TITLE: Ordinal tetration: The issue of ${}^{\epsilon_0}\omega$
QUESTION [15 upvotes]: 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.
$${\tiny⋰}^{\epsilon_n^{\epsilon_n}}={}^{\omega}(\epsilon_n)=\epsilon_{n+1}$$
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
$${}^{{}^{\omega}}{}^{\omega}\omega={}^{\epsilon_0}\omega$$
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)
$$(a^b)^{{}^{n-1}a}=({}^na)^b$$
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):
$${}^{({}^j\omega)}\omega<f({}^{({}^j\omega)}\omega)<{}^{({}^{j+1}\omega)}\omega$$
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)

REPLY [2 votes]: Some time ago I happened to stumble upon an extension of the usual ordinal arithmetic, which I've found to satisfy many properties which may be interesting to consider, including nice relationships to Veblen's functions, a tighter version of the "sandwiching" function you sought, and consequentially identities that allow bases to be changed, reducing things such as ${}^{\omega+\omega}\omega$ to ${}^\omega({}^\omega\omega)$.
Motivation
Rather than Mike Battaglia's approach using left-sided recursion $(\omega^{1+x}=\omega\cdot(\omega^x))$, I sought an approach which used right-sided recursion $(\omega^{x+1}=(\omega^x)\cdot\omega)$ which unified the desired tetration $(\omega\uparrow\uparrow3=\omega^{(\omega^\omega)}$ instead of $\omega\uparrow\uparrow3=(\omega^\omega)^\omega)$ with lower hyperoperators.
At the core of this extension was my answer to the question:

Why should $\omega\uparrow\uparrow3=\omega^{\omega\uparrow\uparrow2}$ instead of $(\omega\uparrow\uparrow2)^\omega$, since the usual definitions have $\omega^3=(\omega^2)\cdot\omega$ and more generally $\omega^{\alpha+1}=(\omega^\alpha)\cdot\omega$?

Mike Battaglia's approach answers this question by suggesting the recursion is left-sided, but if we wish to stay true to the right-sided recursion, then we would end up getting
$$\omega\uparrow\uparrow(\alpha+1)=(\omega\uparrow\uparrow\alpha)\uparrow\omega$$
and if we were to use this and its higher hyperoperator extensions, there wouldn't actually be any issues. Note that $\varepsilon_0^\omega>\varepsilon_0$ as well, so this does not end up becoming constant eventually. It also aligns with lower hyperoperators.

So what is the issue with this? Why do we desire the recursion to work out on the other side?

My answer is that it's simply not as big as it could be. With lower hyperoperations, we can get away with saying that using
$$\omega\cdot(\omega^\alpha)\le(\omega^\alpha)\cdot\omega$$
so choosing the right side for $\omega^{\alpha+1}$ guarantees the result is as large as it could be while staying true to being recursively defined, but tetration is the first point where this makes a difference.
Definition
As such my extension of the usual ordinal arithmetic relies on simply taking the larger of $\alpha\uparrow(\alpha\uparrow\uparrow\beta)$ and $(\alpha\uparrow\uparrow\beta)\uparrow\alpha$. In full, my definition is given by
$$\alpha\uparrow^\mu\beta:=\begin{cases}\alpha\cdot\beta,&\mu=0\\1,&\mu\ne0\land\beta=0\\\sup\{\alpha\uparrow^{\mu'}(\alpha\uparrow^\mu\beta'),(\alpha\uparrow^\mu\beta')\uparrow^{\mu'}\alpha~|~\beta'<\beta\land\mu'<\mu\}&\mu\ne0\land\beta\ne0\end{cases}$$
The above definition is meant to encompass all limit cases, such as $\alpha\uparrow^2\omega$ and $\alpha\uparrow^\omega2$. This can be broken down a bit using the facts that my hyperoperators are increasing (strictly in all arguments other than $\alpha$ except for trivial cases such as $\alpha,\beta=0,1$):
$\alpha\uparrow^{\mu+1}(\beta+1)=\max\{\alpha\uparrow^\mu(\alpha\uparrow^{\mu+1}\beta),(\alpha\uparrow^{\mu+1}\beta)\uparrow^\mu\alpha\}$
$\alpha\uparrow^\mu(\beta+1)=\sup\{\alpha\uparrow^{\mu'}(\alpha\uparrow^\mu\beta),(\alpha\uparrow^\mu\beta)\uparrow^{\mu'}\alpha~|~\mu'<\mu\}$
$\alpha\uparrow^\mu\beta=\sup\{\alpha\uparrow^\mu\beta'~|~\beta'<\beta\}$, $\beta$ is a limit ordinal.
Furthermore, one can prove that the only times that right-sided recursion $(\alpha\uparrow\uparrow(\beta+1)=(\alpha\uparrow\uparrow\beta)\uparrow\alpha)$ occurs for $\alpha\uparrow^\mu\beta$ are when $\mu=0,1,$ or is a limit ordinal, or when $\beta$ is the successor of a non-zero limit ordinal. Otherwise left-sided recursion $(\alpha\uparrow\uparrow(\beta+1)=\alpha\uparrow(\alpha\uparrow\uparrow\beta))$ is always used.
$\rm\large Examples$
Below we present some examples in base $\omega$, but we point out that there are interesting examples in bases other than $\omega$ around the base changing formulae section.
Tetration
Here are some examples for base $\omega$:
$\omega\uparrow^2\omega=\varepsilon_0$
$\omega\uparrow^2(\omega+1)=(\omega\uparrow^2\omega)\uparrow\omega=\varepsilon_0^\omega=\omega^{\omega^{\varepsilon_0+1}}$ (note this is the first point where tetration "swaps" directions)
$\omega\uparrow^2(\omega+2)=\omega\uparrow(\omega\uparrow^2(\omega+1))=\omega^{\omega^{\omega^{\varepsilon_0+1}}}$ (and then we return to the usual tetration recursion)
$\omega\uparrow^2(\omega+\omega)=\omega\uparrow^2(\omega\cdot2)=\varepsilon_1$
$\omega\uparrow^2(\omega\cdot2+1)=\varepsilon_1^\omega=\omega^{\omega^{\varepsilon_1+1}}$ (swap after limit ordinals)
$\omega\uparrow^2(\omega\cdot2+2)=\omega^{\omega^{\omega^{\varepsilon_1+1}}}$ (normal tetration recursion otherwise)
$\omega\uparrow^2(\omega\cdot3)=\varepsilon_2$
$\omega\uparrow^2(\omega^2)=\varepsilon_\omega$
Higher hyperoperators
Some more examples for base $\omega$:
$\omega\uparrow^32=\omega\uparrow^2\omega=\varepsilon_0$
$\omega\uparrow^33=\omega\uparrow^2(\omega\uparrow^32)={}^{{}^\omega\omega}\omega=\varepsilon_{\varepsilon_0}$
$\omega\uparrow^34=\varepsilon_{\varepsilon_{\varepsilon_0}}$
$\omega\uparrow^3\omega=\zeta_0$
$\omega\uparrow^3(\omega+1)=\zeta_0\uparrow^2\omega=\varepsilon_{\zeta_0+1}$ (swapped direction)
$\omega\uparrow^3(\omega+2)=\varepsilon_{\varepsilon_{\zeta_0+1}}$
$\omega\uparrow^3(\omega\cdot2)=\zeta_1$
$~$
$\omega\uparrow^42=\omega\uparrow^3\omega=\zeta_0$
$\omega\uparrow^43=\zeta_{\zeta_0}$
$\omega\uparrow^4\omega=\eta_0$
$\omega\uparrow^4(\omega+1)=\zeta_{\eta_0+1}$
Transfinite hyperoperators $(\uparrow^{\ge\omega})$
We note that our approach differs from Mike Battaglia's approach for these cases. We do not use continuity in $\mu$ but rather just apply lower hyperoperations and take the supremum of the result. This aligns nicely with the behavior of the $\omega$-th Veblen function, which is equivalent to the supremum over applying all $n$-th Veblen functions.
Examples for base $\omega$:
$\omega\uparrow^\omega2=\varphi_\omega(0)$
$\omega\uparrow^\omega3=\varphi_\omega(1)$
$\omega\uparrow^{\omega+1}2=\omega\uparrow^\omega\omega=\varphi_\omega(\omega)$
$\omega\uparrow^{\omega+1}3=\varphi_\omega(\varphi_\omega(\omega))$
As with Mike Battaglia's, most other bases will correspond to changing the initial base used in $\varphi_0$, and the above still applies.
Relationship to Veblen's functions
It can be shown that my hyperoperators satisfy
$\omega\uparrow^{1+\mu+1}(\omega+\omega\cdot\beta)=\varphi_{\mu+1}(\beta)$
and for limit ordinals $\mu$,
$\omega\uparrow^\mu(2+\beta)=\varphi_\mu(\beta)$
It is interesting to see that the differences between the successor and limit cases for $\mu$ can be seen from fundamental sequences of $\varphi_\mu$. For successor $\mu=\gamma+1$, one usually iterates over $\varphi_\gamma$, which corresponds to $\omega$ steps in hyperoperators. For limit $\mu$, one can simply take the limit over applying $\varphi_\gamma$ for $\gamma<\mu$, which is what we've done.
In this sense, our hyperoperator can be interpreted as a way of breaking Mike Battaglia's approach down further, giving us ordinals "between" the fixed-points.
It is also interesting to note that the initial "step" between these fixed-points uses right-sided recursion, which causes
$\omega\uparrow^2(\omega+\omega\cdot\beta+1)=\omega\uparrow\omega\uparrow(\varepsilon_\beta+1)$
$\omega\uparrow^{1+\mu+1}(\omega+\omega\cdot\beta+1)=\varphi_\mu(\varphi_{\mu+1}(\beta)+1)$ when $\mu$ is a successor ordinal.
$\omega\uparrow^{1+\mu+1}(\omega+\omega\cdot\beta+1)=\varphi_\mu(\varphi_{\mu+1}(\beta)+\omega)$ when $\mu$ is a limit ordinal.
When attempting to work out these relations further, some peculiar identities arise:
\begin{align}\varepsilon_0&=\omega\uparrow^2\omega=\omega^{\omega^{\omega^{.^{.^.}}}}\\\varepsilon_1&=\varepsilon_0\uparrow^2\omega=\varepsilon_0^{\varepsilon_0^{\varepsilon_0^{.^{.^.}}}}\\\zeta_0&=\omega\uparrow^3\omega={}^{{}^{{}^{{}^{{}^..}.}\omega}\omega}\omega\\\zeta_1&=\zeta_0\uparrow^3\omega={}^{{}^{{}^{{}^{{}^..}.}\zeta_0}\zeta_0}\zeta_0\end{align}
These seemingly coincidential identities which look pleasing are more than coincidence. They are actually consequences of some interesting properties that are satisfied by our definition.
Base Changing Formulae
Just as there are ways to change the base for multiplication $((\alpha\cdot\beta)\cdot\gamma=\alpha\cdot(\beta\cdot\gamma))$ and exponentiation $((\alpha^\beta)^\gamma=\alpha^{\beta\cdot\gamma})$, there are ways to change between bases in my extension, at least to an extent. As it would turn out, these hyperoperations satisfy a generalization of the Knuth arrow theorem which prescribes a bound on the rate at which hyperoperators should grow, analogous to your attempt at "sandwiching".
The proof of the following double inequality is not trivial. Intuitively, the proof relies on the idea that $\alpha\uparrow^\mu(\beta+\gamma)$ is like starting with $\alpha\uparrow^\mu\beta$ and then iterating $\gamma$ steps more. The upper bound can be interpreted as what would happen if those $\gamma$ additional steps were base $\alpha\uparrow^\mu\beta$ instead of base $\alpha$. The lower bound can interpreted as loose enough that a tighter bound (though significantly less pleasing one) can be proven by induction, similar to the proof of the usual Knuth arrow theorem.
For $\alpha,\mu,\beta\ge2$, the following inequalities hold:
$$(\alpha\uparrow^\mu\beta)\uparrow^\mu\gamma\le\alpha\uparrow^\mu(\beta+\gamma+1)\le(\alpha\uparrow^\mu\beta)\uparrow^\mu(\gamma+2)$$
For limit ordinals $\gamma$, this tightens into an equality by continuity/sandwiching:
$$(\alpha\uparrow^\mu\beta)\uparrow^\mu\gamma=\alpha\uparrow^\mu(\beta+\gamma)$$
This gives an interesting interpretation to examples you provide:
\begin{align}{}^\omega\omega&={}^\omega\omega=\varepsilon_0\\{}^{\omega+\omega}\omega&={}^\omega({}^\omega\omega)=\varepsilon_1\\{}^{\omega+\omega+\omega}\omega&={}^\omega({}^\omega({}^\omega\omega))=\varepsilon_2\\{}^{\omega+\omega+\omega+\dots}\omega&={}^\ldots({}^\omega({}^\omega({}^\omega\omega)))=\varepsilon_\omega\end{align}
Quirks
Due to the way this extension is defined, we fail to keep left-sided identities which cause fixed-point behavior. Instead, we have right-sided recursion and attain fixed-points at limit heights.
Many of the results of this extension involved edge cases occuring at limit ordinals or successors of limit ordinals. On its own, it may seem unnatural for tetration to change directions immediately after limit ordinals. We remark, however, that for higher hyperoperators, this tends to give more pleasing results. For example,
$$\omega\uparrow^3(\omega+1)=\zeta_0\uparrow^2\omega=\varepsilon_{\zeta_0+1}$$
And from there, one has
$$\omega\uparrow^3(\omega+n)=\varepsilon_{\varepsilon_{._{._{._{\zeta_0+1}}}}}=\varphi_1^n(\zeta_0+1)$$
Similarly,
$$\omega\uparrow^2(\omega+n)=\varphi_0^{n+1}(\varepsilon_0+1)$$
The behavior of $\uparrow^\omega$ and other such limit cases may also seem unnatural, but appear less so when looking at fundamental sequences of $\varphi_\omega$. This does however cause a shift in $\uparrow^{\omega+1}$'s relation to the Veblen function.
Feferman–Schütte Ordinal
Just as with most other extensions of ordinal arithmetic, the Feferman–Schütte ordinal is the limit of the ordinal notation and can be given by
$$\Gamma_0=\omega\uparrow^{\omega\uparrow^{\omega\uparrow^\vdots\omega}\omega}\omega=\sup\left\{\omega,\omega\uparrow^\omega\omega,\omega\uparrow^{\omega\uparrow^\omega\omega}\omega,\dots\right\}$$
Extensions of these to larger ordinals such as the small Veblen ordinal can probably be made, though this seems to veer far from the point of the question.
Summary
Two major ways to extend the ordinal arithmetic to tetration and beyond often rely on either identities based around $\omega\uparrow\uparrow(1+\beta)$ or $\omega\uparrow\uparrow(\beta+1)$. In this post, we explored an extension following the latter, showing that it not only reproduces the Veblen function at limits but also satisfies desireable bounds which have reasonable interpretations and allow for interesting arithmetic to be performed.
Although our approach to extending ordinal arithmetic using $\omega\uparrow\uparrow(\beta+1)$ is not unique, we point out that it allows a unification between the desired strictly increasing tetration with lower hyperoperations while preserving the original recursive nature of their definitions: a recursion over lower hyperoperations.