conv(A) equals to the intersection of all convex sets containing A

Question

Let A be a finite set, and let $ conv(A)=\{ \sum _{i=1}^n\lambda_i v_i|\sum _{i=1}^n\lambda_i=1, \space \lambda_i\ge 0,v_i\in A\}$.

I wish to prove that, as written in the title, conv(A) equals to the intersection of all convex sets containing A.

It's very intuitive why it is true, but i have no clue how to prove it.

I'm trying to prove via containment in both directions, But i've encountered the following difficulties -

1. proving that a given x in the intersection is in conv(A): ok, so i have a x. If x is in A, we're done. If x is not in A, we have to show some how that it's imposible? How?
2. proving that a given x in conv(A) is in the intersection : ok, so i have a x as a convex combination of the members of A. How do i prove that x must be in the intersection?

Answer 1

This is probably too late but let me answer it anyway.

Assume some set $S$ is a convex set containing all the $v_i$'s in $A$ and assume further the element $\sum_i t_i v_i$ is not in $S$. If $t_0=1$, we get a contradiction. If $t_0\neq 1$ we can write $\sum_{i=0}^n t_i v_i=t_0 v_0+(1-t_0)\sum_{i=1}^n\frac{t_i}{1-t_0}v_i$ which means $\sum_{i=0}^n t_i v_i$ is in the line segment connecting $v_0$ and $\sum_{i=1}^n\frac{t_i}{1-t_0}v_i$. If $\sum_{i=1}^n\frac{t_i}{1-t_0}v_i\in S$, by definition of a convex set,$\sum_{i=0}^n t_i v_i$ must also be in $S$ so our assumption indicates $\sum_{i=1}^n\frac{t_i}{1-t_0}v_i\notin S$. Proceed inductively, we will end up claiming one of the $v_i$'s is not in $S$, which is a contradiction.

Answer 2

Prove that it is convex. And prove that it is a subset of every convex set that contains $A$.

Both steps are quite straightforward.