[Math] Is the convex hull of closed set in $\mathbb R^{n}$ is closed
convex-analysisconvex-hulls
Is convex hull of closed set in $\mathbb R^{n}$ closed?
Best Answer
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.
Generally, the convex hull $\{\lambda_1x_1+\dots+\lambda_nx_n\colon0\le\lambda_1,\dots,\lambda_n\le1,\lambda_1+\dotsb+\lambda_n=1\}$ of $x_1,\dotsc,x_n\in\mathbb R^d$ is compact, hence closed, since it's the continuous image of the closed simplex $\{(\lambda_1,\dotsc,\lambda_n)\colon0\le\lambda_1,\dots,\lambda_n\le1,\lambda_1+\dotsb+\lambda_n=1\}\subseteq\mathbb R^n$.
To fill in the details, you can prove by induction that the closed simplex is really compact as Hamcke mentioned in the comment, but you can do it simpler: as a subspace of $\mathbb R^n$, it's obviously bounded and it could be written as an intersection of a (in fact finite) collection of closed subsets of $\mathbb R^n$.
Best Answer
Source: Wikipedia.