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.
=== Corollary ===
== Proof ==
=== 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.
Category:Topological Vector Spaces
Category:Birkhoff-Kakutani Theorem