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We have a Convex hull of a set $X\subseteq R^{n}$ defined as $C$, we need to prove that $C$ can we written as the following: $$\bar{C}=\sum_{i=1}^mt_ix_i$$

where $m\geq 1,t_i\geq0, x_1,x_2,....,x_m\in X$ and $\sum_{i=1}^mt_i=1$.

So, we need to prove that $C=\bar{C}$.

We do that by proving $\bar{C}\subseteq C$ and $C\subseteq \bar{C}$.

I understood the first part, but got stuck at the second one.

So, in order to prove $C\subseteq \bar{C}$, the proof says, we need to prove that $\bar{C}$ is convex.

That actually makes sense, since we know $C$ is the convex hull and a convex hull is the intersection of all convex sets that contain $X$. That is, $C$ is contained in all convex sets that contain $X$.

But I don't see how $\bar{C}$ could contain $X$. If we take $m=1$, it could only contain a subset $\{x_1,x_2,.....,x_m\}$ of $X$ and not the whole set.

Am I missing something ? Kindly help !

User9523
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    In this sort of problem, you usually have either $X={x_1,\ldots,x_m}$, or $\bar C = { \sum_{i=1}^m t_ix_i \mid m\ge 1,\ t_i\ge 0,\ x_i\in X,\ \sum_{i=1}^m t_i=1}$. If neither of those apply, you can very well have an extremal point of $C$ that doesn't belong to $\bar C$, as you pointed out. Edit: what I meant in the second case is that $m\ge 1$ can be freely chosen in $\mathbb N$, which makes $\bar C$ the set of all convex combinations of $X$. – N.Bach Oct 19 '18 at 22:26
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    The $x_k$ are not a fixed collection of points in $X$. They are an arbitrary selection of points in $X$. So $X \subset \overline{C}$ is immediate. (Aside, the notation conflicts with the usual notation for set closure.) – copper.hat Oct 19 '18 at 22:58

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$\overset {-} C$ contains all points of the form $\sum_{i=1}^{m} t_i x_i$ with $m \geq 1, t_i \geq o, x_i \in X$. All quantities here are variables, nothing is fixed. It is obvious that every $x \in X$ can be written in above form by taking $m=1,x_1=x,t_1=1$ so $\overset {-} C$ contains every point of $X$. ($x_1,x_2,..,x_m$ are not fixed elements of $X$).

Kavi Rama Murthy
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