A convex set in $\mathbb{R}^n$ whose closure is $\mathbb{R}^n$ Let $S$ be a convex set in $\mathbb{R}^n$. Prove that if a closure of S is $\mathbb{R}^n$ then $S=\mathbb{R}^n$
 A: I'm assuming you mean the standard topology on $\mathbb{R}^n$. Note that what you want to prove is equivalent to the following: 

If $S$ is convex and different of $\mathbb{R}^n$, then it's closure also is different of $\mathbb{R}^n$.

To prove this, let $p\in\mathbb{R}^n$ be such that $p\notin S$. Now we use the Hyperplane Separating Theorem to conclude that there is a point $y\in\mathbb{R}^n$, $y\neq 0$, and $\lambda\in\mathbb{R}$, such that $$\langle x,y\rangle < \lambda < \langle p,y\rangle,\quad \forall x\in S. $$
It means there is a hyperplane separating $p$ and $S$. So $S$ is totally contained in a half plane, therefore it's closure can't be $\mathbb{R}^n$.
A: Here's my (not very formal) idea, which I hope is useful and clear: assume otherwise, that there is some $a\in\mathbb{R}^n\setminus S$. Consider all of the lines which pass through $a$.


*

*Since $S$ is convex and $a\notin S$, we get that for each line passing through $a$ one of its parts (before $a$ or after $a$) cannot intersect $S$ (otherwise, $a\in S$).

*In addition, if $\ell_1$ and $\ell_2$ intersect $S$, then every line between them also intersects $S$.


Assuming my intuition is correct, we get that there is a half plane which does not intersect $S$ -- and thus the closure of $S$ cannot be $\mathbb{R}^n$.
