Interior of Balanced Set containing Origin in Topological Vector Space is Balanced
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Theorem
Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a topological vector space over $\Bbb F$.
Let $B \subseteq X$ be a balanced set such that:
- ${\mathbf 0}_X \in B^\circ$
where $B^\circ$ is the interior of $B$.
Then $B^\circ$ is balanced.
Proof
Let $\lambda \in \Bbb F$ have $0 < \cmod \lambda \le 1$.
Then, we have:
- $\lambda B \subseteq B$
From Interior of Subset, we have:
- $\paren {\lambda B}^\circ \subseteq B^\circ$
Then from Dilation of Interior of Set in Topological Vector Space is Interior of Dilation we have:
- $\lambda B^\circ \subseteq B^\circ$
for $\lambda \in \Bbb F$ with $0 < \cmod \lambda \le 1$.
By hypothesis, we have:
- ${\mathbf 0}_X \in B^\circ$
and so:
- $\set { {\mathbf 0}_X} \subseteq B^\circ$
So we have:
- $\lambda B^\circ \subseteq B^\circ$
if $\lambda = 0$.
So:
- $\lambda B^\circ \subseteq B^\circ$
for all $\lambda \in \Bbb F$ with $\cmod \lambda \le 1$, so $B^\circ$ is balanced.
$\blacksquare$
Sources
- 1991: Walter Rudin: Functional Analysis (2nd ed.) ... (previous) ... (next): $1.13$: Theorem