Sequential compactness of $S_\Omega$ and unbounded sequences.

Question

I'm currently trying to prove that

The space $S_\Omega$ is sequentially compact.

where $S_\Omega$ is the first uncountable ordinal (i.e. it is well ordered, uncountable and all proper sections are countable).

So far, I've managed to see the following

1. By an argument similar to the one done in the reals, sequences in $S_\Omega$ have monotone subsequences: we can say that a point of a sequence is a 'viewpoint' if it is greater than all posterior terms. If there are infinitely many viewpoints, we can construct a decreasing sequence. If on the contrary there are only finitely many, this implies that at a certain point, every term is eventually preceded by a greater one, thus allowing the construction of an increasing sequence.
2. If a sequence is decreasing, it converges in $S_\Omega$. In effect, it will converge to $\min_n x_n$.

In the same fashion, if a sequence is increasing and the set $S = \{x : x \geq x_n \ (\forall n)\}$ is non empty, its first element $s$ will be the sequence's limit point. However, it may a priori be that the original sequence is unbounded. My intuition is that this cannot occur precisely because such a scenario in itself would contradict the sequential compactness of $S_\Omega$, which I'm trying to prove. So, any ideas on how to prove, if true, that no sequence is $S_\Omega$ is unbounded?

Edit: as Henno Brandsma has referenced, theorem 10.3 of Munkres proves that $S$ is non empty, thus the sequence has a supremum (to which it converges): sections of each term $S_{x_n}$ are countable so by cardinality,

$$ S_{\Omega} \setminus \bigcup_{n \geq 1}S_{x_n} \neq \emptyset. $$

Answer 1

Your proof idea can be made to work: Let $(x_n)$ be any sequence in $S_\Omega$.

In any linearly ordered set we have a subsequence determined by indices $n_1 < n_2 < n_3 < \ldots$, such that either

1. $\forall k: x_{n_k} \le x_{n_{k+1}}$ or
2. $\forall k: x_{n_k} \ge x_{n_{k+1}}$.

In the latter case we cannot have the sequence be strictly decreasing because $S_\Omega$ is well-ordered and the sequence stabilises at some $x$ such that all terms beyond a finite initial segment are equal to $x$. This clearly defines a (constant thus convergent) subsequence.

In the former case $C= \sup\{x_{n_k}: k=0,1,2,\ldots\}$ exists and lies in $S_\Omega$ (thm 10.3 Munkres (2nd ed.)) and by the definition of the order topology one can check that this subsequence converges to $C$.

Answer 2

If $(x_n)$ is a sequence in $\Omega$ then let $C = \sup \{x_n: n \in \omega\}$. This is a countable ordinal. $C+1$ is a compact countable subspace in which $(x_n)$ "lives" and $C+1$ is first countable, hence second countable hence metrisable. So $C+1$ is sequentially compact and hence there is a convergent subsequence of the original sequence converging in $C+1$ and thus also in $\Omega$.

Answer 3

Take any sequence $\{x_i\in \omega_1:i\in \omega\}$ in $\omega_1$. Since every sequence is countable and any countable subset of $\omega_1$, which is first uncountable ordinal, has an upper bound then every sequence has an upper bound in $\omega_1$. So $\{\alpha \in \omega_1: \alpha\geq x_i, i\in \omega\}\neq \emptyset $. Then it has minimum $\alpha_0$. If the minimum in the sequence we can take constant sequence which consists of the minimum as a converges subsequence. Now assume the minimum is not in the sequence. So every $x_i<\alpha_0$ then there must be some non-decreasing $x_{k_i}$'s such that $\alpha_0 > x_{k_i}> x_i$ for every $x_i$ and we can take the $\{x_{k_i}\}$ as a converges subsequence.