Let X be the product of [0 , $\Omega$) with the interval topology and $I^I$ with the cartesian product topology , where $I$ is unit interval.

  • $\begingroup$ $X$ is weakly countably compact: every countably compact space is. $X$ is not sequentially compact, since it contains a closed subspace homeomorphic to $I^I$, which is a well-known example of a compact space that is not sequentially compact. $\endgroup$ Apr 17 at 3:18
  • $\begingroup$ Which closed subspace of X is homeomorphic to $I^I$ $\endgroup$ Apr 18 at 2:30
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    $\begingroup$ Every subspace of the form $\{\alpha\}\times I^I$, among many others. $\endgroup$ Apr 18 at 3:12

$[0,\Omega)$ is countably compact (a countable open cover has a finite subcover), and a product of a countably compact space and a compact space (i.e. $I^I$) is still countably compact. So $X$ is countably compact, and so also weakly countably compact (every infinite set has a limit point in $X$); this implication always holds (see e.g. here). $X$ is however not sequentially compact, because $\{0,1\}^I$ is a closed subset of it, and would then also be sequentially compact, which it is not (see e.g. here).

Of course $\{0,1\}^I$ (or $I^I$) by itself is already an example of a (countably) compact space that is not sequentially compact, but $X$ is also non-compact, which might have been the "point" of this example.

  • $\begingroup$ sir ...Give me a proof of above example which is not sequentially compact $\endgroup$ Apr 18 at 2:27
  • $\begingroup$ @MukeshSuthar i gave the proof that it was not sequentially compact didn’t I ? What do you mean? $\endgroup$ Apr 18 at 6:13
  • $\begingroup$ Hello sir..in your proof $I$ is a product of two point set but in my question $I$ is closed unit interval [ 0 , 1 ] $\endgroup$ Apr 18 at 8:39
  • $\begingroup$ @MukeshSuthar it’s a closed subset of $I^I$ so that makes no difference. E.g. If $X$ is seq. compact, so is $I^I$ then so is $\{0,1\}^I$, and this is a product of continuum many two point spaces (for which I show in the link it is not seq. compact). $\endgroup$ Apr 18 at 8:42
  • $\begingroup$ Hello sir...If you have any other simple proof for that $X=[0,\Omega) \times I^I $ is not sequentially compact ...please share with me.. $\endgroup$ Apr 18 at 8:48

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