set theory - "Proof" that $text{cof}(omega_lambda)

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:

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


2. 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.

Answer

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

$$\lambda=\sup_n\lambda_n=\sup_n\omega_{\lambda_n}=\omega_{\sup_n\lambda_n}=\omega_\lambda\;.$$