A set $\displaystyle E$ in $\displaystyle R^n$ is called convex if for every pair of points $\displaystyle x$ and $\displaystyle y$ in $\displaystyle E$ and every real $\displaystyle \lambda \in (0, 1)$,we have $\displaystyle [\lambda x + (1- \lambda)y] \in E$

1. Prove that the interior of convex sets is convex.

My attempt: Let $\displaystyle S^{\circ}$ be the set of interior points of $\displaystyle S$. Choose $\displaystyle x \in S^{\circ}$, this means that $\displaystyle \exists r > 0$ such that $\displaystyle B(x;r) \subseteq S$. Now choose two points in this open ball, $\displaystyle y$ and $\displaystyle z,$ since $\displaystyle y, z \in B(x;r)$ they are also interior points of $\displaystyle S$. And since every open ball is convex, $\displaystyle \lambda y + (1-\lambda)z \in B(x;r)$ and thus $\displaystyle \lambda y + (1-\lambda)z \in S^{\circ}$, as required.

I have a problem with that "proof" since I never used the convexity of S, I basically proved that any interior set is convex. Or that every open set is convex. But I can't find where I'm going wrong, or a different proof that makes the use of the convexity of S.

2. Prove that the closure of a convex set is convex.

My attempt, What I have so far: Let $\displaystyle \overline{E}$ be the closure of $\displaystyle E$. Pick arbitrary $\displaystyle x, y \in \overline{E}$, I need to show $\displaystyle \lambda x + (1- \lambda)y \in \overline{E}$.

3 cases:

(i) $\displaystyle x,y \in E$; I showed the implication from here very easily.

(ii) $\displaystyle x,y \in E'$, where $\displaystyle E'$ is the set of all the limit points of $\displaystyle E$.

(iii) Without loss of generality, $\displaystyle x \in E$ and $\displaystyle y \in E'$.

I'm having problem with the last "2" cases as I don't know how to proceed from the fact that something is a limit point to showing that $\displaystyle \lambda x + (1- \lambda)y \in \overline{E}$.

Any suggestions would be appreciated. Thanks, in advance.