Closure of Convex Set in Topological Vector Space is Convex

Theorem

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

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

Let $C \subseteq X$ be convex.

Then the closure $C^-$ of $C$ is convex.

Proof

Let $t \in \closedint 0 1$.

Since $C$ is convex, we have:

$t C + \paren {1 - t} C \subseteq C$

We show that:

$t C^- + \paren {1 - t} C^- \subseteq C^-$

We have:

\(\ds t C^- + \paren {1 - t} C^-\)

\(=\)

\(\ds \paren {t C}^- + \paren {\paren {1 - t} C}^-\)

Dilation of Closure of Set in Topological Vector Space is Closure of Dilation

\(\ds \)

\(\subseteq\)

\(\ds \paren {t C + \paren {1 - t} C}^-\)

Sum of Closures is Subset of Closure of Sum in Topological Vector Space

\(\ds \)

\(\subseteq\)

\(\ds C^-\)

applying Topological Closure of Subset is Subset of Topological Closure since $t C + \paren {1 - t} C \subseteq C$

$\blacksquare$

Also see

• Closure of Convex Subset in Normed Vector Space is Convex

Sources

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