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. 


<https://math.stackexchange.com/questions/2835590/inequivalence-of-compactness-and-sequential-compactness-especially-for-the-weak>



Inequivalence of Compactness and Sequential Compactness for the Weak* Topology?
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 $X = C[0,\omega\_1]\oplus \ell\_\infty\cong C( [0,\omega\_1]\sqcup \beta \mathbb N)$.


