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.
Proof
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:
\(\ds t C + \paren {1 - t} C\) | \(=\) | \(\ds \set {t x + \paren {1 - t} y : x, y \in C}\) | Definition of Linear Combination of Subsets of Vector Space | |||||||||||
\(\ds \) | \(\supseteq\) | \(\ds \set {t x + \paren {1 - t} y : x, y \in C^\circ}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds t C^\circ + \paren {1 - t} C^\circ\) | Definition of Linear Combination of Subsets of Vector Space |
From Set Interior is Largest Open Set we have that $C^\circ$ is open.
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.
From Sum of Set and Open Set in Topological Vector Space is Open, we have that $t C^\circ + \paren {1 - t} C^\circ$ is 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.
$\blacksquare$
Sources
- 1991: Walter Rudin: Functional Analysis (2nd ed.) ... (previous) ... (next): $1.13$: Theorem