Existence of a $\sup$ and local compactness in the countable ordinals $\Omega$

Question

I understand how the set of countable ordinals $\Omega$ (with the order topology) is not compact, but how is it locally compact? Also, how can the existence of a least upper bound ($\sup$) for a non-decreasing sequence $\{\alpha_i\} \in \Omega$ be shown in order to verify convergence?

Answer 1

To see that $\Omega$ (and indeed every ordinal space) is locally compact, just recall that all closed ordinal spaces $[0,\alpha] = \{ \xi : 0 \leq \xi \leq \alpha \}$ are themselves compact. (The proof of this relies on the fact that there are no infinite strictly decreasing sequences of ordinals.) Next note that $[0,\alpha] = [0, \alpha+1)$ is an open neighbourhood of $\alpha$ when $\alpha \in \Omega$.

Assuming that you want to show that every sequence in $\Omega$ has a convergent subsequence, note that every sequence in a linearly ordered set $\langle X , \preceq \rangle$ has a subsequence which is either strictly decreasing, constant, or strictly increasing. For $\Omega$ there cannot be infinite strictly decreasing sequences, so we're left with the other two options. Clearly every constant sequence converges. If $\langle \xi_i \rangle_i$ is strictly increasing, then the supremum $\sup_i \xi_i$ is the limit of the sequence.