General Topology: Is $S_{\Omega} \times \bar{S_{\Omega}} $ limit point compact? 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!
 A: 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 $A$.
