Convex Set with Empty Interior Lies in an Affine Set In Section 2.5.2 of the book Convex Optimization by Boyd and Vandenberghe, the authors mentioned without proving that "a convex set in $\mathbb{R}^n$ with empty interior must lie in an affine set of dimension less than $n$." While I can intuitively understand this result, I was wondering how it can be proved formally?
 A: Look at $d+1$,  the largest number of affinely independent points from $C$. Let $x_0$, $\ldots$, $x_d$ one such affinely independent subset of largest size. Note that every other point is an affine combination of the points $x_k$, so lies in the affine subspace generated by them, which is of dimension $d$.  
If $d < n$ then this subspace is contained in an affine hyperplane. 
If $d=n$, then $C$ contains $d+1$ affinely independent points. Since $C$ is convex, it will also contain the convex hull of those $n+1$ points. Now, in an $n$-dimensional space the convex hull of $n+1$ affinely independent points has non-empty interior. So the interior of $C$ is non-empty. 
A: Let $C$ be a convex with empty interior of $R^n$. Suppose that $C$ contains two points $x_1,x_2$ then it contains the segment $[x_1,x_2]$, if $C$ is not contained in the affine line $D_2$ which contains $x_1,x_2$ then there exists an element $x_3$ of $C$ not in $D_2$, thus the 2 simplex $[x_1,x_2,x_3]$ is contained in $C$, recursively suppose constructed a $i-1$-simplex $[x_1,...,x_i]\subset C$, if $C$ is not contained in the affine $i-1$-plane $D_i$ which contains $[x_1,...,x_i]$, you have $x_{i+1}\in C$ not in the $i-1$-plan $D_i$, thus $C$ contains $[x_1,....,x_{i+1}]$. This shows that if $C$ is not contained in an affine subspace of $R^n$ distinct of $R^n$, then it contains a $n$ simplex $[x_1,...,x_{n+1}]$ so its interior is not empty.
