Assume we have a non empty, open convex set $ K \subseteq \mathbb{R}^d $. Then $\overline{K}$ is also convex.
How can we prove that statement?
I thought that breaking it into cases would help.
If $x,y \in int(K)=K^o$ then from hypothesis the statement holds.
If $x \in K^o $ and $y \in \partial K$ then from a theorem of my notes :P means $[x,y) \in K $.
I have stacked how to show it with details that for each $x,y \in \overline{K}$ then $[x,y] \in \overline{K}$
Well, let $x, y \in \overline{K}$. By definition there exist sequences $(x_n)_{n \in \mathbb{N}},~(y_n)_{n \in \mathbb{N}} \subseteq \mathrm{int}(K)$ such that $x_n \rightarrow x$ and $y_n \rightarrow y$. Let $\lambda \in [0, 1]$. As $\mathrm{int}(K)$ is convex, we conclude $$ \lambda x_n + (1- \lambda)y_n \in \mathrm{int}(K). $$ But $\lambda x_n + (1- \lambda)y_n \rightarrow \lambda x + (1- \lambda)y$. We know that limits of sequences lie in the closure, so $\lambda x + (1- \lambda)y \in \overline{K}$. This proves that $\overline{K}$ is convex.
Note that the assumption of $K$ being open is unnecessary.
