The Banach-Alaoglu says that the closed unit ball in a dual space $X^*$ of a normed space $X$ is compact in the weak-star topology.

The sequential Banach-Alaoglu says that the closed unit ball in a dual space of a separable normed space is sequentially compact in the weak-star topology.

According to page 71, example 5.3.1 ii) here, the assumption of separability cannot be dropped in the sequential Banach-Alaoglu. The proof makes sense to me. However I also reason that the (strong) Banach-Alaoglu covers the sequential Banach-Alaoglu as a special case. So by Banach Alaoglu, shouldn't the sequential Banach Alaoglu theorem still hold true for $X$ inseparable?

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    $\begingroup$ Not all compact spaces are sequentially compact (and not all sequentially compact spaces are compact). See here for an example of a sequence without weak$^{\ast}$-convergent subsequence. $\endgroup$ – Daniel Fischer Jan 15 '18 at 11:32
  • $\begingroup$ Thanks! I was under the impression tt compact => sequentially compact which requires first countability. But why then does wikipedia consider the sequential version of the theorem to be a special case? $\endgroup$ – Zhanfeng Lim Jan 15 '18 at 13:06
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    $\begingroup$ Because in the separable case, the dual unit ball with the weak$^*$ topology is metrizable and compactness and sequential compactness are equivalent in metric spaces. $\endgroup$ – Jochen Jan 15 '18 at 13:21

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