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.
A: This does hold in $\mathbb R^n$, as we can show by induction (the base case $n=0$ being trivial). Let $x=\sum_{i=1}^\infty a_ix_i$, and let
$$
  V=\mathbb R_{>0}(C-x)=\{y\in\mathbb R^n\mid x+\alpha y\in C\text{ for some }\alpha>0\}.
$$
Since $C$ is convex, $V$ is a convex cone. If $V=\mathbb R^n$ then $0\in V$ implies $x\in C$, and we are done. Suppose $V\neq\mathbb R^n$. As wikipedia notes, as a consequence of Farkas' lemma this implies $V$ is contained in a half-space
$$
  H=\{y\in\mathbb R^n\mid v\cdot y\leq0\}
$$
where $v\in\mathbb R^n$ is nonzero. Hence $v\cdot c\leq\lambda$ for all $c\in C$ where $\lambda=v\cdot x$. In particular
$$
  \lambda=\sum_{i=1}^\infty a_i(v\cdot x_i)\leq\left(\sum_{i=1}^\infty a_i\right)\lambda=\lambda,
$$
so we must have $v\cdot x_i=\lambda$ for each $i$ with $a_i>0$. Now the result follows from the inductive hypothesis applied to the hyperplane
$$
  \{y\in\mathbb R^n\mid v\cdot y=\lambda\}.
$$
