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
Note that ${\mathbf 0}_X \in \map A r$ for all $r \in D$.

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

Therefore we have $\map d {x, x} = \map f {x - x} = \map f { {\mathbf 0}_X} = 0$.

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:

Proof of Translation Invariance
Let $x, y, z \in X$.

Then, we have:


 * $\map f {x - y} = \map f {\paren {x + z} - \paren {y + z} } = \map d {x + z, y + z}$

Hence $d$ is an invariant pseudometric.

Proof of $(1)$ and $(2)$
For each $x \in X$ and $r > 0$, let $\map {B_r} x$ be the open ball of radius $r$ and centered at $x$ in the pseudometric space $\struct {X, d}$.

Note that for each $\delta > 0$, we have:
 * $\ds \map {B_\delta} { {\mathbf 0}_X} = \set {x \in X : \map f x < \delta} = \bigcup_{r \in D, \, 0 < r < \delta} \map A r$

We argue that $\map A r$ is open for each $r \in D \setminus \set 0$.

Let $r \in D \setminus \set 0$.

Since $r \ne 0$, there exists $n \in \N$ such that $\map {c_n} r \ne 0$.

Then, we have:
 * $\ds \map A r = \map {c_n} r V_n + \sum_{j \mathop \ne n} \map {c_n} r V_n$

From Dilation of Open Set in Topological Vector Space is Open, $\map {c_n} r V_n$ is open.

From Sum of Set and Open Set in Topological Vector Space is Open, $\map A r$ is therefore open for each $r \in D \setminus \set 0$.

So $\map {B_\delta} { {\mathbf 0}_X}$ is open for each $\delta > 0$.

From Union of Balanced Sets in Vector Space is Balanced, we have that $\map {B_\delta} { {\mathbf 0}_X}$ is balanced, establishing $(2)$.

Note that for each $\delta < 2^{-n}$, we have $\map A \delta \subseteq \map A {2^{-n} }$, so that:
 * $\map {B_{4^{-n} } } { {\mathbf 0}_X} \subseteq \map A {2^{-n} } = V_n$ for each $n \in \N$.

So $\sequence {\map {B_{4^{-n} } } { {\mathbf 0}_X} }_{n \in \N}$ is a local basis for ${\mathbf 0}_X$.

We now move to deduce that every open set in $\struct {X, \tau}$ is open in $\struct {X, d}$.

Let $U$ be an open set in $\struct {X, \tau}$.

From Translation of Local Basis in Topological Vector Space, $\sequence {V_n + x}_{n \mathop \in \N}$ is a local basis for $x$.

For each $y \in U$, there exists $n_y \in \N$ such that $V_{n_y} + y \subseteq U$.

Then, we have:
 * $\ds \bigcup_{y \mathop \in U} \paren {V_{n_y} + y} \subseteq U$

We then have:
 * $\ds \bigcup_{y \mathop \in U} \paren {\map {B_{4^{-n_y} } } { {\mathbf 0}_X} + y} \subseteq U$

From Translation of Open Ball in Invariant Pseudometric on Vector Space, we then obtain:
 * $\ds \bigcup_{y \mathop \in U} \map {B_{4^{-n_y} } } y \subseteq U$

so that:
 * $\ds U = \bigcup_{y \mathop \in U} \map {B_{4^{-n_y} } } y$

Hence $U$ is the union of open sets in $\struct {X, d}$.

Hence we have proved that $\tau$ is precisely the topology induced by $d$, now establishing $(1)$.

Necessary Condition
Suppose that $\struct {X, \tau}$ is pseudometrizable.

Then from Pseudometric Space is First-Countable, $\struct {X, \tau}$ is first-countable.

As shown in the sufficient condition, there then exists invariant pseudometric $d$ on $X$ such that:
 * $(1): \quad$ $d$ induces $\tau$
 * $(2): \quad$ the open balls in $\struct {X, d}$ are balanced.