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It is well-known that compactness has nothing to do with sequential compactness in most cases. For example, the unit ball in $(\ell^\infty)'$ is compact in weak$^*$ topology but it is NOT sequentially compact in this topology. In particular, if we set $e_n=(0,0,\cdots, 1, \cdots)$ (attains 1 at n-th term), then $\{e_n\}$ does not have any convergent subsequence even it is contained in a compact set.

However, a topological space $X$ is compact if and only if every net has a convergent subnet with a limit in $X$ and any sequence is a net. So my question is as follows: $\{e_n\}$ is a sequence in a compact space in $(\ell^{\infty})'$ but has no convergent subsequence. How to construct a convergent subset out of $\{e_n\}$

I know that a subnet of a sequence is not necessarily a sequence. But, I really want an explicit construction in the problem mentioned above, or at least give some hints on why it is possible.

Yuhang
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    I don't think explicit is really possible. The answer I'd give starts "Let $\mathscr{U}$ a free ultrafilter on $\mathbb{N}$". And I don't know of a way to make that explicit. – Daniel Fischer Aug 15 '17 at 20:28
  • @DanielFischer Thank you. It is just too hard to ``accept'' it. – Yuhang Aug 15 '17 at 20:33
  • I couldn't do an example of a compact space that isn't sequentially compact, but I did find an explicit example (not requiring the axiom of choice) of a sequence with a convergent subnet but no convergent subsequence in the first part of the answer here: https://math.stackexchange.com/a/3495609 – Robert Furber Aug 15 '21 at 17:39
  • Did you try doing it in the style of Mazur's lemma, taking convex combinations of the sequence? https://en.wikipedia.org/wiki/Mazur%27s_lemma – Zatrapilla Feb 27 '23 at 23:42

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This question and answer give a concrete example of a sequence $(\delta_n)$ in a compact space that has no convergent subsequence (so this compact space is not sequentially compact). The answer gives a "concrete" (if you believe ultrafilters are concrete) subnet $(x_d), d \in D$ that converges to some $f_\mathcal{U}$. This subnet cannot have a subsequence (that is also a subnet!) that converges, because this would be a subsequence of the original sequence that would converge (and this cannot be). It is possible to find $d_n$ that are increasing in the index set $D$ of the subnet, but this is not a subnet of the subnet (as they will not be cofinal).

A subsequence of a sequence is a subnet as well, but a subnet need not have a cofinal subsequence.

Henno Brandsma
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