In all the counterexamples to the assertion $\text{Sequentially Compact} \implies \text{Compact}$, they involve ordinals in some form of the other, such as
- Ordinal Space
- Long Ray/Line
- 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?
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.