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In all the counterexamples to the assertion $\text{Sequentially Compact} \implies \text{Compact}$, they involve ordinals in some form of the other, such as

  1. Ordinal Space
  2. Long Ray/Line
  3. Altered Long Ray/Line

to name the ones I could find.

Since I have no knowledge of ordinals, I was wondering if there were counterexamples to the above assertion which does not involve ordinals. Also, if there does not exist such counterexamples, is there a deeper reason for this? Or is it just that such examples are not known?

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    $\begingroup$ Try $$X = \Bigl\{ f \in \{0,1\}^{\mathbb{R}} : \{t \in \mathbb{R} : f(t) \neq 0\} \text{ is countable}\Bigr\}\,.$$ It's dense in $K = \{0,1\}^{\mathbb{R}}$, hence not compact. But every sequence in $X$ is contained in a compact metrisable subset of $K$, hence has a convergent subsequence. $\endgroup$ Jul 27, 2020 at 19:58
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    $\begingroup$ @Ishan Part of the reason that these counterexamples look a little weird and arbitrary is that for metric spaces, "compact" and "sequentially compact" are equivalent, which rules out many typical spaces (including subsets of $\Bbb{R}^n$ in the subspace topology) from serving as counterexamples. $\endgroup$ Jul 27, 2020 at 23:06

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The $\Sigma$-product is another good one (also suggested by Daniel in the comment)

$$X=\{f \in \{0,1\}^{\Bbb R}: \operatorname{supp}(f):=\{x: f(x) \neq 0\} \text{ is at most countable }\}$$

where in fact any uncountable index set (as $\Bbb R$, as we chose) will do.

This is countably compact, sequentially compact, pseudocompact but not first countable nor compact, and a topological group as well (under the mod 2 pointwise addition, so a subgroup of $\{0,1\}^{\Bbb R}$). If $\{f_n: n \in \Bbb N\}$ is any countable subset of $X$, then the "total support" $A:= \bigcup_n \operatorname{supp}(f_n)$ is also at most countable and $\{0,1\}^A$ is a compact metric space in which this set essentially lives. A limit point, or subsequence limit there can then be found, stuffed with $0$'s on the other coordinates and we get one for the sequence/set we started with in $X$. Non-compactness is obvious as $X$ is dense in $\{0,1\}^{\Bbb R}$, making it also a ccc (but non-separable) space as well. It's a very nice example in many ways.

So we only need to know about countable sets and that uncountable sets exist. No ordinals or long line required.

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  • $\begingroup$ We do need to know (from AC) that a countable union of countable sets is countable. $\endgroup$ Jul 27, 2020 at 22:33
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    $\begingroup$ @DanielWainfleet ACC is enough for that, I think, and I think most people will it more elementary than ordinals (though those are very nice and should be known by all mathematicians, really, IMHO). $\endgroup$ Jul 27, 2020 at 22:38

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