# Does every unbounded countable ordinal have an unbounded sequence?

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 $S$?

• Are you asking if every countable ordinal does, or if there is some countable ordinal which does? Mar 17, 2018 at 19:18
• Thanks, I meant does every countable ordinal...? Mar 17, 2018 at 19:20
• Take $\omega + 1$. Mar 17, 2018 at 19:23
• Thanks for the example but indeed this is not what I wanted. Archimedean was not a proper term that I used here... Mar 17, 2018 at 19:27
• $\omega + 1$ is a countable ordinal, and equivalently $0, 1, 2, \dotsc, \omega$ is well-ordered but does not contain an unbounded increasing sequence since $\omega$ is an upper bound for the whole sequence. Mar 17, 2018 at 19:29

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.

• Nice solution. It works for any totally ordered countable set. Mar 17, 2018 at 19:54

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"

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.

• It is not necessarily unbounded. It is possible that some element of $S$ is greater than all the elements of the sequence. Mar 18, 2018 at 11:20