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Let $X= \prod_{n\in \omega} X_n\subset \prod_{n\in \omega}\aleph_n$, where $X_n \subset \aleph_n$ (where $\aleph_n$ is the space with order topology) is Lindelöf for each $n\in \omega$. My question is this:

Must $X$ be Lindelöf?

Thanks for any comment.

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    $\begingroup$ Which topology does each $\aleph_n$ have? If it's the order topology, then your claim that each $X_n$ must be countable is wrong. If it's the discrete topology then the answer to your question is easily yes, because $X$ would then be homeomorphic to the irrational numbers. $\endgroup$
    – Santi Spadaro
    Commented Jun 2, 2019 at 10:34
  • $\begingroup$ Yes, $X_n$ may not be countable, it only has countable cofinality. $\endgroup$
    – user1
    Commented Jun 2, 2019 at 11:35
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    $\begingroup$ Wouldn't that imply that the product of countably many copies of $\omega$ is Lindelof? $\endgroup$
    – Péter Komjáth
    Commented Jun 2, 2019 at 12:43
  • $\begingroup$ "Countable cofinality" does not tell the whole story. For a set of ordinals to be Lindelöf, it is necessary that the set contain all its limit points of uncountable cofinality. But that is not sufficient, since if it omits uncountably many limit points (of countable cofinality) on a scattered set, then we can also make a violation of the Lindelöf property. $\endgroup$
    – Joel David Hamkins
    Commented Jun 2, 2019 at 16:32
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    $\begingroup$ Isn't it the case that if $X$ consists of all isolated points of $\omega_1 + 1$ along with the last point, then $X$ is Lindel\"{o}f? $\endgroup$
    – Anonymous
    Commented Jun 3, 2019 at 17:07

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