Since using transfinite induction it's possible to demonstrate that for every ordinals $\zeta$, we have $\zeta<\omega_\zeta$, it seems that it isn't possible that there exist an infinite cardinal $\alpha$ (which can also be an initial ordinal) such that $\alpha=\aleph_\alpha$: infact if it's true, it results that there exist two different ordinal ($\alpha$ and $\omega_\alpha$) such that $\alpha<\omega_\alpha$ and $|\alpha|=|\omega_\alpha|$, so while $\omega_\alpha$ is an initial ordinal it must be $\alpha=\omega_\alpha$ and for the hypothesis this is impossible.
If $\zeta<\omega_\zeta$ how is it possible there exist a cardinal $\alpha$ such that $\alpha=\aleph_\alpha$
cardinalsordinalsset-theory
Related Solutions
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.
I have an idea and hope I haven't made a mistake anywhere.
Suppose that exists such $\beta$. There are 2 cases:
- Let $\beta$ $-$ limit ordinal. We can find ordinal $\alpha$ such that $\aleph_\alpha > \beta$. Then we have the following:
$$\aleph_\alpha > \beta \geq cf(\beta) = cf(\alpha\beta) = cf(\aleph_{\alpha\beta}) = cf(2^{\aleph_\alpha}) > \aleph_{\alpha}.$$ Thus we have a contradiction.
- If $\beta$ is not a limit ordinal, then let’s choose arbitrary limit ordinal $\alpha > \omega$. Then we have the following:
$$\aleph_\alpha \geq \alpha \geq cf(\alpha) = cf(\alpha\beta) \geq cf(\aleph_{\alpha\beta}) = cf(2^{\aleph_\alpha}) > \aleph_{\alpha}.$$ Contradiction again.
Best Answer
No, it isn't. What you can show is that $\zeta\le\omega_\zeta$ for all $\zeta$, but that's quite different.
Any attempted inductive argument to prove the claim above will run into trouble at limit stages. For example, just because $\alpha_n<\beta_n$ for all $n<\omega$ does not mean that $\sup\{\alpha_n:n\in\omega\}<\sup\{\beta_n:n\in\omega\}$; consider $$\alpha_n=n, \beta_n=2n.$$
So when $\lambda$ is a limt you cannot argue that $$\forall\zeta<\lambda(\omega_\zeta<\omega_\lambda)\quad\implies\quad\lambda<\omega_\lambda.$$