Is a subset of $\ell^\infty$ metrizable?

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.

• Subsets of metric spaces are metrizable: simply restrict the metric to the subspace. Definition of $X$ is not clear. What is $b$? – Kavi Rama Murthy Dec 30 '18 at 12:15
• To talk about metrizability and compactness you have to fix a topology first. These two questions are dependent, they cannot be treated separately. $a_n$ is a sequence, what does $b^ma_n\ge 0$ mean? – A.Γ. Dec 30 '18 at 12:17
• @Kavi Rama Murthy great answer thanks a lot! $b^m$ is just another sequence. You can interpret $X$ as the set of sequences $a_n\in\ell^\infty$ that aren't negative. Can we use something like Banach-Alaoglu for the compactness ? – ThomasMuller Dec 30 '18 at 19:30

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.