TITLE: Convex dense subset of $\Bbb{R}^n$ is the entire space
QUESTION [7 upvotes]: 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 ?

REPLY [2 votes]: Using induction over the dimension $n$, it is easy to show that, given any set of $2^n$ points meeting every quadrant of $\Bbb R^n$, the zero vector is in their convex hull.
Let $Q_i$ denote the $i$-th quadrant for $i=1,\dots,2^n$. Assume $a$ is a point in $\Bbb R^n$. Since $a+Q_i$ is open, we can pick a point $x_i$ from $X\cap(a+Q_i)$. Now since $a$ is in the convex hull of these $x_i$ and $X$ is convex, $a$ must be in $X$.