Q:

How to prove that a convex set is closed under infinite convex combinations? 

Suppose $C$ is a convex set, and $x_1, x_2, ... \in C$.   Suppose further that $\sum_{i=1}^\infty a_i=1$, where $a_i\ge 0$.  How do we show that $\sum_{i=1}^\infty a_ix_i \in C$?   (Suppose the series converges.)   Thank you! 

A:

This is not true.  For instance, if we work in the spare of square-summable sequences $\ell^2(\mathbb N)$, we note that it has as a subspace the set of sequences which have only finitely many non-zero terms.  Being a subspace, this set is obviously convex.  However, if we let $e_i$ be the sequence with a $1$ at the $i^{th}$ position and a $0$ everywhere else, we find that $\frac{1}2e_1+\frac{1}4e_2+\frac{1}8e_3+\ldots$ converges in $\ell^2$, but does not converge within the given subspace as it has infinitely many non-zero terms, despite every term $e_i$ being in the subspace. 
If you restrict that $C$ is closed as well, this holds rather trivially, as one can see that the sequence
$$x_n=\frac{\sum_{i=1}^na_ix_i}{\sum_{i=1}^na_i}$$
converges to $\sum_{i=1}^n a_ix_i$ whenever that exists and is a sequence of convex combinations.