# 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 $\omega_{\delta}$?

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.
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