Open Neighborhood of Origin in Topological Vector Space contains Balanced Open Neighborhood

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

Let $X$ be a topological vector space over $\Bbb F$.

Let $U$ be an open neighborhood of ${\mathbf 0}_X$.

Then there exists a balanced open neighborhood $W$ of ${\mathbf 0}_X$ with $W \subseteq U$.

Proof
Equip $\Bbb F \times X$ with the product topology.

From Box Topology on Finite Product Space is Product Topology, this is precisely the box topology.

Define $M : \Bbb F \times X \to X$ by:


 * $\map M {\lambda, x} = \lambda x$

for each $\tuple {\lambda, x} \in \Bbb F \times X$.

From the definition of a topological vector space, $M$ is continuous.

In particular, $M$ is continuous at $\tuple {0, {\mathbf 0}_X}$.

So there exists an open neighborhood $A$ of $\tuple {0, {\mathbf 0}_X}$ such that:


 * $M \sqbrk A \subseteq U$

From Basis for Box Topology and the definition of a topology generated by a synthetic basis, there exists an open neighborhood $D$ of $0$ and an open neighborhood $V$ of ${\mathbf 0}_X$ such that:


 * $D \times V \subseteq A$

Using:


 * the definition of an open set in $\R$ if $\Bbb F = \R$
 * the definition of an open set in $\C$ if $\Bbb F = \C$

there exists $r > 0$ such that if $\alpha \in \Bbb F$ has $\cmod \alpha < r$ we have $\alpha \in D$.

Then for $\alpha \in \Bbb F$ with $\cmod \alpha < r$ we have:


 * $\set \alpha \times V \subseteq D \times V$

so that:


 * $M \sqbrk {\set \alpha \times V} \subseteq U$

from Image of Subset under Mapping is Subset of Image.

We have:


 * $M \sqbrk {\set \alpha \times V} = \alpha V$

so we have:


 * $\alpha V \subseteq U$

for each $\alpha \in \Bbb F$ with $\cmod \alpha < r$.

Let:


 * $\ds W = \bigcup_{\cmod \alpha < r} \alpha V$

From Dilation of Open Set in Topological Vector Space is Open, $\alpha V$ is open for each $\alpha \in \Bbb F$.

Since $W$ is the union of open sets, $W$ is open.

From Union of Subsets is Subset, we have $W \subseteq U$.

Since $0 \in \alpha V$ for all $\alpha \in \Bbb F$ with $\cmod \alpha < r$, we also have $0 \in W$.

So $W$ is an open neighborhood of $0$.

It remains to show that $W$ is balanced.

Let $x \in W$ and $\lambda \in \Bbb F$ have $\cmod \lambda \le 1$.

Then there exists $\alpha \in \Bbb F$ with $\cmod \alpha < r$ such that $x \in \alpha V$.

Then we have $\lambda x \in \paren {\alpha \lambda} V$.

We have $\cmod {\alpha \lambda} < r$, so $\paren {\alpha \lambda} V \subseteq W$.

So we have $\lambda x \in W$.

So we have $\lambda W \subseteq W$ for all $\lambda \in \Bbb F$ with $\cmod \lambda \le 1$, so $W$ is balanced.