Is convex hull of a finite set of points in $\mathbb R^2$ closed? Is the convex hull  of a finite set of points in $\mathbb R^2$ closed? Intuitively, yes. But not sure how to show that. Thanks!
 A: I'd like to generalize this to the statement that if $K$ is a convex compact subset of a topological vector space $V$ and $z$ is some vector, then the convex hull $H$ of $K$ and $z$ is compact.
Proof: Each convex combination $x=\sum_{i=1}^n t_i x_i+sz$ of points in $K\cup\{z\}$, where $x_i\in K$, $t_i,s\ge 0,\ s+t_1+\cdots +t_n=1$, can be written as
$$x=sz+(1-s)k,\qquad k=\sum \frac{t_i}{\sum t_i}x_i$$ where $k\in K$. If we define 
$$h:K\times[0,1]\to V,\qquad (k,s)\mapsto sz+(1-s)k$$
then $h$ is continuous and its image is $H$. Since the domain is compact, $H$ must be compact, too. Hence $H$ is closed.
Now assume by induction that $K$, the convex hull of $n-1$ points is compact, and it follows that the hull of all $n$ points, which is just $H$, is compact.
A: Although the fact needs a proof, the hull of a finite set is also the intersection of all closed half-spaces containing the set. So closed.
A: 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 $\mathbb R^n$.
