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?
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.