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I would like to know whether tetration and quintation functions are well defined for infinite cardinals, thus, for example, $$\aleph_0 \text{ [tet] } \aleph_0 = \aleph_0 ^ {\aleph_0 ^ {\aleph_0^{\dots}}}$$ taken $\aleph_0$ times and compounded from top to bottom, to produce a higher cardinality power set upon every recursive iteration.

In an analogous manner, $$\aleph_0 \text{ [quint] }\aleph_0 = \aleph_0 \text{ [tet] } (\aleph_0 \text{ [tet] } (\aleph_0 \text{ [tet] } \dots))$$ taken recursively $\aleph_0$ times, again compounded from top to bottom.

Since as one moves to ever higher compouding operations like tetration, quintation, sextation etc. the series of new cardinals so producded would cleary increase MUCH faster than under an operation as "slow" as "mere" exponentially recursive power set taking, do we perhaps get the production of new huge cardinals that are of a different order and character than those generated by applying infinite power set operartions? Would all of these much larger cardinals, including even $\aleph_0\, [\aleph_0\text{ compounding operation}] \,\aleph_0$, itself compounded $\aleph_0$ times etc. still be smaller than other better known large cardinals defined in some other manner?

I hope these questions are fairly easy to comprehend. They hardly go much beyond elementary set theory. Thanks for reading and let me know if you can answer any of them!

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    $\begingroup$ Please use 'MathJax' for equations $\endgroup$
    – Eddy
    Nov 21, 2017 at 21:37
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    $\begingroup$ As far as I know there is no natural way to define an infinite tower of cardinal exponentiation. You could define it as a limit, but that is inconsistent with how infinite products are defined as distinct from a limit of finite products. $\endgroup$ Nov 21, 2017 at 21:38
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    $\begingroup$ I once asked a similar question about ordinal arithmetic; the Op might find it interesting. math.stackexchange.com/questions/1359801/… $\endgroup$
    – Sheldon L
    Nov 22, 2017 at 1:33
  • $\begingroup$ Thanks for the reference, Sheldon! I did find it very interesting. It reminded me of how I once mentioned the ordinal "epsilon omega" to one of my math colleagues only to hear him stammer back emotionally; "What?! Not that damn thing again!! I was up for three days straight trying to prove to myself that it must still be countable!!" $\endgroup$
    – Wd Fusroy
    Nov 25, 2017 at 1:26

2 Answers 2

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dot dot dots are problematic in expressions because we require all expressions to be finite in length. A reasonable way to interpret the first part of your question is $$\aleph_0 \text{ [tet] } \aleph_0 = \sup\{\aleph_0, \aleph_0 ^ {\aleph_0},\aleph_0 ^ {\aleph_0 ^ {\aleph_0}},\dots\}$$ This is less than or equal to the first strong inaccessible (assuming it exists). A standard result is that if the exponent is infinite we can replace any base less than or equal to the cofinality of the exponent with $2$ so now we have $$\aleph_0 \text{ [tet] } \aleph_0 = \sup\{\aleph_0, 2 ^ {\aleph_0},2 ^ {2 ^ {\aleph_0}},\dots\}$$ If $\kappa$ is strongly inaccessible, we have $2^\alpha \lt \kappa$ whenever $\alpha \lt \kappa$ so the $\sup$ is at most $\kappa$. It must be lower because this is an increasing chain of length $\aleph_0$ so the cofiniality of the $\sup$ is $\aleph_0$ and the $\sup$ is not inaccessible.

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    $\begingroup$ Hmm, why can't we take $\cup_n A_n$ with $A_n = 2^{A_{n-1}}$, $A_1 = \mathbb{N}$ within ZFC? Doesn't this show the $\sup$ is an accessible cardinal? $\endgroup$ Nov 22, 2017 at 6:12
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    $\begingroup$ @JairTaylor: You are right. It has to have cofinality $\aleph_0$. $\endgroup$ Nov 22, 2017 at 6:23
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    $\begingroup$ That makes sense, thanks! I was not previously familiar with cofinality. $\endgroup$ Nov 22, 2017 at 6:29
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    $\begingroup$ @JairTaylor . You can't "reach" an inaccessible cardinal by recursion from "below" it. For one thing, if ZF (or ZFC) is consistent then it cannot prove that an inaccessible exists. $\endgroup$ Nov 23, 2017 at 2:16
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    $\begingroup$ @DanielWainfleet I know, I was showing that it was accessible by showing you could prove its existence in ZFC. Or did I misunderstand you? $\endgroup$ Nov 23, 2017 at 19:43
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Assuming the GCH, the tetration example is tantamount to $\aleph_\omega$. If you rewrite the GCH as a tetration formula, then the pentation gets you the sequence $\aleph_\omega$, $\aleph_{\omega_\omega}$... until you get the first fixed point of the aleph function. (This is because $\aleph_0 \uparrow\uparrow \aleph_0$ gets you $\aleph_{\aleph_0}$ by replacing the index with the given factor, so then $\aleph_0 \uparrow\uparrow \aleph_{\aleph_0}$ would be $\aleph_{\aleph_{\aleph_0}}$, etc.)* But even waiving the GCH, the other posters are right, you never get higher than various $\beth$ numbers and their like, based on higher operations, without presupposing an inaccessible number (as the index of your operator) in the first place.

Major edit: even a compounded operation like $\aleph_0 \uparrow^{\aleph_0 \uparrow^{\aleph_0}\aleph_0} \aleph_0$ or an infinite such tower would give a cardinal cofinal with $\aleph_0$ and hence not an inaccessible cardinal.

Second major edit: consider $2 \uparrow^n m$, for $m$ greater than 2. This tends towards $\omega$ as $n$ increases, but without presupposing $\omega$ in place of $n$, this sequence never actually reaches it. Likewise, the smallest uncountable inaccessible must be presupposed as the index of the operation that "accesses" it. Let $\kappa$ be the smallest uncountable inaccessible. Then write $\aleph_{\alpha<\kappa} \uparrow^{\beta<\kappa} \lambda < \kappa$, when $\lambda$ is < $\kappa$ also. This expresses how without starting from $\kappa$ somewhere in one's arithmetic, one will not reach it. So one might parse "inaccessible" as "inaccessible without presupposition," hence the need for an axiom of inaccessible cardinality.

*Start with the fact that $2^{\aleph_0} = \aleph_0^{\aleph_0} = \aleph_0 \uparrow\uparrow 2$, or $\aleph_0 \uparrow\uparrow 1+1$. Then the GCH (for the $\aleph_n$ anyway) in tetration format would be $\aleph_n \uparrow\uparrow 1 + m = \aleph_{n+m}.$ For example, $2^{\aleph_2} = \aleph_2^{\aleph_2} = \aleph_2 \uparrow\uparrow 2$, so the formula gives the successor cardinal in this case again, and so on. In the infinite case, $1 + \aleph_0 = \aleph_0$, so $\aleph_0 \uparrow\uparrow \aleph_0 = \aleph_{0+\aleph_0}$. Note that $n+\aleph_0 = \aleph_0$, for all natural numbers $n$. So here $\aleph_2 \uparrow\uparrow \aleph_0$ also gives $\aleph_{\omega}$, and so on for all the $\aleph_n$. But $\aleph_0 \uparrow\uparrow \aleph_2 = \aleph_{\omega_2}$, here, etc.

Last edit: it turns out, though, that these formulae break down at the first fixed point of the aleph function, $\aleph_{\omega_{\omega_{\omega...}}}$. I will call this $\aleph_f$ or just then $f$ since $\aleph_f = f$ (this is its fixed-point character). Even in ZFC, we can establish some hypothetical bounds on powersets, like Shelah's $2^{\aleph_n} < \aleph_{\omega} \rightarrow 2^{\aleph_{\omega}} < \aleph_{\omega_4}$. We can also express failures of the GCH at various points. See Moti Gitik's "No bound for the first fixed point" for a model where the GCH holds up to $f$ but there is no bound as such on the powerset of $f$. Analogously, on our tetration model, $f \uparrow\uparrow f = f \uparrow^5 2$; I will leave to you the exercise of showing that our tetration formula gives a directly impossible result thereby. Note that here $\aleph_0 \uparrow\uparrow \aleph_f = \aleph_{\aleph_f} = \aleph_f = f$. The GCH would have $f^f = f^+$, which I think would be $\aleph_{f+1}$, which I think isn't a fixed point. But all this gives us that $f \uparrow\uparrow 2 = \aleph_{f+1}$. Now what happens then if our formula adds $f$ to itself?

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  • $\begingroup$ Thanks, Kristian. That strikes me as both very interesting and hardly trivial since it essentially means that no kind or number of compounded arithmetic operations will get you beyond the entire non-inaccessible indexed Beth series, so presumably there is no series ordered by end parts, "bep"s, however rapid it may be, that "breaks through" to the smallest inaccessible. That doesn't surprise me, but it is hardly trivial. Has anyone explicitly studied this "arithmetical inaccessibility" in detail. Usually they only go as far as the power set operation. $\endgroup$
    – Wd Fusroy
    Feb 5, 2021 at 2:32
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    $\begingroup$ Usually the axiom of replacement covers this. Without it, $\aleph_\omega$ cannot be shown to exist in ZFC, but replacement lets us construct functional images of lower sets to get higher ones (put roughly). So infinite tetration, etc. all fall under replacement. Hint: look into descriptions of the first inaccessible that compare it to $\aleph_0$ for the analysis you're looking for. $\endgroup$ Feb 5, 2021 at 3:46
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    $\begingroup$ I added some details to my answer, hopefully well-formed and informative. $\endgroup$ Feb 6, 2021 at 2:58
  • $\begingroup$ Thanks for the additional information, Kristian! I noticed that someone voted your latest comment as "unhelpful." I am baffled by that since I found it VERY helpful to me as it allows me to see much more clearly and simply why no arithmetical compounding operation can ever get one to some higher level cardinals than are obtainable -- albeit more slowly -- through the power set operation. Whether some compounding operation[s] existed beyond exponentiation that also led to whole new cardinals has bugged me since I was in the U of Chicago Math Dept. 45 years ago! $\endgroup$
    – Wd Fusroy
    Feb 6, 2021 at 11:27
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    $\begingroup$ You're welcome! Like you, my initial analysis of compounding, here, led me to wonder how large these cardinals would be relative to inaccessible, measurable, etc. cardinals. It wasn't until I learned about cofinality that my wonder was answered. $\endgroup$ Feb 6, 2021 at 14:59

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