Definitions:
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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 $\nu<\mu<\lambda$.
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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.
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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
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$\alpha_\xi < \kappa$ for all $\xi < \lambda$
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$\lambda<\kappa$
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$\lambda$ is a limit ordinal
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$\kappa=\lim_{\xi \to\lambda}\alpha_\xi$
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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:
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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?
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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?
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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!
Best Answer
Questions 1 and 2 do highlight slight inconsistencies in the approach, but these can be easily fixed. Question 3 follows immediately from the definition of $\omega_\alpha$ when $\alpha$ is a limit.
One easy fix is to view this as a proof by contradiction: we begin by supposing that $\omega_\gamma$ is uncountable, singular, and so large that no singular $\alpha\ge\omega_\gamma$ has $\omega_\alpha=\alpha$. From this we then know that the sequence $(\omega_\gamma,\omega_{\omega_\gamma},\omega_{\omega_{\omega_\gamma}},...)$ is increasing (specifically this uses the fact that if $\kappa$ is singular then $\omega_\kappa$ is singular), and hence we can define its limit $\alpha$ and proceed unworried. Since $\alpha$ has countable cofinality (and is clearly uncountable) by construction, we'll be done if we can show that $\omega_\alpha=\alpha$.
That $\alpha=\lim_{n\rightarrow\omega}\alpha_n\implies \omega_\alpha=\lim_{n\rightarrow\omega}\omega_{\alpha_n}$ then follows immediately from the definition of the ordinal $\omega_\theta$: recall that by definition, if $\theta$ is a limit then $\omega_\theta=\sup_{\beta<\theta}\omega_\beta$. Now since $\alpha$ is a limit ordinal (the limit of any increasing sequence is a limit ordinal) we have $\omega_\alpha=\sup_{\beta<\alpha}\omega_\beta$, but since the set $\{\alpha_n: n\in\omega\}$ is cofinal in $\alpha$ this implies $$\omega_\alpha=\sup_{\beta<\alpha}\omega_\beta=\sup_{\beta\in\{\alpha_n:n\in\omega\}}\omega_\beta,$$ and this is just $\sup_{n<\omega}\omega_{\alpha_n}.$
We could also modify the construction of the sequence: let $\alpha_0=\omega_\gamma$ and let $(\alpha_{i+1}=\omega_{\alpha_i})^+$. Then the sequence $(\alpha_i)_{i\in\omega}$ is clearly increasing, and its limit is uncountable and has cofinality $\omega$, hence is singular.
Not entirely related, but worth mentioning: we can also modify the definition of "limit" to apply to a broader class of sequences. Specifically, whenever $(\alpha_\eta)_{\eta<\lambda}$ is a sequence of ordinals which is nondecreasing - that is, which satisfies $\eta<\delta<\lambda\implies\alpha_\eta\le\alpha_\delta$ - then $\sup_{\eta<\lambda}\alpha_\eta$ exists, and we can call this the limit of that sequence (note that this matches up with the notion of limit of a $\lambda$-indexed sequence coming from topology).
With this modification, everything works nicely: it's clear that the sequence $\omega_\gamma,\omega_{\omega_\gamma},\omega_{\omega_{\omega_{\gamma}}}$,... is nondecreasing, and so we can define its limit.
Ignoring singularity for a moment, it's a good exercise to show that even with this modified definition, the following remains true:
HINT: if the sequence $(\alpha_\eta)_{\eta<\lambda}$ is not eventually constant, WLOG assume it is increasing (if necessary, pass to a subsequence) and use the definition of $\omega_\theta$ for limit $\theta$ as above; otherwise, things are fairly trivial ...
So this gives that there are arbitrarily large fixed points of the map $\alpha\mapsto\omega_\alpha$ (note that any such fixed point must be an uncountable cardinal). It doesn't give singularity, of course, but it's still worth understanding. In fact, here's an important principle you should really prove and understand:
So this really isn't about sequences at all, just how the map $\alpha\mapsto\omega_\alpha$ behaves with respect to suprema.