I need to apply Choquet's Theorem for an exercise. But Choquet has two versions of the theorem (a version for metrizable subsets and a version for non-metrizable subsets). In the metrizable case he says:
"Suppose that $X$ is a metrizable compact convex subset of a locally convex space $E$ and that $x_0$ is an element of X. Then there is a probability measure $\mu$ [...]"
Here's the setting: $E=\ell^\infty$ and $X=\{a_n\in\ell^\infty |\ n,m\in\mathbb{N}\ b^ma_n\geq0\}$
My Question: Is $X$ metrizable and why is it compact ?
My first approach was using the facts that $\ell^\infty$ is not separable and $\ell^\infty=(\ell^1)^\ast$, but I didn't help so much.
As mentioned in my comment above $X$ is metrizable. If $(a_n) \in X$, not all $a_n$'s $0$, then $\{(a_n),(2a_n),(3a_n),\cdots\}$ is a sequence in $X$ which has no convergent subsequence. Hence $X$ is not compact.