"Proof" that $\text{cof}(\omega_\lambda)<\omega_\lambda$ if $\lambda$ is a nonzero limit ordinal The following lemma is from this introduction to cardinals.

Lemma 2.7. Let $\omega_\alpha$ be a limit cardinal. Then $\alpha$ is a limit ordinal and $\text{cof}(\omega_\alpha)=\text{cof}(\alpha)$.

I understand the proof of this lemma, including the proof of relevant lemma 2.4 (beware, there is a typo in that proof: $g$ instead of $h$).
Now I want to use it to make the following conclusion:
$$\text{cof}(\omega_\lambda)\stackrel{2.7}=\text{cof}(\lambda)\leq \lambda < \omega_\lambda$$
However, in the same paper I'm being told that the existence of regular limit cardinals is independent of ZFC! So there must be a flaw in my conclusion.
I can think of two things that would undermine my argument:


*

*$\lambda < \omega_\lambda$ is not always true.

*For a nonzero limit ordinal $\lambda$, $\omega_\lambda$ need not be a limit cardinal.


But I don't see how either 1. or 2. is possible.
 A: It’s the first alternative: it’s not always true that $\lambda<\omega_\lambda$. Let $\lambda_0=\omega$, and for $n\in\omega$ let $\lambda_{n+1}=\omega_{\lambda_n}$. Let $\lambda=\sup_n\lambda_n$; then 
$$\lambda=\sup_n\lambda_n=\sup_n\omega_{\lambda_n}=\omega_{\sup_n\lambda_n}=\omega_\lambda\;.$$
A: Adding to Brian's answer: We can prove (in $\operatorname{ZF}$) that $C := \{ \lambda \in \operatorname{On} \mid \lambda = \omega_{\lambda} \}$ is a unbounded proper class. (When I talk about proper classes in $\operatorname{ZF}$, I always mean virtual classes.)
A slight variation of Brian's answer shows that $C$ is unbounded. Given some $\gamma \in \operatorname{On}$, let $\lambda_{0} = \gamma$ and then $\lambda_{n+1} = \omega_{\lambda_{n}}$. It follows that $\lambda := \sup \{ \lambda_{n} \mid n < \omega \}$ satisfies $\lambda = \omega_{\lambda}$. I leave it to you to prove that $C$ is closed as well.
You may also want to check that for any uncountable and regular cardinal $\kappa$, $C_{\kappa} := C \cap \kappa$ is club in $\kappa$.
