# Inequivalence of Compactness and Sequential Compactness, Especially for the Weak* Topology?

I know that for topological spaces, compactness and sequential compactness are generally not equivalent. From wikipedia: "there exist sequentially compact spaces that are not compact (such as the first uncountable ordinal with the order topology), and compact spaces that are not sequentially compact (such as the product of ${\displaystyle 2^{\aleph _{0}}={\mathfrak {c}}}$ copies of the closed unit interval [e.g., the space of functions $f:\mathbb{R} \to [0,1]$ with pointwise convergence])."

For the weak topology on a normed space, compactness and sequential compactness are equivalent:

Theorem (Eberlein-Smulian) Let $X$ be a normed space and let $A$ be a subset of $X$. Then $A$ is weakly compact if and only $A$ is weakly sequentially compact.

I'm interested in examples where compactness and sequential compactness are not equivalent in the weak* topology.

Example. The closed unit ball $B_{*}$ of $X^{*} = (\ell^{\infty})^{*}$ is weak* compact (by Banach-Alaoglu) but is not weak* sequentially compact (the sequence of functionals $f_n(x_1,x_2,\ldots,)=x_n$ has no weak* convergent subsequence).

Question 1. What is an example of a normed space $X$ and an $A \subseteq X^{*}$ that is weak* sequentially compact but not weak* compact?

Question 2. Is there a normed space $X$ such that (i) there is a set $A \subseteq X^{*}$ that is weak* sequentially compact but not weak* compact and (ii) the closed unit ball $B_{*} \subseteq X^{*}$ is weak* compact but not weak* sequentially compact.

Question 3. (Easier than Question 2). Is there a topological space $X$ such that (i) there is a set $A \subseteq X$ that is sequentially compact but not compact and (ii) there is a set $B \subseteq X$ that is compact but not sequentially compact.

Edit: Question 3 turned out to be much easier than I thought (see Henno Brandsma's answer below).

• The $(l^\infty)^\ast$ example is sort of the example $[0,1]^I$ for large $I$ in disguise, as the weak-star topology looks a lot like the product topology; evaluations = projections etc. – Henno Brandsma Jun 29 '18 at 9:45
• For 2. we need a non-WCG space, by the Amir-Lindenstrauss theorem, see this answer – Henno Brandsma Jun 29 '18 at 14:08

Q3 is simple: $X = \{0,1\}^I$ where $I$ is size continuum (or more), e.g. $I = 2^\mathbb{N}$, will work.
$B=X$ is compact but not sequentially compact (Wikipedia mentions it, and I also showed it here).
$A = \{x \in X: \{i \in I: x_i \neq 0\} \text{ is at most countable}\}$ is a subset that is sequentially compact but not compact (it's dense in $X$).