If $S \subseteq \mathbb{R}^n$ is finite, show that conv(S) is a closed set...

If $S \subseteq \mathbb{R}^n$ is finite, show that conv(S) is a closed set. Is the statement still true if S is not finite? Where conv(S) is the convex hull of S.

From what I've read, the convex hull's aren't necessarily closed. Why would finiteness be a sufficient condition for this?

• By the way, this is true for compact $S$, not just finite; this is usually proven as a corollary of Carathéodory's theorem.
– user21467
Nov 17, 2014 at 22:00

Take $S$ to be an open half-plane. Then $\operatorname{Conv}(S) = S$ isn't closed.
Now assume $S \subseteq \mathbb{R}^n$ is finite, say $S = \{s_1,\dots,s_p\}$. Let $\Delta = \{(x_1,\dots,x_p) \mid x_1, \dots, x_p \geq 0, x_1 + \dots + x_p = 1\}$, and define $f: \mathbb{R}^p \to \mathbb{R}^n$ by $f(x_1,\dots,x_p) = x_1s_1 +\dots + x_ps_p$. Then $\operatorname{Conv}(S) = f(\Delta)$. Since $\Delta$ is compact and $f$ continuous (in fact linear), $\operatorname{Conv}(S)$ must be compact.
• Thanks! Should $f: \mathbb{R}^p \rightarrow \mathbb{R}$ ? Nov 13, 2014 at 2:40
• No, $s_1, \dots, s_p$ are points in $\mathbb{R}^n$.
For an infinite set whose convex hull is not closed, consider the set of points $$S=\left\{\left(x,e^{-x^2}\right):x\in\Bbb R\right\}$$ in $\Bbb R^2.$ This example can be readily generalized to $\Bbb R^n$ for any $n\ge 2.$