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.
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$X$ 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. – Brian M. Scott Apr 17 '21 at 03:18
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Which closed subspace of X is homeomorphic to $I^I$ – Mukesh Suthar Apr 18 '21 at 02:30
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1Every subspace of the form ${\alpha}\times I^I$, among many others. – Brian M. Scott Apr 18 '21 at 03:12
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$[0,\Omega)$ 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.
Henno Brandsma
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sir ...Give me a proof of above example which is not sequentially compact – Mukesh Suthar Apr 18 '21 at 02:27
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@MukeshSuthar i gave the proof that it was not sequentially compact didn’t I ? What do you mean? – Henno Brandsma Apr 18 '21 at 06:13
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Hello sir..in your proof $I$ is a product of two point set but in my question $I$ is closed unit interval [ 0 , 1 ] – Mukesh Suthar Apr 18 '21 at 08:39
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@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). – Henno Brandsma Apr 18 '21 at 08:42
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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.. – Mukesh Suthar Apr 18 '21 at 08:48
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@MukeshSuthar The closed set argument, reducing to ${0,1}^I$, is the simplest one IMO. – Henno Brandsma Apr 18 '21 at 08:49
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@MukeshSuthar What's your point? That's the whole proof. You just need to know that a closed subset of a sequentially compact space is again sequentially compact and reason by contradiction, as I showed. – Henno Brandsma Apr 18 '21 at 08:58
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That means in your proof..First we assume that $X=[0,\Omega) \times I^I $ is sequentially compact and after we will use the cosed subspace of sequentially compact space is sequentially compact and we get contradiction . So original space X is not sequentially compact – Mukesh Suthar Apr 18 '21 at 09:04
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@MukeshSuthar ${0,1}^I$ is a closed subspace of $[0,1]^I$ which is homeomorphic to ${0} \times I^I$ which is a closed subspace of $X$. – Henno Brandsma Apr 18 '21 at 12:47
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Under which homeomorphism ${0,1/}^{I}$ and ${ 0 /} /times I^{I} $ are homeomorphic. – Mukesh Suthar Apr 20 '21 at 10:07
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But under which homeomorphism both are homeomorphic..give me a homeomorphism – Mukesh Suthar Apr 20 '21 at 10:18
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$I^I$ is homeomorphic to ${0} \times I^I \subseteq X$ by $f(x)=(0,x)$ of course. – Henno Brandsma Apr 20 '21 at 11:11