The problem
Given a convex set $S \subset \mathbb{R}^n$. Prove that $\overline{S}$ is also convex.
What i've tried:
To solve this problem i've tried using the definition that $\overline{S}$ is composed by the interior set and the frontier set of $S$. Hence, $S$ is convex, so, the interior set of $S$ is already convex. to prove that $\overline{S}$ is convex we just need to prove that the frontier set of $S$ is convex.
Problem
As you guys can see, i've came here because i can't prove this statement. So i ask you guys: is my way of approaching this problem right?
I assume that $\overline{S}$ denotes the topological closure of $S$. Any element $p$ of $\overline{S}$ can be written as $p=\lim_n x_n$ where $x_n\in S$. Now, let $p=\lim_n x_n$ and $q=\lim_n y_n$ be elements of $\overline{S}$, where $x_n, y_n\in S$. We want to show that $\lambda p+(1-\lambda) q\in \overline{S}$, for $\lambda\in [0,1]$. Now, $\lambda p+(1-\lambda) q=\lim_n \lambda x_n+(1-\lambda)y_n$. By convexity of $S$, we have $\lambda x_n+(1-\lambda)y_n\in S$ for each $n$. Thus $\lambda p+(1-\lambda) q$ is a limit of a sequence of points in $S$, so by definition of the closure it lies in $\overline{S}$, as desored.