Say we have a convex dense set $X\subseteq \Bbb{R}^n$, does it follow that $X=\Bbb{R}^n$ ?
For $n=1$ it's true because convex sets of real numbers are intervals, and if it's dense then it must be $\Bbb{R}$. 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.
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 $X$.
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
