Prove that the following two definitions of the convex hull are equivalent. I was wondering if a topology expert could help me solve this proof, as I have no idea but want to understand these concepts.  This is not for homework.
Let X be a point set, not necessarily finite, in R^d.  Prove the following two definitions of the convex hull are equivalent. 


*

*The set of all points that are convex combinations of all points in X.

*The intersection of all convex sets that include X.
http://en.wikipedia.org/wiki/Convex_hull 
http://en.wikipedia.org/wiki/Convex_combination
I can visualize in my head why this is true, but cannot convert it to mathematical terms.  I would really appreciate any expert's help on this, as it is very simple but I have taking it away from just geometry, to a general proof.
 A: There is a standard procedure when showing equalities like these: Let $A$ be the set, and define $\text{Conv}(A)$ as the set of all convex combinations of $A.$ You want to show that $\text{Conv}(A)=\bigcap\mathcal C$ where $\mathcal C=\{C\mid C\text{ is a convex superset of }A\}.$ So you should show that $\text{Conv}(A)$ is itself convex and contains $A,$ that tells you that $\bigcap\mathcal C\subseteq \text{Conv}(A).$ Then show that $\text{Conv}(A)$ is in each convex superset of $A,$ that tells you that $\text{Conv}(A)\subseteq\bigcap\mathcal C.$
The second inclusion is harder than the first. Let $B$ be a convex set containing $A.$ Use induction: Assume that for $n\in\mathbb N$ all convex combinations $\sum_{i=0}^n t_i x_i,\ t_i\ge 0,\ \sum t_i=1$ are in $B$, then try to express $\sum_{i=1}^{n+1}s_i x_i$ as $tx_{n+1}+(1-t)\sum_{i=0}^n t_i x_i$ for some $t\in[0,1]$.
A: I'm not an expert, but still hope to be capable of answering.
The intersection of all convex supersets is the same as the smallest convex superset (why is it convex?). So it is enough to see that the set from the first definition is convex (obvious), "superset" (obvious) and "the smallest" (for example by induction on the number of nontrivial elements in the convex combination).
