Tetration of Alephs

Question

Does this even mean anything?

$\underbrace{x^{x^{x^{...^{x^x}}}}}_n$

Where $n = \aleph_0$?

Because I "know" it converges when (say) $x = .5$ and $n \to \infty$

You probably want $n=\omega$ as an ordinal, rather than $\aleph_0$ as a cardinal. In either case, cardinals and ordinals and $\infty$ are not the same thing. — Asaf Karagila, Jun 06 '13 at 05:37
The answer is in the question: it makes sense when the sequence $u_n=\underbrace{x^{x^{x^{...^{x^x}}}}}_n$ converges. — Jean-Claude Arbaut, Jun 06 '13 at 05:37
See also my answer here: http://math.stackexchange.com/questions/90758/the-aleph-numbers-and-infinity-in-calculus/ — Asaf Karagila, Jun 06 '13 at 05:38
ok, but you can do $1 + \aleph_0 = \aleph_0$, right? why is that allowed? why must that not be an ordinal number? — Tom Lehman, Jun 06 '13 at 05:40
why is it that you "need" an ordinal in the tetration case and not in the addition case? — Tom Lehman, Jun 06 '13 at 05:42
@arbautjc: Yes, if it's an initial ordinal, but even then, cardinal operations and ordinal operations are not the same in general. — Cameron Buie, Jun 06 '13 at 05:44

A.S · Answer 1 · 2013-06-06T05:45:19.923

2

This is meaningful for ordinal numbers only. This operation has no meaning in cardinal arithmetic (explained below). We can define an operation $\uparrow \uparrow$ using transfinite recursion as follows:

$$\alpha \uparrow \uparrow 0 = 1$$

$$\alpha \uparrow \uparrow \beta + 1 = \alpha^{ \alpha \uparrow \uparrow \beta} $$

$$\alpha \uparrow \uparrow \gamma = \sup \{ \alpha \uparrow \uparrow \beta : \beta < \gamma \} \text{ where $\gamma$ is a limit ordinal}$$

EDIT: This is only meaningful if we are talking about this as an operation with ordinal numbers. This has no meaning for cardinal numbers. However, we can talk about this operation with the ordinal number $\omega$, for example.

The problem is that we want our hypothetical operation $2\uparrow \uparrow \omega$ or $2 \uparrow \uparrow \aleph_0$ to be a fixed point so that $2^{ 2 \uparrow \uparrow \omega } = 2 \uparrow \uparrow \omega$. But if $\aleph_\kappa$ is an infinite cardinal, then $2^{\aleph_\kappa} \ne \aleph_\kappa$ so our operation would have no fixed points.

edited Jun 06 '13 at 05:45

answered Jun 06 '13 at 05:38

A.S

9,016
2
28
63

2

Note that $\aleph_0$ is not an ordinal. It's a cardinal. – Asaf Karagila Jun 06 '13 at 05:38
ok, but you can do $1 + \aleph_0 = \aleph_0$, right? why is that allowed? why must that not be an ordinal number? – Tom Lehman Jun 06 '13 at 05:41
3

@AsafKaragila, note that a cardinal is also an ordinal. – Jean-Claude Arbaut Jun 06 '13 at 05:42
@arbautjc How so? – Loki Clock Jun 06 '13 at 05:44
1

@HoraceLoeb There's addition of quantities, and addition of positions. If you were adding the position 1 to the position $\omega$, you would expect it to be distinct from $\omega$. That is, ordinal addition is affine. – Loki Clock Jun 06 '13 at 05:46
1

Well, actually $1 + \omega = \omega \ne \omega + 1$. – A.S Jun 06 '13 at 05:47
@AndrewSalmon That makes sense. Then you can also talk about the index to sequences in the limit. – Loki Clock Jun 06 '13 at 05:48
2

@arbautjc: Cardinals are not ordinals. The arithmetical systems are completely different in every aspect when it comes to infinite ordinals and infinite cardinals. – Asaf Karagila Jun 06 '13 at 05:50
4

@Asaf Many people identify cardinals with initial ordinals. I would go so far as to say this is the standard definition in $\mathsf{ZFC}$. – Trevor Wilson Jun 06 '13 at 05:59
1

@TrevorWilson I believe Asaf is referring to the difference between ordinal and cardinal arithmetic: the two systems have completely different rules for addition, multiplication, and exponentiation. – A.S Jun 06 '13 at 06:01
@Trevor: What Andrew said. Also, pack the book! – Asaf Karagila Jun 06 '13 at 06:02
2

Well, he also said that. But twice he said that cardinals are not ordinals, which is what I am disagreeing with (in the $\mathsf{ZFC}$ context.) – Trevor Wilson Jun 06 '13 at 06:03
@Asaf Thanks for the remainder about the book – Trevor Wilson Jun 06 '13 at 06:03
@Trevor: I meant as arithmetical systems, as my third clarification. Third time the charm and all that... – Asaf Karagila Jun 06 '13 at 06:05

bof · Answer 2 · 2020-04-02T01:38:04.003

1

I don't know about cardinal numbers, but hyperoperations on ordinal numbers have been studied, e.g. by John Doner and Alfred Tarski, An extended arithmetic of ordinal numbers, Fund. Math. 65 (1969), 95–127. That paper is freely available here:

http://yadda.icm.edu.pl/mathbwn/element/bwmeta1.element.bwnjournal-article-fmv65i1p10bwm

edited Apr 02 '20 at 01:38

answered Apr 02 '20 at 01:24

bof

78,265

user76284 · Answer 3 · 2020-04-02T15:19:00.007

For a cardinal/combinatorial perspective, note that $2 \uparrow \uparrow n = | V_{n+1} | = 2^{|V_n|}$ for finite $n$, where $V_n$ is stage $n$ of the cumulative hierarchy of sets. We could extend this to transfinite $n$. There might also be a sensible way to extend it to bases other than 2.

A problem here is that the arithmetic we would expect seems to be inconsistent for transfinite cardinals. For instance, \begin{align} 2 \uparrow \uparrow \aleph_0 &= 2 \uparrow \uparrow (\aleph_0 + 1) \\ &= 2^{2 \uparrow \uparrow \aleph_0} \\ &> 2 \uparrow \uparrow \aleph_0 & \text{(Cantor's theorem)} \end{align}

Kristian Berry · Answer 4 · 2021-04-26T21:46:59.350

Since $2^{\aleph_0} = \aleph_0^{\aleph_0}$, we can generalize and have that $\wp(κ) = κ ↑↑ 2$. Technically, $\aleph_0 ↑↑ \aleph_0 = \beth_{\omega}$ (I think someone else on this site noted this elsewhere, but I can't find the posting). You can also use Wolfram Alpha to iterate self-exponentiation of $\aleph_0$ and get the $\beth_n$ from this.

Tetration also shows up in accounts of ultrafilters, via the expression "2^{2^κ}." See e.g. page 4 of this article.

As for 2 ↑↑ $\aleph_0$, this would depend on what $2^{\aleph_0}$ (as 2 × 2 × ...) is (note that 2 + 2 + ... as 2 × $\aleph_0$ = $\aleph_0$). Presumably, it would be some $\kappa$ greater than $\aleph_0$, but how much so? We wouldn't know...

Tetration of Alephs

4 Answers4

Linked