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A space $X$ is

  • sequential if every sequentially closed set is closed,
  • compact, if every open cover has a finite subcover,
  • sequentially compact, if every sequence has a convergent subsequence.

Any first-countable space is sequential. Any first-countable compact space is sequentially compact.

The typical example $I^I$ with $I = [0,1]$ for a compact not sequentially compact space is not sequential. For such a counterexample it is necessary that $X$ is not first-countable.

The database version of Steen-Seebach $\pi$-base does not provide such an example.

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  • $\begingroup$ You mean covering compact but not sequentially compact? $\endgroup$ – Brandon Thomas Van Over Jan 13 '16 at 10:06
  • $\begingroup$ No, $X$ should be sequential, covering compact and not sequentially compact. I add the definitions. $\endgroup$ – yadaddy Jan 13 '16 at 10:08
  • $\begingroup$ Being covering compact is equivalent to being sequentially compact. $\endgroup$ – Brandon Thomas Van Over Jan 13 '16 at 10:09
  • $\begingroup$ What is your definition of "covering compact"? $\endgroup$ – yadaddy Jan 13 '16 at 10:11
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    $\begingroup$ $\omega_1 = [0, \omega_1)$ is first-countable, sequentially compact and not compact. (Also, $\omega_1 + 1 = [0, \omega_1]$ is compact and sequentially compact but not sequential.) $\endgroup$ – yadaddy Jan 13 '16 at 10:25
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If $X$ is compact and sequential, it is in particular countably compact. And countable compactness and sequential compactness are equivalent for sequential spaces (I do believe we need at least $T_1$ here, but if your compact space is Hausdorff this is no problem).

As noted in the comments, a sequential and sequentially compact Hausdorff space need not be compact (we only get to countable compactness; the Lindelöfness has to be assumed to get compactness), as shown by $\omega_1$. But from compactness we do get sequential compactness for sequential spaces.

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  • $\begingroup$ Thanks for the hint for the equivalence of countable compactness and sequential compactness for sequential spaces ($T_1$ is not necessary). $\endgroup$ – yadaddy Jan 13 '16 at 17:12
  • $\begingroup$ For a proof of "countably compact + sequential $\Rightarrow$ sequentially compact" (without $T_1$) see here. $\endgroup$ – yadaddy Apr 8 '16 at 7:10

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