# There are arbitrary large singular cardinals $\aleph_\alpha$ such that $\aleph_\alpha=\alpha$

Definitions:

1. 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 $$\nu<\mu<\lambda$$.

2. 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.

3. 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.

1. 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?

2. 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?

3. 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!

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:

If $$(\alpha_\eta)_{\eta<\lambda}$$ is a nondecreasing sequence of ordinals with limit $$\alpha$$, then the sequence $$(\omega_{\alpha_\eta})_{\eta<\lambda}$$ is also nondecreasing and has limit $$\omega_\alpha$$.

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:

Suppose $$A$$ is any set of ordinals. Then $$\omega_{\sup A}=\sup\{\omega_\alpha:\alpha\in A\}.$$

So this really isn't about sequences at all, just how the map $$\alpha\mapsto\omega_\alpha$$ behaves with respect to suprema.

• Now it is too late for me to carefully read your answer. I will read it tomorrow. I am very surprised at how comprehensive and well-organized your answers are. They are all very high quality. Commented Dec 8, 2018 at 16:10
• It seems to me that your definition of nondecreasing is exactly the same as that of increasing from the authors. Commented Dec 8, 2018 at 16:15
• @LeAnhDung Whoops, fixed. But the point should be clear - limits make sense as long as the sequence doesn't "oscillate," so the nondecreasing condition (+ the fact that every set of ordinals is bounded above, and all nonempty sets of ordinals have least elements) is enough to ensure that limits always exist. Commented Dec 8, 2018 at 16:29
• Thank you for your comment! Please check if my underatanding correct! 1. For the limit of the increasing/nondecreasing sequence $(\alpha_\xi \mid \xi < \lambda)$ of ordinals to be meaningful, we require $\lambda$ to be limit. 2. For any sequence $(\alpha_\xi \mid \xi < \lambda)$ of ordinals, $\sup\{\alpha_\xi \mid \xi <\lambda\}$ always exists. Commented Dec 8, 2018 at 16:35
• @LeAnhDung Sure, although we don't actually have to require $\lambda$ to be a limit - it's just that if $\lambda$ isn't a limit, the limit is just the last term. More importantly, though, the subtleties around sequences aren't what this result is about; see the very last point of my answer: really, what's going on here is that for any ordinal $\gamma$ we can define the set $\{\omega_\gamma,\omega_{\omega_\gamma},\omega_{\omega_{\omega_\gamma}}, ...\}$; the supremum of this set exists, and we can show that it is a fixed point for the function $\alpha\mapsto\omega_\alpha$. Commented Dec 8, 2018 at 16:37