Q:

If $\alpha$ is a limit ordinal, then $\operatorname{cf}(\alpha)$ is a limit ordinal

In the textbook Introduction to Set Theory by Hrbacek and Jech, Section 9.2, the authors first introduce the definition of increasing sequence of ordinals:

Then they introduce cofinality:

My question: how do we prove that $\operatorname{cf}(\alpha)$ is a limit ordinal?
If I take a sequence $\langle \alpha_\nu \mid \nu<1 \rangle$ where $\alpha_0=\omega$. It is clear that the limit of this sequence is $\omega$ and thus $\operatorname{cf}(\omega)=1$, which is a successor ordinal.
I don't know what's wrong with my reasoning.

A:

As Martin Sleziak observed, the definition of limit the text uses applies only to sequences of limit-ordinal length - so $\langle\omega\rangle$ does not count as an increasing sequence of ordinals whose limit is $\omega$.
Personally, I think this isn't quite optimal; another approach is to define $cf(\alpha)$ as the smallest $\lambda$ (limit or not) such that there is a sequence of ordinals $<\alpha$ of length $\lambda$ with every ordinal $<\alpha$ being $\le$ some term in the sequence. Then the cofinality of a successor ordinal is $1$ (consider the sequence $\langle \beta\rangle$ in $\alpha=\beta+1$), and the cofinality of a limit ordinal is a limit ordinal. 

Note that if we replace "$\le$" with "$<$" in the definition above, the cofinality of a successor ordinal becomes undefined since the limit of a sequence of ordinals $<\beta+1$ is at most $\beta$. This isn't really a problem, since nobody talks about the cofinality of successor ordinals, but it is in my opinion a bit annoying.
Note also that the definition above does apply to $0$, and gives the "correct" answer $cf(0)=0$, via the empty sequence.

