# Interior of a convex set is convex [duplicate]

A set $S$ in $\mathbb{R}^n$ is convex if for every pair of points $x,y$ in $S$ and every real $\theta$ where $0 < \theta < 1$, we have $\theta x + (1- \theta) y \in S$.

I'm trying to show that the interior of a convex set is convex.

If $x, y \in$ int $S$, then I know there exists open balls such that $B(x) \subseteq S$ and $B(y) \subseteq S$. I need to show that there exists a ball $B(\theta x + (1- \theta) y) \subseteq S$.

Other then writing down the definitions, I don't really see how to proceed. Could someone give me a hint?

• Consider the map $\lambda \colon [0,\,1]\times B(x)\times B(y) \to \mathbb{R}^n;\;\: \lambda(t,\xi,\eta) = (1-t)\xi + t\eta$. You have $\lambda([0,\,1]\times B(x)\times B(y)) \subset S$ by convexity. You have $\lambda([0,\,1]\times B(x)\times B(y)) = \bigcup\limits_{t\in[0,\,1]} \left((1-t)\cdot B(x) + t\cdot B(y)\right)$. Commented Aug 19, 2013 at 22:54
• @DanielFischer: That would also work for topological vector spaces, wouldn't it? Commented Aug 19, 2013 at 23:04
• Try proving it in $\mathbb R^2$ first. Draw two circles with lines tangent to them. Then you'll probably see the magic behind the proofs given here. Commented Aug 20, 2013 at 1:06

Let $U= S^\circ$. Fix $0<t<1$. We have $tS + (1-t)S \subseteq S$ by convexity, so $tU + (1-t)U \subseteq S$. But $tU$ is open, so $tU+(1-t)U$ is open (exercise, sum of an open set and any set is open), and thus $tU + (1-t)U \subseteq S^\circ = U$, and hence $U$ is convex.
• A set $S$ is open if every point $a$ can be surrounded by a ball $B(a)$ such that $B(a) \subseteq S$. Since adding two sets in $\mathbb{R}^n$ just shifts the points, each point can still be surrounded by an open ball and be contained by the sum. Commented Aug 19, 2013 at 23:21
• Worth mentioning that this proof extends naturally to Topological Vector Spaces. Continuity of scalar mupltiplication implies $tU$ is open and continuity of addition shows $tU+(1-t)U$ is open.
• In the case of topological vector spaces: Note, that $tU+(1-t)U=\bigcup_{x\in (1-t)U}(tU+x)$ is open as union of open sets. Commented Jul 25, 2023 at 10:25