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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?

p-value
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2 Answers2

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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.

orangeskid
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  • It is not letting me edit the above messed up comment. Anyway, I agree that $C\subseteq\text{aff}({x_0,\dots, x_d})$. However, I believe that the statement that $C$ lies in the subspace generated by them is not correct. This is to say that $C$ is a subset of the $d$-dimensional subspace $\text{aff}({x_0,\dots, x_d})-x_0=:V$, which I don't think is true per se. However, I believe you can get around this, for if $d<n$ then $V$ is contained in some half space $H$, so that $C\subseteq V+x_0\subseteq H+x_0$, and the latter object $H_x_0$ is still a halfspace. – Satana Jun 02 '17 at 21:41
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    @Atsana I believe the original proof works as intended; "subspace" here refers to the "affine subspace" $\mathrm{aff}{x_0,\dots,x_d}$, not a "linear subspace" – Benjamin Bray Jun 12 '17 at 18:20
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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.

Tsemo Aristide
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