Does every unbounded countable ordinal have an unbounded sequence?

Question

Let $S$ be a countable well-ordered set which is unbounded (i.e. it has no maximum). Does there exist an unbounded increasing sequence in $S$?

Answer 1

Hint: let $f : \omega \to S$ enumerate $S$. Starting at $1$, remove $i$ from the domain of $f$ if $f(i) < f(j)$ for some $j < i$. The result is an unbounded partial function from $\omega$ to $S$, which you can convert into a sequence by renumbering the arguments.

Answer 2

I found a related question which answers this question too.

A generalization of "any countable limit ordinal is the union of a sequence of increasing ordinal"

Answer 3

Yes.

Let $s_1$ be an element of $S$. Because $S$ is unbounded, $s_1$ can't be the maximum element. Therefore, there exists at least one element of $S$ greater than $s_1$. Call this element $s_2$. $S$ also can't have $s_2$ as its maximum element, so there must be an element of $S$ (call it $s_3$) greater than $s_2$.

Repeat this process ad infinitum, and you have an unbounded increasing sequence.