Is the convex hull of closed set in $\mathbb R^{n}$ is closed?

Solution 1:

Topologically, the convex hull of an open set is always itself open, and the convex hull of a compact set is always itself compact; however, there exist closed sets that do not have closed convex hulls. For instance, the closed set $$ \left\{(x,y):y\geq\frac{1}{1+x^2}\right\}\subset\mathbb R^2 $$ has the open upper half-plane as its convex hull.

Source: Wikipedia.

Solution 2:

Here is a simple counterexample: Let $S=(\{0\} \times [0,1] ) \cup ([0,\infty)\times \{0\})$. $S$ is closed since it is the union of two closed sets, however, $\operatorname{co} S = ([0,\infty) \times [0,1)) \cup \{ (0,1)\}$, which is clearly not closed.

If the set $S \subset \mathbb{R}^n$ is compact, however, the convex hull is closed. This follows from Carathéodory's theorem which states that any point in $\operatorname{co} S $ can be written in terms of at most $n+1$ elements of $S$ (this is the convex analogue of the fact that any $n+1$ elements of $\mathbb{R}^n$ must be linearly dependent). Hence we can write $\operatorname{co} S = f(S^{n+1} \times \Sigma)$, where $f:\mathbb{R}^n \times \cdots \times \mathbb{R}^n \times \Sigma \to \mathbb{R}^n $ is defined by $f((x_1,...,x_{n+1}, \mu)) = \sum_{k=1}^{n+1} \mu_k x_k$, and $\Sigma$ is the simplex $\Sigma = \{\mu| \mu_k \ge 0, \sum_k \mu_k = 1 \}$. Since $f$ is continuous, $S^{n+1}$ and $\Sigma$ are compact, it follows that $f(S^{n+1} \times \Sigma)$ is compact, hence closed.