I am having trouble understanding the proof for:
4.3.1 Lemma. Convex hull $C$ of a set $X \subseteq \Re $ equals the set:
$$D= \left\{ \sum_{i=1}^{m}{t_i x_i} : m \geq 1, x_1…x_m \in X, t_1,…,t_m \geq 0, \sum_{t=1}^{m}{t_i} = 1 \right\}$$
of all convex combinations of finitely many points of X.
For the direction $C\subseteq D$, the authors say:
For the reverse inclusion it suffices to prove that $D$ is convex, that is, to verify that whenever $x,y \in D$ are two convex combinations and $t\in (0,1)$, then $tx + (1-t)y$ is again a convex combination.
Why does it suffice to prove that $D$ is convex? Shouldn't we prove that any point in $C$ is in $D$?
P.S. And what would it mean for a point to be in $C$ – which is convex hull? I find that confusing…
Best Answer
On drozzy's request I'm posting my comment as an answer:
By definition the convex hull $C$ of $X$ is the intersection of all convex sets $C'$ containing $X$: $$C = \bigcap_{\substack{C' \text{ convex} \\ C' \supset X}} C'.$$ If you know that $D$ is convex (that's what the authors show later on and needs some argument) and contains $X$ (that's obvious by taking $m = 1$, $t_1 = 1$ and $x_1 = x$ for each $x \in X$) then you know that $D$ will appear as some $C'$ in the intersection on the right hand side, so that $C \subset D$.
For the sake of completeness, the reasoning for the other inclusion $D \subset C$ is this: since all $C'$ appearing in the intersection are convex and contain $X$, they must contain all convex combinations of points of $X$. But $D$ is by its very description the set of all convex combinations of points of $X$, so $D$ is contained in all $C'$ appearing in the intersection and thus $D \subset C$.