# Order type of final segment of countable limit ordinal

Suppose $\gamma$ is a countable limit ordinal, with only finitely many limit ordinals less than it. If $\zeta$ is the greatest limit ordinal less than $\gamma$, does this necessarily imply that the final segment $\gamma\setminus\zeta$ is of order-type $\omega$?

My attempt: $$f(n) = \zeta + n$$

Is the same true for uncountable ordinals? i.e if $\alpha$ is of cardinality $\aleph_{\delta}$, with finitely many limit ordinals less than it, then the final segment $\alpha\setminus\zeta$ is of order type $\omega_{\delta}$?

## 1 Answer

For the first question, the answer is yes. Simply because the only ordinals with finitely many limit ordinals below them are $\omega\cdot n$ for $n\in\omega$. So it is a finite sequence of copies of $\omega$ stacked on top of one another.

In the case of uncountable ordinals your question should be different and it should involve cofinality. More specifically, there is no uncountable ordinal with only finitely many limit ordinals below it.

Even if you only count limit ordinals starting $\omega_\delta$, you still don't have the wanted result, since you have $\omega_\delta+\omega\cdot n$ as these ordinals and none of them have an end segment of order type $\omega_\delta$ (well, except $n=0$).

However, if $\alpha$ is a limit ordinal of cofinality $\omega_\delta$ with only finitely many ordinals below it with cofinality $\omega_\delta$, then it has to be of the form $\omega_\delta\cdot n$ for some $n<\omega$ and the result follows again.

• so is my function a good isomorphism to prove that the final segment is of order type $\omega$? – Joshhh Jan 6 '16 at 6:49
• That is correct. – Asaf Karagila Jan 6 '16 at 7:08