Interior 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 a convex set.

Then the interior of $C$, $C^\circ$, is convex.

Let $t \in \closedint 0 1$.

Since $C$ is convex, we have:

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

Since $C^\circ \subseteq C$, we have:

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

$=$

$\ds \set {t x + \paren {1 - t} y : x, y \in C}$

$\ds $

$\supseteq$

$\ds \set {t x + \paren {1 - t} y : x, y \in C^\circ}$

$\ds $

$=$

$\ds t C^\circ + \paren {1 - t} C^\circ$

From Dilation of Open Set in Topological Vector Space is Open, we have that $t C^\circ$ and $\paren {1 - t} C^\circ$ are open.

So $t C^\circ + \paren {1 - t} C^\circ$ is an open set contained in $C$, so we have:

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

from the definition of interior.

Since this holds for all $t \in \closedint 0 1$, it follows that $C^\circ$ is convex.

$\blacksquare$