Weak* topologies: compactness vs. sequential compactness 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?
 A: 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 ;)
