# We know that the space given below is countably compact . But how can i prove that X is not sequentially compact?

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.

• $X$ 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. Apr 17 at 3:18
• Which closed subspace of X is homeomorphic to $I^I$ Apr 18 at 2:30
• Every subspace of the form $\{\alpha\}\times I^I$, among many others. 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.
• Hello sir..in your proof $I$ is a product of two point set but in my question $I$ is closed unit interval [ 0 , 1 ] Apr 18 at 8:39
• @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). Apr 18 at 8:42
• 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.. Apr 18 at 8:48