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Can anybody answer me is $S_{\Omega} \times \bar{S_{\Omega}} $ limit point compact ?
Edit: $S_{\Omega}$ is minimal uncountable well-ordered set in order topology. $\bar{S_{\Omega}}$ is the set $S_{\Omega}$ together with the largest element $\Omega$ such that the set $\{x| x < \Omega \}$ (this set is call section of $\bar{S_{\Omega}}$ by $\Omega$) is uncountable while the other section of $\bar{S_{\Omega}}$ is countable.

Thanks in advance!

Huy Nguyen
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    Can you give the definition of $S_\Omega$ and $\overline S_\Omega$? –  May 18 '21 at 07:28
  • Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or closed. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – SolubleFish May 18 '21 at 08:16
  • I just edited it. Thank you – Huy Nguyen May 18 '21 at 09:16

1 Answers1

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Revised, moving and drastically expanding one paragraph that was far too terse.

Yes, it is. Let $A\subseteq S_\Omega\times\overline{S_\Omega}$ be infinite. Let $A_0=\pi_0[A]$ and $A_1=\pi_1[A]$, where $$\pi_0:S_\Omega\times\overline{S_\Omega}\to S_\Omega$$ is the projection map to the first factor, and $$\pi_1:S_\Omega\times\overline{S_\Omega}\to\overline{S_\Omega}$$ is the projection map to the second factor.

If $A_0$ is finite, there must be an $\alpha\in A_0$ such that $A\cap\left(\{\alpha\}\times\overline{S_\Omega}\right)$ is infinite. And $\{\alpha\}\times\overline{S_\Omega}$ is homeomorphic to $\overline{S_\Omega}$ and is therefore compact, so $A\cap\left(\{\alpha\}\times\overline{S_\Omega}\right)$ has a limit point $p\in\{\alpha\}\times\overline{S_\Omega}$, and it’s easy to check that $p$ is a limit point of $A$ in $S_\Omega\times\overline{S_\Omega}$.

Similarly, if $A_1$ is finite, there must be an $\alpha\in\overline{S_\Omega}$ such that $A\cap\big(S_\Omega\times\{\alpha\}\big)$ is infinite, and we can argue similarly, using the fact that $S_\Omega$ is limit point compact, to conclude that $A$ has a limit point in $S_\Omega\times\overline{S_\Omega}$.

Now suppose that $A_0$ and $A_1$ are infinite; then there is a sequence $\langle x_n:n\in\Bbb N\rangle$ in $A$ such that $\langle\pi_0(x_n):n\in\Bbb N\rangle$ is strictly increasing in $S_\Omega$. Every countable subset of $S_\Omega$ has a least upper bound in $S_\Omega$, so let $\alpha=\sup_{n\in\Bbb N}\pi_0(x_n)$; then $\langle\pi_0(x_n):n\in\Bbb N\rangle$ converges to $\alpha$ in $S_\Omega$.

Now consider $\{\pi_1(x_n):n\in\Bbb N\}$; if this set is finite, we can argue as in the preceding paragraph to get a limit point of $A$, so we may assume that it is infinite. Then there is a subsequence $\langle x_{n_k}:k\in\Bbb N\rangle$ such that $\langle\pi_1(x_{n_k}):k\in\Bbb N\rangle$ is strictly increasing in $\overline{S_\Omega}$. (The proof is just like the proof that every sequence of real numbers has a monotone subsequence, combined with the fact that a well-ordered set has no infinite decreasing sequence.) Let $\beta=\sup_{k\in\Bbb N}\pi_1(x_{n_k})$; then $\langle\pi_1(x_{n_k}):k\in\Bbb N\rangle$ converges to $\beta$, $\langle\pi_0(x_{n_k}):k\in\Bbb N\rangle$ converges to $\alpha$, and $\langle x_{n_k}:k\in\Bbb N\rangle$ converges to $\langle\alpha,\beta\rangle$, which is therefore a limit point of $A$.

Brian M. Scott
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  • Nice solution! I have also solved this problem already by involving the countably compactness. In $T_1$ space, countably compactness is equivalent of limit point compactness – Huy Nguyen May 18 '21 at 09:17
  • @HuyNguyen: I wasn’t sure whether you were familiar with the equivalence of countable compactness and limit point compactness in $T_1$ spaces, so I decided to stick with limit point compactness, but yes, countable compactness works nicely too, especially if you’ve already proved that the product of a countably compact space with a compact space is countably compact. – Brian M. Scott May 18 '21 at 17:01
  • @BrianM.Scott The reasoning in the paragraph where $A_0$ and $A_1$ are infinite would go through for the product of two countably compact spaces. But the product of two such spaces need not be countably compact. So it seems there would need to be additional argumentation using the fact that one of the factors is compact. To the OP: for countably compact times compact, see https://math.stackexchange.com/questions/3486708 for example. – PatrickR May 23 '21 at 07:09
  • @PatrickR: The argument that I had in mind does not go through for the product of two countably compact spaces, since it relies on the specific structure of $\omega_1$. However, I see now that I didn’t say enough to make clear what I was thinking. I’ll expand it. – Brian M. Scott May 23 '21 at 20:13