Prove that the convex hull of a set is the smallest convex set containing that set

Question

How do you prove that the convex hull of A is the smallest convex set containing A?

edit: definition of a convex hull: Given a set A ⊆ ℝⁿ the set of all convex combinations of points from A is called the convex hull of A.

What is your definition of the convex hull? And don’t you mean the smallest convex set containing A? — Brian M. Scott, Oct 5 '11 at 2:54

Nathan Portland · Accepted Answer · 2011-10-05 02:57:43Z

4

Young, when you wrote

How do you prove that the convex hull of A is the smallest set containing A?

You meant that convex hull of A is the minimal convex set containing A, right?

To show this, which part is your definition? The linear-algebraic characterization?

You can see that any intersection of convex sets containing A is also a convex set containing A.

answered Oct 5 '11 at 2:57

Nathan Portland

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$\begingroup$ Yea, the convex hull of A is the minimal convex set containing A. I guess what I need to prove is that the convex hull of A is itself a convex set containing A. How do I prove that or is it just by definition of a convex hull? $\endgroup$ – jeremy Oct 5 '11 at 3:55
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$\begingroup$ @Young: Let $\mathscr{C}$ be the collection of all convex sets containing $A$, and let $K=\bigcap\limits_{C\in\mathscr{C}}C$; certainly $K\supseteq A$, and you need to show that $K$ is convex. Let $x,y\in K$ and $t\in[0,1]$; you need to show that $tx+(1-t)y\in K$. Use the facts that (1) $x,y\in C$ for every $C\in\mathscr{C}$ and (2) every $C\in\mathscr{C}$ is convex. $\endgroup$ – Brian M. Scott Oct 5 '11 at 4:05

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