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I find quite difficult to work with weak and weak* topologies. Let's consider the following problem (I don't know where my Professor found it out).

Let $\Gamma$ be an infinite uncountable set. Consider \begin{equation} Z := \{ x \in \ell_\infty (\Gamma) \colon \| x \|_\infty \le 1 \text{ and the support of x is at most countable} \} \end{equation} equipped with the $w^*$-topology on $\ell_\infty (\Gamma)$ when considered as the dual of $\ell_1 (\Gamma)$. Decide whether $Z$ is compact and/or sequentially compact. (Hint. You may look whether $Z$ is $w^*$-closed in the closed ball of radius one $B_{ \ell_\infty (\Gamma) }$

I know that in a general topological space there is no equivalence between compactness and sequential compactness and I know the basis of the weak and weak* topologies (Mazur's Theorem, Banach-Alaoglu Theorem, Goldstine Theorem, $\dots$), but I don't know how to use them. Maybe it would be easier to look whether $B_{ \ell_\infty (\Gamma) } \backslash Z$ is $w^*$-open, but how?

Lorenzo
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Paul Bear
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This set is weak-star sequentially compact but not weak-star compact.

Let $(x_k)$ be a sequence in $Z$. Define $I:=\cup_k supp(x_k)$. This set is countable. So $(x_k)$ can be interpreted as a sequence in $l^\infty(I)$. This space is the dual of $l^1(I)$, which is separable. So $(x_k)$ has subsequence that converges weak-star in $l^\infty(I)$ by Banach-Alaoglu. It is now easy to check that this implies weak-star convergence in $l^\infty(\Gamma)$. Note that Banach-Alaoglu cannot be applied to the sequence directly as $l^1(\Gamma)$ is not separable.

Let me show that the weak star closure of $Z$ is the unit ball. Let $x_0\in l^\infty(\Gamma)$ with $\|x_0\|_\infty\le 1$. Let $U$ be open such that $x\in U$. Then there are finitely many $y_1\dots y_n\in l^1(\Gamma)$ and $\epsilon>0$ such that $$ \{ x\in l^\infty(\Gamma): \ y_i(x-x_0)<\epsilon \ \forall i=1\dots n\} \subset U. $$ Let $I:=\cup_{i=1}^n supp(y_i)$. Define $x$ by $x_k=x_{0,k}$ for $k\in I$ and $x_k=0$ for $k\not\in I$. Then $x\in Z$, and $x$ belongs to the open set $U$. That is, each neighborhood of $x_0$ contains a point from $Z$. Hence the unit ball is contained in the closure of $Z$. The reverse inclusion is trivial.

To get an intuition about this, note that general topological spaces do not go well with countability requirements (as in the definition of $Z$). Then one gets the idea that the set under question is not weak-star compact. As usual for such kind of exercises, the other property might be true ;)

daw
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  • I guess it should be mentioned that in the third paragraph you are using Goldstine's Theorem and that $l^\infty \subset (l^1)^*$. – Jonathan Hole Nov 16 '22 at 15:38
  • Did you prove the weak star closure of Z is the unit ball in order to use the corollary of the Banach-Alaoglu theorem which says that weak star closed and bounded implies weak star compact (so Z cannot be weak star closed)? – Paul Bear Nov 16 '22 at 15:58
  • @JonathanHole In the third paragraph, I used $l^\infty = (l^1)^*$ (as claimed in the OP) and the fact that sets of this type are a basis of the weak-star topology. (I might be wrong with both claims). – daw Nov 16 '22 at 17:18
  • @PaulBear I did prove that the w-star closure of $Z$ is the unit ball, so it cannot be w-star closed and hence not weak-star compact (because compact sets are closed in Hausdorff spaces) – daw Nov 16 '22 at 17:23
  • @daw, I see. I think the argument goes through just by knowing that $\ell^\infty\subseteq (\ell_1)^$, but indeed $\ell^\infty = (\ell_1)^$ by https://math.stackexchange.com/questions/405357/when-exactly-is-the-dual-of-l1-isomorphic-to-l-infty-via-the-natural-map. – Jonathan Hole Nov 17 '22 at 15:34