Convex hull of the union of compact and convex sets

Question

Let X be a normed space and $A,B \subset X$.

I need to prove that if $A$ and $B$ are compact and convex then $conv(A \cup B)$ is compact. (Here $conv(A \cup B)$ is the convex hull of $A \cup B$)

If $U_{i \in I}$ is a finite cover for $A$ and $V_{j \in J}$ is a finite cover for $B$, how do I cover the elements of $z \in conv(A \cup B)$ such that $z \notin A$ and $z \notin B$?

I tought to pick one element of each $U_{i}$ and $V_{j}$ and for each pair $(u_{i},v_{j})$, the number of pairs $(u_{i},v_{j})$ is finite, create a line segment. Then at each midpoint create an open ball that covers the entire line. But I can't prove that this will cover all that points in the convex hull and I think that this will not be the right path.

Answer 1

I think you can consider the function $f : A \times B \times [0,1] \rightarrow \mathrm{Conv}(A \cup B)$ defined by $$\forall (a,b,t) \in A \times B \times [0,1], \quad f(a,b,t)=ta+(1-t)b$$

I think that $f$ is surjective (because $A$ and $B$ are convex), and that it is continuous on the compact $A \times B \times [0,1]$ (for the natural topology), so its image is compact.

Answer 2

First notice that $$\operatorname{Conv}(A \cup B) = \{(1-t)a+tb : t \in [0,1], a\in A, b \in B\}$$ Clearly $\supseteq$ holds. Conversely, take $x \in \operatorname{Conv}(A \cup B)$ and write it as $$x = \sum_{i=1}^n \alpha_ia_i + \sum_{j=1}^m \beta_jb_j$$ for some $a_1, \ldots, a_n \in A$, $b_1, \ldots, b_m\in B$, $\alpha_1, \ldots, \alpha_n, \beta_1,\ldots\beta_m \ge 0$ and $\sum_{i=1}^n \alpha_i + \sum_{j=1}^n \beta_j = 1$. We have $$x = \sum_{i=1}^n \alpha_ia_i + \sum_{j=1}^m \beta_jb_j = \left(1 - \sum_{j=1}^m \beta_j\right)\underbrace{\sum_{i=1}^n \frac{\alpha_i}{1 - \sum_{j=1}^m \beta_j}a_i}_{\in A} + \underbrace{\left(\sum_{j=1}^m \beta_j\right)}_{\in[0,1]}\underbrace{\sum_{j=1}^m \frac{\beta_j}{\sum_{j=1}^m \beta_j}b_j}_{\in B}$$ since $A$ and $B$ are convex. Hence $\subseteq$ follows.

Now you can easily show that $\operatorname{Conv}(A \cup B)$ is sequentially compact: let $(x_n)_n$ be a sequence in $\operatorname{Conv}(A \cup B)$. Hence $$x_n = (1-t_n)a_n + t_nb_n, \text{ for some } t_n \in [0,1], a_n \in A, b_n \in B$$ Since $[0,1],A,B$ are all compact, we can extract subsequences $(t_{p(n)})_n, (a_{p(n)})_n, (b_{p(n)})_n$ which converge to $t_0 \in [0,1], a_0\in A, b_0 \in B$.

Therefore $x_{p(n)} \to (1-t_0)a_0 + t_0b_0 \in \operatorname{Conv}(A \cup B)$ which proves the claim.

Answer 3

A more general version of this claim is proved in Aliprantis-Border (2006, Lemma 5.29, page 183, Infinite Dimensional Analysis) for an arbitrary topological vector space.