Weak-star compact does not imply weak-star sequentially compact

For each $n\in \mathbb{N}$ define ${\mathrm{\Lambda }}_n\in ({\mathrm{\ell }}^{\mathrm{\infty }}{)}^{\ast }$ by ${\mathrm{\Lambda }}_n(x)=x_n$. Then $\mathrm{\mid }{\mathrm{\Lambda }}_n(x)\mathrm{\mid }\le \parallel x{\parallel }_{\mathrm{\infty }}$ and equality holds at the $n$th coordinate vector, so $\parallel {\mathrm{\Lambda }}_n\parallel =1$.

Consider any subsequence $({\mathrm{\Lambda }}_{n_k})$, with the indices $n_k$ strictly increasing. Define $x\in {\mathrm{\ell }}^{\mathrm{\infty }}$ by $x_{n_k}=(-1{)}^k$ and $x_j=0$ off the range of $(n_k)$. This is well defined because the indices are distinct and has $\parallel x{\parallel }_{\mathrm{\infty }}=1$ by [F3].

Its evaluations are ${\mathrm{\Lambda }}_{n_k}(x)=(-1{)}^k$, which do not converge in $\mathbb{R}$ or $\mathbb{C}$. By [F2], the chosen subsequence is not weak-star convergent. Since the subsequence was arbitrary, $({\mathrm{\Lambda }}_n)$ has no weak-star convergent subsequence.

Nevertheless [F1] makes the closed unit ball containing this sequence weak-star compact under the ultrafilter lemma. It is therefore a compact space with a sequence having no convergent subsequence, as claimed.