Definitions:
Let $(\alpha_\xi \mid \xi < \lambda)$ be a transfinite sequence of ordinals of length $\lambda$. We say that the sequence is increasing if $\alpha_\nu < \alpha_\mu$ whenever $\nu<\mu<\lambda$.
If $\lambda$ is a limit ordinal and if $(\alpha_\xi \mid \xi < \lambda)$ is an increasing sequence of ordinals, we define $$\alpha=\lim_{\xi \to\lambda}\alpha_\xi:=\sup\{\alpha_\xi \mid \xi <\lambda\}$$ and call $\alpha$ the limit of the increasing sequence.
An infinite cardinal $\kappa$ is called singular if there exists an increasing transfinite sequence $(\alpha_\xi \mid \xi < \lambda)$ of ordinals $\alpha_\xi$ such that
$\alpha_\xi < \kappa$ for all $\xi < \lambda$
$\lambda<\kappa$
$\lambda$ is a limit ordinal
$\kappa=\lim_{\xi \to\lambda}\alpha_\xi$
Theorem: There are arbitrary large singular cardinals $\aleph_\alpha$ such that $\aleph_\alpha=\alpha$.
My textbook Introduction to Set Theory by Karel Hrbacek and Thomas Jech presents the proof as follows
Proof: Let $\aleph_\gamma$ be an arbitrary cardinal. Consider the sequence $(\alpha_n \mid n\in\omega)$ defined recursively by $\alpha_0=\omega_\gamma$ and $\alpha_{n+1}=\omega_{\alpha_n}$.
Let $\alpha=\lim_{n\to\omega}\alpha_n$. It is clear that the sequence $(\aleph_{\alpha_n} \mid n\in\omega)$ has limit $\aleph_\alpha$. But then we have $$\aleph_\alpha=\lim_{n\to\omega}\aleph_{\alpha_n} = \lim_{n\to\omega}\alpha_{n+1}=\alpha$$
Since $\aleph_\alpha$ is the limit of a sequence of smaller cardinals of length $\omega$, it is singular.
I have some confusion about this proof:
In the authors' definition of limit, the sequence must be increasing. But if we choose $\aleph_\gamma$ such that $\aleph_\gamma=\gamma$, then the sequence $(\alpha_n \mid n\in\omega)$ is certainly constant and thus not increasing. How can $\alpha=\lim_{n\to\omega}\alpha_n$ be well defined?
Similarly, if we choose $\aleph_\gamma$ such that $\aleph_\gamma=\gamma$, then the sequence $(\alpha_n \mid n\in\omega)$ is certainly constant. It follows that the sequence $(\aleph_{\alpha_n} \mid n\in\omega)$ is constant and thus not increasing. How can the limit of the sequence $(\aleph_{\alpha_n} \mid n\in\omega)$ be well defined?
Put the above problems aside, I assume that the limit of sequence $(\aleph_{\alpha_n} \mid n\in\omega)$ is well defined. I am unable to prove $$\alpha=\lim_{n\to\omega}\alpha_n \implies \aleph_\alpha=\lim_{n\to\omega}\aleph_{\alpha_n}$$ Please shed some lights!
Thank you so much!