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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?

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    I'm not sure that you mean $\text{card}(\omega_\alpha)>\text{card}(\alpha)$. If so, then for instance, $\omega_2=3$.2012-11-30
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    While $\omega_\omega$ is the thing that everyone else calls $\omega_1$.2012-11-30
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    Sorry, the correct definition is the one given by Asaf Karagila.2012-11-30
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    You should edit your post then. I should also remark that you can easily prove the existence of $\alpha$ such that $\alpha=\omega_\alpha$, so the definition above is grossly incorrect.2012-11-30
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    I edited my post.2012-11-30

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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.

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    Minor correction: you want $x_{n_1}=\min\{x_n:n\>n_0\}$. You should probably also mention that by *increasing* you mean *weakly increasing, non-decreasing*.2012-12-01
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    Sorry, sometimes I speak like in French... I edited my answer.2012-12-01
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    No reason to apologize: I’ve seen the same usage from native speakers of English. I had a typo in my last comment, though, that made part of it incomprehensible: you want $x_{n_1}=\min\{x_n:n>n_0\}$, with strict inequality, or you’ll simply pick the same term over and over again.2012-12-01
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    Exact, I correct it.2012-12-01
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    Looks good now. I’d give it +1, except that I already did.2012-12-01
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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.