0
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ Subsets of metric spaces are metrizable: simply restrict the metric to the subspace. Definition of $X$ is not clear. What is $b$? $\endgroup$ – Kavi Rama Murthy Dec 30 '18 at 12:15
  • $\begingroup$ 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? $\endgroup$ – A.Γ. Dec 30 '18 at 12:17
  • $\begingroup$ @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 ? $\endgroup$ – ThomasMuller Dec 30 '18 at 19:30
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.