$.

Let $\epsilon > 0$.

Using the definition of infimum to pick $r \in r \in D \cup \hointr 1 \infty$ such that:
 * $\map f x \le r \le \map f x + \epsilon$

and $x \in \map A r$.

Similarly pick $s \in r \in D \cup \hointr 1 \infty$ such that:
 * $\map f y \le s \le \map f y + \epsilon$

and $y \in \map A s$.

Since $\map A r + \map A s \subseteq \map A {r + s}$, we have $x + y \in \map A {r + s}$.

Hence, $\map f {x + y} \le r + s$.

That is, $\map f {x + y} \le \map f x + \map f y + 2 \epsilon$.

Since $\epsilon > 0$ was arbitrary, we obtain:
 * $\map f {x + y} \le \map f x + \map f y$

Now let $x, y, z \in X$.

We have:

Hence we have proved for $d$.

Proof of
Recall that $\map A r$ is balanced for each $r \in D$.

From Balanced Set in Vector Space is Symmetric, $\map A r$ is symmetric for each $r \in D$.

That is, for each $r \in D$ and $x \in X$ we have $x \in \map A r$ $-x \in \map A r$.

Hence we have:
 * $\set {r \in D \cup \hointr 1 \infty : x \in \map A r} = \set {r \in D \cup \hointr 1 \infty : -x \in \map A r}$

so that:
 * $\map f x = \map f {-x}$ for each $x \in X$.

Then, for $x, y \in X$ we have: