I met the following problem in my course on Set Theory:
Prove that there is no ordinal $\beta$ such that $2^{\aleph_\alpha} = \aleph_{\alpha\beta}$ for all ordinals $\alpha > 0$.
There are no ideas on how to approach it. Any hints?
There is no ordinal $\beta$ such that $2^{\aleph_\alpha} = \aleph_{\alpha\beta}$ for all ordinals $\alpha > 0$.
cardinalsset-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.
Having $L_{\alpha+1}\models$ "$\alpha$ is countable" is a strong countability property - this means that we don't want $\alpha$ to be $\omega_1$-like, so we emphatically don't want to look at things like elementary submodels of $L_{\omega_1}$. In particular, a club of countable ordinals will not have the desired property (take e.g. set of Mostowski collapse images of $\omega_1^M$ for $M$ a countable transitive model of $L_{\omega_2}$).
Instead, note that the desired condition is the same as "$L_\alpha$ has a definable bijection between $L_\alpha$ and $\omega$" (since the stuff in $L_{\alpha+1}$ is exactly the stuff definable in $L_\alpha$, and $L_\alpha$ has a bijection between itself and $\alpha$). One nice way for a level of $L$ to see its own countability is for it to$^1$ be the first level of $L$ satisfying some sentence with hereditarily countable parameters:
Suppose $L_\eta$ satisfies enough of $\mathsf{ZFC}$, $a\in \mathsf{HC}^{L_\eta}$, and there is a formula $\varphi$ such that $L_\eta$ is the least level of $L$ containing $a$ as an element and satisfying $\varphi(a)$. Let $f_1,...,f_n$ be definable-over-$L_\eta$ Skolem functions for $\varphi(a)$, and - in $L_\eta$ - let $C$ be the Mostowski collapse of the closure of $tc(\{a\})$ under the Godel operations and the $f_i$s We have that $C$ is definable (scince we've only used bounded-quantifier-rank "tools" to build $C$ - regardless of how complicated $\varphi$ is, there are only finitely many Skolem functions we need), $a\in C$ (since we folded in the transitive closure), and $L_\eta\models$ "$C$ is countable" (since $a\in\mathsf{HC}^{L_\eta}$), so by assumption we get $C=L_\eta$.
And we can now apply this in a very silly way:
For each $a\in L_{\omega_1}$ consider the least level of $L$ which sees that $a$ is hereditarily countable.
$^1$Actually, we also need $L_\eta$ to satisfy a small fragment of $\mathsf{ZFC}$ as well. But I do mean a small fragment - we just need $L_\eta$ to be able to perform basic recursive constructions, a la Lowenheim-Skolem and Mostowski. So in partcular, $\mathsf{KP}$ is enough.
Best Answer
I have an idea and hope I haven't made a mistake anywhere.
Suppose that exists such $\beta$. There are 2 cases:
$$\aleph_\alpha > \beta \geq cf(\beta) = cf(\alpha\beta) = cf(\aleph_{\alpha\beta}) = cf(2^{\aleph_\alpha}) > \aleph_{\alpha}.$$ Thus we have a contradiction.
$$\aleph_\alpha \geq \alpha \geq cf(\alpha) = cf(\alpha\beta) \geq cf(\aleph_{\alpha\beta}) = cf(2^{\aleph_\alpha}) > \aleph_{\alpha}.$$ Contradiction again.