Closure of Balanced Set in Topological Vector Space is Balanced

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.

Then the closure $B^-$ of $B$ is balanced.

Proof

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

Then, we have:

$\lambda B \subseteq B$

So, from Topological Closure of Subset is Subset of Topological Closure we have:

$\paren {\lambda B}^- \subseteq B^-$

From Dilation of Closure of Set in Topological Vector Space is Closure of Dilation we have $\paren {\lambda B}^- = \lambda B^-$ and so:

$\lambda B^- \subseteq B^-$

So $B^-$ is balanced.

$\blacksquare$

Sources

• 1991: Walter Rudin: Functional Analysis (2nd ed.) ... (previous) ... (next): $1.13$: Theorem