Is the convex hull operator continuous? Is the convex hull operator continuous?
I am trying to prove that the CONVEX HULL OF a finite union of non-empty convex compact sets is compact. It is easy to prove that the union of compact sets is compact, but less so to prove that the convex hull operator is continuous...
Could someone help?
EDIT: Also, does this result hold only in finite-dimensional spaces?
 A: It follows from Carathéodory's theorem on convex hulls that the convex hull of a compact subset of $\Bbb R^n$ is itself compact (and thus automotically closed.)
EDIT : it's always encouraging to be downvoted without explanation, especially so when the post contains material relevant to the question at hand. So let me expand on why 

Carathéodory's theorem : let $X$ be a subset of $\Bbb R^d$, then for every point $x\in\mathrm{conv}(X)$ in the convex hull of $X$, there are $d+1$ points (not necessarily distinct) $x_0,\dots,x_d\in X$ with $x\in\mathrm{conv}(\lbrace x_0,\dots,x_d\rbrace)$.

is relevant to the problem the OP raises : "I am looking for a proof that that the convex hull (not the closed convex hull) of a compact set is compact"
(see his/her comment on ncmathsadist's answer.)
Let $X\subset \Bbb R^d$ be a compact subset : since we are working in some finite dimensional real vector space, $\mathrm{conv}(X)$ is compact iff it is closed and bounded. Since $X$ is compact, it is bounded, i.e. contained in some ball $B$ (for some norm.) Balls are convex, and the alternate descritpion of $\mathrm{conv}(X)$ as the intersection of all convex sets containing $X$ shows that $\mathrm{conv}(X)\subset B$, and thus
$$\mathrm{conv}(X)\text{ is bounded.}$$
Now suppose $y_n,~n\in\Bbb N$ is a sequence of points in $\mathrm{conv}(X)$ that converges to $y\in\Bbb R^d$. By Carathéodory's theorem, there are, for every $n\in\Bbb N$, $d+1$ points $x_{n,0},\dots,x_{n,d}\in X$ with $y_n\in\mathrm{conv}(\lbrace x_{n,0},\dots,x_{n,d}\rbrace)$ and thus real numbers $t_{n,0},\dots,t_{n,d}\in[0,1]$ with 
$$\sum_{i=0}^d t_{n,i}=1\qquad\text{and}\qquad\sum_{i=0}^d t_{n,i}x_{n,i}=y_n$$
By compactness of $X$ and $[0,1]$ we may find an extraction (i.e. a strictly increasing map $\phi:\Bbb N\to\Bbb N$) such that for all $i\in\lbrace 0,\dots,d\rbrace$ the sequences $\big(x'_{n,i}=x_{\phi(n),i}\big)_{n\in\Bbb N}$ and $\big(t'_{n,i}=t_{\phi(n),i}\big)_{n\in\Bbb N}$ converge to $x'_i\in X$ and $t'_i\in[0,1]$ respectively. It follows then from the continuity of the vector space operations that 
$$\sum_{i=0}^d t_{i}'=1\qquad\text{and}\qquad y'_n=y_{\phi(n)}=\sum_{i=0}^d t'_{n,i}x'_{n,i}\xrightarrow{\:n\to+\infty\:}\sum_{i=0}^d t'_{i}x'_{i}.$$
By uniqueness of limits, $y=\sum_{i=0}^d t'_{i}x'_{i}\in\mathrm{conv}(X)$, and $$\mathrm{conv}(X)\text{ is closed.}$$
A: This result holds even in the case of an arbitrary Banach space, but for the closure of the convex hull of the compact set. The convex hull of a compact set is not necessarily compact:
Counterexample. Let $K=\{e_n/n : n\in\mathbb N\}\cup\{0\}\subset \ell^2(\mathbb N)$, where $e_n=\{\delta_{kn}\}_{k\in\mathbb N}$. Clearly $K$ is compact but $\mathrm{co}(K)$ is not compact!
Closure does take care of compactness:
Theorem. ${\,}$  Let $X$ be a Banach space, $K\subset X$ compact and 
$L$ be the closure of the convex hull of $K$, i.e., $L=\overline{\mathrm{co}(K)}$. Then
$L$ is also compact. 
Proof. Let $\{x_n\}_{n\in\mathbb N}\subset L$. We need to show that $\{x_n\}_{n\in\mathbb N}$ possesses a convergent subsequence, with a limit in $L$. Since
$X$ is a Banach space, $L$ is a complete metric space, as a closed subset of a complete metric space, and thus it suffices to prove that $\{x_n\}_{n\in\mathbb N}$ possesses a Cauchy subsequence. Also $\mathrm{co}(K)$ is dense in $L$,  thus it suffices to show that every sequence $\{x_n\}_{n\in\mathbb N}\subset\mathrm{co}(K)$ possesses a Cauchy subsequence. 
In particular, we shall show that for every $\varepsilon>0$, 
$\{x_n\}_{n\in\mathbb N}$ possesses a subsequence $\{y_n^\varepsilon\}_{n\in\mathbb N}$ 
with the property
$$
\|y_m^\varepsilon-y_n^\varepsilon\| \,<\, \varepsilon,  \qquad\qquad \mathrm{(P)}
$$
for every $\,m,n\in \mathbb N$, and once this is achieved, using a suitable diagonal argument, we can establish the existence of Cauchy subsequence of $\{x_n\}_{n\in\mathbb N}$. 
Let now $\varepsilon>0$. Then due to the compactness of $K$ there exist $z_1,\ldots,z_m\in K$, such that
$$
K \subset \bigcup_{j=1}^m B_{\varepsilon/3}(z_j).
$$
It is not hard to see that for every $x\in\mathrm{co}(K)$, there exists 
$y\in \mathrm{co}\{z_1,\ldots,z_m\}$, such that 
$$
\|x-y\| < \frac{\varepsilon}{3}.
$$ 
Thus, for every term $x_n$ of $\{x_n\}_{n\in\mathbb N}$, there exists
an $y_n\in \mathrm{co}\{z_1,\ldots,z_m\}$, such that 
$\|x_n-y_n\|<\varepsilon$. Meanwhile, $y_n$ can be expressed as a convex combination of
$z_1,\ldots,z_m$, i.e.,
$$
y_n = \lambda_{n,1}z_1+\cdots+\lambda_{n,m}z_m,
$$
where $\lambda_{n,j}\ge 0$ and $\sum_{j=1}^m\lambda_{n,j}=1$.
We can construct a convergent subsequences
$\{\lambda_{n_k,j}\}_{k\in\mathbb N}$, $j=1,\ldots,m$, with
$$
\lim_{k\to\infty}\lambda_{n_k,j} = \lambda_j^*\ge 0,\,\,j=1,\ldots,m,
\quad\text{and}\quad \sum_{j=1}^m \lambda_j^*=1.
$$
Therefore 
$y_{n_k}\to\sum_{j=1}^m \lambda_j^*z_j=y^* \in  
\mathrm{co}\{z_1,\ldots,z_m\}$, ka'i sunep~wc 
$\|y_{n_k}-y^*\|<\varepsilon/3$, for every $k\ge k_0$, for a suitable 
$k_0\!\in\!\mathbb N$. This implies that for every
$k\ge k_0$:
$$
\|x_{n_k}-x_{n_\ell}\| \,\le\, \| y_{n_k}-y_{n_\ell}\|
+\|x_{n_k}-y_{n_k}\|+\|x_{n_\ell}-y_{n_\ell}\| \,<\, 
\frac{\varepsilon}{3}+\frac{\varepsilon}{3}+\frac{\varepsilon}{3} \,=\, \varepsilon.
$$
Indeed, the subsequence $\{y_{n_k}\}_{k\ge k_0}$ 
has the propery $\mathrm{(P)}$. $\Box$ 
A: To expand on my comment:
Let $V$ be a Banach space (not necessarily finite dimensional). (The main properties I will use are that $V$ is a normed vector space complete in the induced metric.) Let $C_1,C_2\subseteq V$ be compact and convex sets. 
Let $C$ be the convex hull of $C_1\cup C_2$. By definition 
$$ x\in C \iff x = \sum_{i = 1}^{n_x} a_i x_i $$
where $a_i\geq 0$ with $\sum a_i = 1$ and $x_i\in C_1\cup C_2$. Without loss of generality we can assume that for $i \in \{ 1,\ldots, N\}$, $x_i\in C_1$, and for the rest $x_i\in C_2$. Since $C_1$ and $C_2$ are convex, we note that
$$ \frac{1}{\sum_{i = 1}^{N} a_i} \sum_{i = 1}^N a_i x_i = c_1$$
represent a convex combination of finitely many points from $C_1$, and so belongs to $C_1$ by assumption of convexity. Similarly 
$$ \frac{1}{\sum_{i = N+1}^{n_x} a_i} \sum_{i = N+1}^{n_x} a_i x_i = c_2 \in C_2 $$
And hence 
$$ x = \lambda_1 c_1 + \lambda_2 c_2 $$
where 
$$ \lambda_1 = \sum_{i = 1}^{N} a_i \qquad \lambda_2 = \sum_{i = N+1}^{n_x} a_i $$
satisfies $\lambda_1 + \lambda_2 = 1$. 
In other words, for every $x \in C$ there exists $t\in [0,1]$ and $c_1\in C_1$ and $c_2\in C_2$ such that $$x = tc_1 + (1-t)c_2 $$
Proposition $C$ is compact. 
Proof: Since $V$ is a metric space, it suffices to show that every sequence in $C$ has a converging subsequence. Let $(x^{(i)})$ be a sequence in $C$. By the construction above there exists sequences
$$ c_1^{(i)} \in C_1 \qquad c_2^{(i)} \in C_2 \qquad t^{(i)}\in [0,1] $$
such that
$$ x^{(i)} = t^{(i)} c_1^{(i)} + (1-t^{(i)}) c_2^{(i)} $$
Compactness of $[0,1]$, $C_1$, $C_2$ implies that we can simultaneously refine the three sequences to a subsequence, which we will abuse notation and still index by $i$, such that 
$$ t^{(i)} \to t^{(\infty)}\in [0,1] \qquad c_1^{(i)} \to c_1^{(\infty)}\in C_1\qquad c_2^{(i)} \to c_2^{(\infty)}\in C_2 $$
Furthermore, since $C_1, C_2$ are compact, they are totally bounded, and hence bounded, and hence we have that their elements $\|c_i\| < M$ for some large $M$. 
Therefore writing $x^{(\infty)} = t^{(\infty)} c_1^{(\infty)} + (1-t^{(\infty)})c_2^{(\infty)}$ we have that
$$ x^{(\infty)} - x^{(i)} = t^{(\infty)} (c_1^{(\infty)} - c_1^{(i)}) - (t^{(i)} - t^{(\infty)}) c_1^{(i)} + (1- t^{(\infty)}) (c_2^{(\infty)} - c_2^{(i)}) + (t^{(i)} - t^{(\infty)}) c_2^{(i)} $$
so by triangle inequality we have that
$$ x^{(i)} \to x^{(\infty)} $$
as desired. Q.E.D.

Remark: the condition that $V$ is Banach can be weakened to $V$ being Frechet with basically the same proof (though one has to be a bit more careful with the step where the distance between $x^{(i)}$ and $x^{(\infty)}$ are treated). I think the proof can also be adapted for topological vector spaces with uniform structures, provided that the uniform structure satisfy certain compatibility conditions (we need that it plays well with the scaling operation), and we work with nets instead of sequences. I'll leave that to interested readers. 
A: This question is a moving target.
Suppose you have a finite collection of compact convex sets; it is compact.  This means it is closed and bounded, so its closed convex hull will be closed and bounded, and therefore compact in a finite-dimensional space.
