Is $\omega_{\alpha}$ sequentially compact? For an ordinal $\alpha \geq 2$, let $\omega_{\alpha}$ be as defined here. 
It is easy to show that $\omega_{\alpha}$ is limit point compact, but is it sequentially compact?
 A: Let me return to the usual definition of $\omega_\alpha$, namely the $\alpha$-th infinite initial ordinal (where initial ordinal means not in bijection with any of its members).
If the cofinality of $\omega_\alpha$ is countable then there is an unbounded sequence in which case we have a sequence which does not converge. 
Otherwise every countable set is bounded, and is either finite (if decreasing) or contains a strictly increasing subsequence. It is not hard to see that the supremum of the strictly increasing subsequence is a limit point.
A: I think I finally found a solution:
In a well-ordered set, every sequence admits an non-decreasing subsequence. Indeed, if $(x_n)$ is any sequence, let $n_0$ be such that $x_{n_0}= \min \{ x_n : n\geq 0 \}$, and $n_1$ such that $x_{n_1}= \min \{ x_n : n > n_0 \}$, and so on; here, $(x_{n_i})$ is an non-decreasing subsequence.
Because an non-decreasing sequence is convergent iff it admits a cluster point, limit point compactness and sequentially compactness are equivalent.
