Convex dense subset of $\Bbb{R}^n$ is the entire space Say we have a convex dense set $X\subset\Bbb{R}^n$, does it follow that $X=\Bbb{R}^n$ ? 

For $n=1$ it's true because convex set of real numbers are intervals, and if it's dense then it's $\Bbb{R}$. 
Anyway, it seems more difficult in the general case, perhaps it's false, I don't know.

Any ideas ?
This question (Why does a convex set have the same interior points as its closure?) shows that the interior of $X $ is the same as the interior of the closure of $X $ (which is all of $\Bbb{R}^n $). Since a set contains it's interior,  we are done.
Let $x \in X$ so we can choose three points that $x$ is inside the triangle(Or in hider dimentions $n+1$ points). We can proxsimate dots from $X$ that $x$ is still in the "triangle". So because $X$ is convex $x \in X$. We proved that $X= \mathbb{R}^n$. Q.E.D