Compactness and sequential compactness are not equivalent in a general topological space.

For the weak topology on a normed space, they are:

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 want to know about the weak* topology on the dual space.

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.

Note: I asked this on MSE along with a third easier question (the analog of Q2 for arbitrary topological spaces), but only that easier question produced something fruitful.



1 Answer 1


Let $X=C[0,\omega_1]$. Consider $A = \{\delta_\alpha\colon \alpha < \omega_1\}$ in $X^*$. This is a homeomorphic copy of $\omega_1$ which is a sequentially compact but not compact space. (Here $\delta_x$ stands for the evaluation functional at $x$.)

As for your second question, take $X = C[0,\omega_1]\oplus \ell_\infty\cong C( [0,\omega_1]\sqcup \beta \mathbb N)$.


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