Convex hull approximated from inside by only finite number of elements? In approximating the convex hull "from inside", i.e.

$$ \text{conv}S = \{ x \in \mathbb{R}^n \mid x= \sum_{i=1}^k \lambda_i x^i, x^i \in S, \lambda_i \geq 0, \sum_{i=1}^k \lambda_i= 1 \} \text{,}$$

in the case where $S$ is infinite - why can we restrict ourselves to finite many $x^i \in S$?
 A: I assume you are really asking this: Let $S$ be an infinite subset of $\mathbb{R}^n$. Then why is the convex hull of $S$, denoted $\operatorname{conv}S$, defined as you show it, using only finite convex combinations of elements of $S$?
The answer is that the set defined this way really is convex (a convex combination of two finite convex combination is another finite convex combination), and this is the smallest convex set containing $S$ (since any convex set containing $S$ must contain any finite convex combination of members of $S$).
I guess the main point is that the definition of convexity nowhere speaks of inifinite convex combinations.
As pointed out in the comments, Carathéodory's theorem is worth mentioning: It states that you never need to take a convex combination of more than $n+1$ of the points if you work in $\mathbb{R}^n$.
There is an analogy with linear combinations: The linear span of an infinite set of vectors consists of finite linear combinations of the given vectors.
