So in the past few months when trying to learn about the properties of the fixed points in ordinals as I move from $0$ to $\epsilon_{\epsilon_0}$ I noticed when moving from $\epsilon_n$ to the next one $\epsilon_{n+1}$, the operation that does that is a left associative operation i.e.


Since the exact same rule holds for all epsilons to those that are indexed by successor ordinals, unwrapping the whole thing gives $\epsilon_{n},n < \omega$ in terms of $\epsilon_0$:

$$\epsilon_n=\underbrace{{}^{\omega}(\cdots{}^{\omega}({}^{\omega}(}_{\textrm{n times}}\epsilon_0)))=\underbrace{{}^{\omega}(\cdots{}^{\omega}({}^{\omega}(}_{\textrm{n-1 times}}\epsilon_1)))=\underbrace{{}^{\omega}(\cdots{}^{\omega}({}^{\omega}(}_{\textrm{n-2 times}}\epsilon_2)))=etc.$$

That means, while $\epsilon_0$ is informally speaking the same as ${}^{\omega}\omega=\omega^{\omega^⋰}$ hence $\omega [5] 2=\omega\uparrow^3 2$, a pentation, all subsequent epsilon numbers are not tetration because the iterative operations that define them recursively is left associative but not right associative (as it would be the case for hyperoperators).

This then prompt the question: What prevent us from defining ordinal tetration, do they end up collapsing into $\epsilon_0$ or related terms thus making them unnecessary, or something else happens

To begin, since $\epsilon_0=\omega[5]2$, the next probable candidate for $\omega [5]3$ will be ${}^{{}^{\omega}}{}^{\omega}\omega=\left(\left({}^{({}^{\omega})}{}^{\omega}\right)\omega\right)$ Note the definition of $\epsilon_0$ means


However, because exponentiation already lost associativity, tetration have almost no useful general identities except for finite integers $a,b,n$ (Using exponential identity $\alpha^{\beta\gamma}=(\alpha^\beta)^\gamma$ and that finite integers at the index commute)


Therefore, no known ordinal arithmetic can be used to simplify ${}^{\epsilon_0}\omega$. This prompt me to try a sandwiching approach as follows (where $f$ is some increasing normal function that serves as the iterative operation that climb up this sequence):


If we move through all $j < \omega$ and take the supremum, then we discover that ${}^{\epsilon_0}\omega$ is a fixed point of $f$. But nothing can be deduced further as I have no idea how to derive $f$ explicitly other than it is an operation that is similar to exponentiation, but only applied to the height i.e. $f: {}^j\omega \to \omega^{{}^j\omega}$. Either way, it had said nothing on whether ${}^{\epsilon_0}\omega$ is one of the epsilon number or where it is ordered wrt the usual ordinals.

Wikpedia also has a talk section saying that ordinal tetration is trivial, or that it will collapse into $\epsilon_0$. But as shown with ${}^{\epsilon_0}\omega$, I have so far failed to reproduce their results.

What other squeezing schemes (i.e. where should I put my $j$) I can use (or other properties of ordinals) in order to determine the value of ${}^{\epsilon_0}\omega$ ?

(NB You know my main focus is not large countable numbers since Veblen functions will eventually overshoot any hyperoperation)

  • $\begingroup$ For a definition of ordinal tetration, see the 5th comment under my answer here. See also John Baez's 3 essays on Large Countable Ordinals in June-July 2016: Part 1 AND Part 2 AND Part 3 $\endgroup$ – Dave L. Renfro Jun 7 '17 at 16:09
  • $\begingroup$ Incidentally, using the definition of ordinal tetration I gave in my comment (involves a bottom-up evaluation of an exponential tower; the usual top-down evaluation of exponential towers doesn't work because of collapsing problems), if $\alpha \geq 2$ and $\beta \geq \omega,$ then one can show that $^{\alpha}{\beta} = {\alpha}^{{\alpha}^{\beta}}.$ $\endgroup$ – Dave L. Renfro Jun 7 '17 at 16:16

So while in the main chat room, I devised a way to define tetration, and it works quite nicely.


For finite $\alpha,\beta$, $\alpha\uparrow\uparrow\beta$ is what you expect it to be. Then,


Then we have,


And so forth. Then,


Thus, by this definition,



The following definition may be nicer to use:


And likewise extends to higher operations.

  • $\begingroup$ This definition is pretty good, and though there is no "canonical" way to define ordinal tetration, I think a good property to preserve is $\alpha^{\alpha\uparrow\uparrow\omega}=\alpha\uparrow\uparrow\omega$ and $\alpha^{\alpha\uparrow\uparrow\beta}=\alpha\uparrow\uparrow(\beta+1)$ $\endgroup$ – user477899 Sep 18 '17 at 22:52
  • $\begingroup$ @Zetapology Yeah. However, satisfying both will not allow one to exceed $\varepsilon_0$. In the definition provided above (and in the edit), the second condition is satisfied only for finite $\beta$. $\endgroup$ – Simply Beautiful Art Sep 19 '17 at 0:24
  • $\begingroup$ Good point. Transfinite hyperoperation almost satisfies it though $\endgroup$ – user477899 Sep 19 '17 at 0:28
  • $\begingroup$ @Zetapology You really think so?$$\omega\uparrow\uparrow(\omega+1)=\varepsilon_0^{\varepsilon_0}$$ $\endgroup$ – Simply Beautiful Art Sep 19 '17 at 0:32
  • $\begingroup$ I said almost. By that I meant $\alpha^{\alpha\uparrow\uparrow\omega}=\alpha\uparrow\uparrow (\omega+1)$ $\endgroup$ – user477899 Sep 19 '17 at 0:35

I believe I incidentally answered this question while working on some research recently -- it is possible to extend the usual hyperoperation sequence $\mathcal{H}_\omega$ on $\mathbb{N}$ to a recursive sequence of operations $\mathcal{H}$ on $O_n$, with the following result:

$$\mathcal{H}_{_\Omega}(\alpha,\beta)=\begin{cases} \mathcal{S}\alpha, & \text{if} \ \Omega=0. \\ \alpha, & \text{if} \ \Omega=1 \ \text{and} \ \beta=0. \\ \mathcal{S}\alpha,& \text{if} \ \Omega=1 \ \text{and} \ \beta=1. \\ 0, & \text{if} \ \Omega=2 \ \text{and} \ \beta=0. \\ \alpha, & \text{if} \ \Omega=2 \ \text{and} \ \beta=1. \\ 1, & \text{if} \ \Omega\geq3 \ \text{and} \ \beta=0. \\ \alpha, & \text{if} \ \Omega\geq3 \ \text{and} \ \beta=1. \\ \mathcal{H}_{_{\Omega-1}}\big(\mathcal{H}_{_\Omega}(\alpha,\beta-1),\alpha\big), & \text{if} \ \Omega=\mathcal{S}\bigcup\Omega \ \text{and} \ 1<\beta=\mathcal{S}\bigcup\beta. \\ \bigcup_{\delta<\beta}\mathcal{H}_{_\Omega}(\alpha,\delta), & \text{if} \ \Omega=\mathcal{S}\bigcup\Omega \ \text{and} \ \ 1<\beta=\bigcup\beta . \\ \bigcup_{\rho<\Omega}\mathcal{H}_{_\rho}(\alpha,\beta), & \text{if} \ 0\neq\Omega=\bigcup\Omega \ \text{and} \ 1<\beta=\mathcal{S}\bigcup\beta. \\ \bigcup_{\rho<\Omega}\bigcup_{\delta<\beta}\mathcal{H}_{_\rho}(\alpha,\delta), & \text{if} \ 0\neq\Omega=\bigcup\Omega \ \text{and} \ 1<\beta=\bigcup\beta. \\ \end{cases}$$

This 'transfinite hyperoperation sequence' is a very 'large' object in the sense that each hyperoperation in the sequence is a proper class, but it is well defined under the appropriate reflection principle or under ETR and satisfies nice relations like $\mathcal{H}_4(\omega,\omega)=^\omega\omega=\omega^{\omega^{\omega^{\dots}}}$ in your question, and $\mathcal{H}_5(\omega,\omega)=^{^{^{\dots}\omega}\omega}\omega$, so on and so forth.

We can also generalize $\gamma$, $\delta$ and $\epsilon$ numbers to the $\Omega^{th}$ hyperoperation, finding ordinals that 'absorb' larger and larger recursively defined binary operations. If you would like to see a proof of the above claims, feel free to check out my paper https://arxiv.org/abs/1706.08908 -- the part involving this hyperoperation sequence begins on page 20.

  • 1
    $\begingroup$ I'm assuming here $\bigcup$ is being used in place of $\mathrm{sup}$? $\endgroup$ – user477899 Sep 18 '17 at 23:32
  • $\begingroup$ @Zetapology I'm using the Von-Neumann construction of the orindals. For any class $\mathbb{X}$ in general, $\bigcup\mathbb{X}=\{y:\exists x\big(y\in x\in\mathbb{X}\big)\}$. I'm not familiar with $\sup$. $\endgroup$ – Alec Rhea Sep 19 '17 at 0:49
  • 1
    $\begingroup$ @Simply Beautiful Art, please refrain from making trivial edits. $\endgroup$ – Alec Rhea Sep 19 '17 at 0:54
  • $\begingroup$ Ok, in the Von Neumann construction (when dealing with limit ordinals) infinite unions of previous ordinals are supremums. $\endgroup$ – user477899 Sep 19 '17 at 0:56
  • $\begingroup$ @Zetapology Got it, thanks for the clarification. $\endgroup$ – Alec Rhea Sep 19 '17 at 0:57

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