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Lets define function T as $$T(0) = \aleph_0$$ $$T(1) = \aleph_{\aleph_0}$$ $$T(2) = \aleph_{\aleph_{\aleph_0}}$$ etc

No finite tower of alephs can reach the first inaccessible cardinal

My questions are:

  1. Can we 'feed' infinite ordinal numbers as a parameter to function T? I read somewhere - it was about the inaccessible cardinals - that the size of the tower is countable, but is it just a limitation of the first order ZFC with finitely-sized formulas? Can generalized function T() be expressed in, say, second order ZFC?

  2. So if there is nothing wrong with that definition of T, what is relative size of, say, $$T(\omega+1)$$ $$T(\omega^2)$$ $$T(2^{\aleph_0})$$ Are they still below the first inaccessible cardinal?

  3. Obviously, it leads to the last question: How far can we go with it, where is the fixed point of T relative to the hierarchy of the cardinals?

Thank you

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    $\begingroup$ It's not clear how you would extent $T$ nontrivially to infinite ordinals - if you let $T(\omega)$ be the limit of $T(n),n<\omega$, and try to define $T(\omega+1)$ as $\aleph_{T(\omega)}$, then you will find $T(\omega+1)=T(\omega)$, and by this process you will find $T(\alpha)=T(\omega)$ for all $\alpha\geq\omega$. Do you have some different definition in mind? $\endgroup$
    – Wojowu
    Commented Feb 27, 2021 at 11:08
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    $\begingroup$ Similar things have been asked elsewhere, eg math.stackexchange.com/questions/2689417/… and math.stackexchange.com/questions/1747465/… . See also en.m.wikipedia.org/wiki/… $\endgroup$
    – David Roberts
    Commented Feb 27, 2021 at 12:49
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    $\begingroup$ I'm not aware of any general definitions of "tower of length [whatever]" which would let $\epsilon_0^{\epsilon_0^{\epsilon_0}}$, do you know of a definition which achieves that? I don't think the problem is with limitation of ZFC to only having finite formulas or being first-order, it's about it being unclear what such a tower should mean. To address your last question: I'm not sure what $\theta_0$ is, but $T(\omega)$, the limit of $T(n),n<\omega$, is way smaller than the first inaccessible. $\endgroup$
    – Wojowu
    Commented Feb 27, 2021 at 13:27
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    $\begingroup$ An inaccessible can’t have the same cofinality as any ordinal smaller than it, and these things you’re defining have smaller cofinalities by definition. I don’t think you can reach any really big ordinals this way. $\endgroup$
    – Alec Rhea
    Commented Feb 28, 2021 at 0:15
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    $\begingroup$ @tzimie "Wiki claims that $\epsilon_1$ can be defined as infinite tower of $\epsilon_0$" Not quite - we also need to throw in some "$+1$"s to make it work. Specifically, $$\epsilon_1=\sup\{\epsilon_0+1, \omega^{\epsilon_0+1}, \omega^{\omega^{\epsilon_0+1}}, ...\}.$$ That "$+1$" might look minor but it makes a huge difference. (See e.g. here.) $\endgroup$ Commented Mar 1, 2021 at 17:46

1 Answer 1

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"Transfinite towers" often run into the problem of termination: when $F$ is a "nice" function on ordinals (= monotonic, nondecreasing, and continuous), we get $$F(\sup\{F^n(0): n\in\omega\}=\sup\{F^n(0):n\in\omega\}.$$ That is, the "iterating $F$" tower stops at level $\omega$. Moreover, if the sequence $(F^n(0))_{n\in\omega}$ was strictly increasing, then this least fixed point of $F$ will have cofinality $\omega$ and so not be inaccessible.

The "$+1$"-versions of such towers avoid this termination issue, at the cost of being a bit stranger. For example, consider the function on ordinals defined recursively as follows:

  • $S(0)=\omega+1$.

  • $S(\alpha+1)=\omega_{S(\alpha)}+1$.

  • $S(\lambda)=\sup\{S(\alpha): \alpha<\lambda\}+1.$

This function never "stops," in the sense that we always have $S(\alpha)<S(\alpha+1)$. We can then try to "strip off" the added $+1$s by taking cardinalities: let $$\hat{T}(\alpha)=\vert S(\alpha)\vert.$$ This $\hat{T}$ function isn't quite your $T$ but it's fairly similar: it begins $$\hat{T}(0)=\aleph_0, \quad\hat{T}(1)=\vert \omega_{\omega+1}+1\vert=\aleph_{\omega+1},\quad\hat{T}(2)=\vert\omega_{\omega_{\omega+1}+1}+1\vert=\aleph_{\omega_{\omega+1}+1}, \quad...$$ In particular, we get $$T(0)=\hat{T}(0)<T(1)<\hat{T}(1)<T(2)<\hat{T}(2)<...$$ (Note that $\aleph$s should not be used in subscripts to $\aleph$ numbers.)

The functions $S$ and $\hat{T}$ are provably total in $\mathsf{ZFC}$, and - being "non-silly" (e.g. not having an inaccessible somehow baked in if possible) - don't reach up to an inaccessible before we feed in an inaccessible at the outset. Specifically, the least $\alpha$ such that $S(\alpha)$ is $\ge$ the least inaccessible (in fact, equal to the least inaccessible $+1$) is the least inaccessible itself.


But from the end of your question, it sounds like this isn't quite what you want; rather, you want an iterative process which does have a fixed point, but whose least fixed point is extremely large (e.g. plausibly an inaccessible cardinal). This is going to be tricky: since $\mathsf{ZFC}$ doesn't prove that inaccessibles exist (unless it's inconsistent in the first place!), such a function has to have some aspect which is independent of $\mathsf{ZFC}$.

Here's one example of such a function, albeit in a modified setting. Working in $\mathsf{ZFC}$ + "There is a proper class of inaccessible cardinals," and restricting to uncountable cardinals for simplicity, let $C(\kappa)$ be the smallest inaccessible cardinal $\lambda$ such that every unsatisfiable $\mathcal{L}_{\kappa,\kappa}$-theory of size at most $\kappa$ has an unsatisfiable subset of size $<\lambda$. Obviously $C(\kappa)\le\kappa^+$; fixed points for $C$ are exactly the weakly compact cardinals, whose existence is independent of $\mathsf{ZFC}$ + "There is a proper class of inaccessible cardinals."

But the function $C$ itself isn't really very interesting, and in fact the property of being a $C$-fixed point is more naturally expressed without reference to $C$ itself (namely, "is inaccessible and has the weak compactness property"). In general, large cardinals are rarely best thought of as least fixed points.

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