Ordinal tetration: The issue of ${}^{\epsilon_0}\omega$

Question

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)

Answer 1

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

$$\alpha\uparrow\uparrow\beta=\begin{cases}0,&\beta=-1\\1,&\beta=0\\\alpha,&\beta=1\\\alpha^{\alpha\uparrow\uparrow\zeta},&\alpha<\omega,\beta=\zeta+1>1\\(\alpha\uparrow\uparrow\zeta)^{\alpha\uparrow\uparrow\zeta},&\alpha\ge\omega,\beta=\zeta+1>1\\\sup\{\alpha\uparrow\uparrow(\beta[\eta])|\eta<\operatorname{cf}(\beta)\},&\beta\in\mathbb{Lim}\end{cases}$$

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

$$\alpha\uparrow\uparrow\beta=\omega\forall\alpha<\omega\land\beta\ge\omega$$

Then we have,

$$\omega\uparrow\uparrow1=\omega\\\omega\uparrow\uparrow2=\omega^\omega\\\omega\uparrow\uparrow3=(\omega^\omega)^{\omega^\omega}=\omega^{\omega^\omega}$$

And so forth. Then,

$$\omega\uparrow\uparrow\omega=\varepsilon_0\\\omega\uparrow\uparrow(\omega+1)=\varepsilon_0^{\varepsilon_0}\\\vdots\\\omega\uparrow\uparrow(\omega2)=\varepsilon_1\\\vdots\\\omega\uparrow\uparrow(\omega(1+\beta))=\varepsilon_\beta\forall\beta\le\zeta_0$$

Thus, by this definition,

$$\omega\uparrow\uparrow\varepsilon_0=\varepsilon_{\varepsilon_0}$$

Edit:

The following definition may be nicer to use:

$$\alpha\uparrow^\beta\delta=\begin{cases}\alpha,&\delta=1\\\alpha^\delta,&\beta=1\\\sup\{(\alpha\uparrow^\beta\psi)\uparrow^\gamma(\alpha\uparrow^\beta\psi)|0<\gamma<\beta,0<\psi<\delta\},&\text{else}\end{cases}$$

And likewise extends to higher operations.

Answer 2

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.

Answer 3

CLIMBING METHOD

Jonathan Bowers has a method for ordinal tetration known as climbing method.

It was used by Jonathan Bowers here.

It is also used by Sbiis Saibian.

In climbing method, we pull out the 'electric guitar' - infinity barriers.

$\varepsilon_0=\sup\{1,\omega^1,\omega^{\omega^1},\omega^{\omega^{\omega^1}},...\}$

Using an infinity barrier, $\varepsilon_0$ can be thought of as $\omega^{\omega^{\omega^\cdots}}|1$, where the 1 has breached the infinity barrier.

Let me please tell you about Jack. Jack loves to climb beanstalks.

Jack starts climbing the beanstalk $\omega^{\omega^{\omega^\cdots}}$, and we obtain $\omega^{\omega^{\omega^\cdots}}+1$.

Jack climbs more, and we obtain $\omega^{\omega^{\omega^\cdots}+1}$.

After Jack climbs even more, we obtain $\omega^{\omega^{\omega^\cdots+1}}$.

Jack keeps climbing the beanstalk $\omega^{\omega^{\omega^\cdots}}|1$, until he eventually breaches the infinity barrier, and adds 1 to the 1 that already breached the infinity barrier. After he does that, we obtain $\omega^{\omega^{\omega^\cdots}}|2=\varepsilon_1$.

And then, after Jack breaches the infinity barrier again, we obtain $\omega^{\omega^{\omega^\cdots}}|3=\varepsilon_2$.

Eventually, we obtain $\omega^{\omega^{\omega^\cdots}}|\omega=\varepsilon_\omega=\omega\uparrow\uparrow(\omega+1)$.

Climbing method gives $\varepsilon_\omega$ as $\omega\uparrow\uparrow(\omega+1)$, the $\omega$'th fixed point of $\alpha\mapsto\omega^\alpha$, which is the ordinal that Mike Battaglia said they thought might make more sense to equal $\omega\uparrow\uparrow(\omega+1)$ than $\varepsilon_1$ in their answer.

Jack keeps climbing, and eventually, we obtain $\omega^{\omega^{\omega^\cdots}}|\omega^\omega=\varepsilon_{\omega^\omega}=\omega\uparrow\uparrow(\omega+2)$.

In fact, for any countable ordinals $\alpha$ and $\beta$, $\omega\uparrow^{\alpha+1}(\omega+\beta)$ is the $(\omega\uparrow^{\alpha+1}\beta)$'th fixed point of $\gamma\mapsto\omega\uparrow^\alpha\gamma$.

VEBLEN FUNCTION AND HIGHER HYPEROPERATIONS

We can continue with ordinals like $\omega\uparrow\uparrow(\omega+3)=\varepsilon_{\omega^{\omega^\omega}}$, until we reach $\omega\uparrow\uparrow(\omega*2)=\varepsilon_{\varepsilon_0}$.

After that, comes $\omega\uparrow\uparrow(\omega*2+1)=\varepsilon_{\varepsilon_\omega}$.

And then, after that, comes $\omega\uparrow\uparrow(\omega*2+2)=\varepsilon_{\varepsilon_{\omega^\omega}}$.

We can continue with that, and reach $\omega\uparrow\uparrow(\omega*3)=\varepsilon_{\varepsilon_{\varepsilon_0}}$.

We see that $\omega\uparrow\uparrow(\omega^2)=\zeta_0$.

We also see that $\omega\uparrow\uparrow(\omega^2+1)=\zeta_\omega$.

The ordinal $\omega\uparrow\uparrow(\omega^2+2)$ is equal to $\zeta_{\omega^\omega}$.

Eventually, we reach $\omega\uparrow\uparrow(\omega^2+\omega)=\zeta_{\varepsilon_0}$.

We could keep going, until we eventually reach $\omega\uparrow\uparrow(\omega^2*2)=\zeta_{\zeta_0}$.

After $\omega\uparrow\uparrow(\omega^2*2)$, we reach $\omega\uparrow\uparrow(\omega^2*3)=\zeta_{\zeta_{\zeta_0}}$.

At $\omega\uparrow\uparrow(\omega^3)$, we reach $\eta_0$.

Please notice that $\omega\uparrow\uparrow(\omega^\alpha)=\varphi_\alpha(0)\forall\alpha>0\in\textrm{Ord}$.

This means that $\omega\uparrow\uparrow(\omega^\omega)=\varphi_\omega(0)$.

It also means that $\omega\uparrow\uparrow(\omega^{\omega^\omega})=\varphi_{\omega^\omega}(0)$.

To answer the question of what $\omega\uparrow\uparrow\varepsilon_0$ is equal to, it is equal to $\varphi_{\varepsilon_0}(0)$.

After that, there is $\omega\uparrow\uparrow\uparrow4=\varphi_{\varphi_{\varepsilon_0}(0)}(0)$.

To do ordinal pentation, we will use super infinity barriers - things I made up to extend infinity barriers. I have not seen Jonathan Bowers use super infinity barriers.

$\omega\uparrow\uparrow\uparrow\omega=\sup\{1,{}^1\omega,{}^{^1\omega}\omega,{}^{^{^1\omega}\omega}\omega,{}^{^{^{^1\omega}\omega}\omega}\omega,...\}=\sup\{1,\omega,\varepsilon_0,\varphi_{\varepsilon_0}(0),\varphi_{\varphi_{\varepsilon_0}(0)}(0),...\}=\Gamma_0$

$\Gamma_0$ is the first fixed point of $\alpha\mapsto\omega\uparrow\uparrow\alpha$.

$\Gamma_0$ can be thought of as $^{^{^\cdots\omega}\omega}\omega||1$, where the one has breached the super infinity barrier.

Jack starts climbing the super beanstalk $^{^{^\cdots\omega}\omega}\omega$, and we obtain $^{^{^\cdots\omega}\omega}\omega+1$.

After Jack climbs more, we obtain $^{^{^\cdots\omega}\omega+1}\omega$.

After Jack climbs even more, we obtain $^{^{^\cdots\omega+1}\omega}\omega$.

Jack keeps climbing the super beanstalk, and eventually breaches the super infinity barrier, and adds 1 to the 1 that already breached the super infinity barrier. After he does that, we obtain $^{^{^\cdots\omega}\omega}\omega||2=\Gamma_1$.

And then, after he breaches the super infinity barrier again, we obtain $^{^{^\cdots\omega}\omega}\omega||3=\Gamma_2$.

Eventually, we obtain $^{^{^\cdots\omega}\omega}\omega||\omega=\Gamma_\omega=\omega\uparrow\uparrow\uparrow(\omega+1)$.

After that, comes $\omega\uparrow\uparrow\uparrow(\omega+2)=\Gamma_{\omega\uparrow\uparrow\omega}=\Gamma_{\varepsilon_0}$.

And then, even after that, comes $\omega\uparrow\uparrow\uparrow(\omega+3)=\Gamma_{\omega\uparrow\uparrow\omega\uparrow\uparrow\omega}=\Gamma_{\varphi_{\varepsilon_0}(0)}$.

And we can keep going with that, until we reach $\omega\uparrow\uparrow\uparrow(\omega*2)=\Gamma_{\Gamma_0}$.

And at $\omega\uparrow\uparrow\uparrow(\omega^2)$, we obtain $\varphi(1,1,0)$.

And then, at $\omega\uparrow\uparrow\uparrow(\omega^2+1)$, we obtain $\varphi(1,1,\omega)$.

At $\omega\uparrow\uparrow\uparrow(\omega^2+2)$, we obtain $\varphi(1,1,\omega\uparrow\uparrow\omega)=\varphi(1,1,\varepsilon_0)$.

And then, after that, at $\omega\uparrow\uparrow\uparrow(\omega^3)$, we obtain $\varphi(1,2,0)$.

And then, even after that, we obtain $\omega\uparrow\uparrow\uparrow(\omega^4)=\varphi(1,3,0)$.

Please notice that $\omega\uparrow\uparrow\uparrow(\omega^{1+\alpha})=\varphi(1,\alpha,0)\forall\alpha\in\textrm{Ord}$.

This would mean that $\omega\uparrow\uparrow\uparrow\omega\uparrow\uparrow\uparrow\omega=\varphi(1,\omega\uparrow\uparrow\uparrow\omega,0)=\varphi(1,\Gamma_0,0)$.

$\omega\uparrow\uparrow\uparrow\uparrow\omega=\sup\{\omega\uparrow\uparrow\uparrow\uparrow0,\omega\uparrow\uparrow\uparrow\uparrow1,\omega\uparrow\uparrow\uparrow\uparrow2,\omega\uparrow\uparrow\uparrow\uparrow3,\omega\uparrow\uparrow\uparrow\uparrow4,...\}=\sup\{1,\omega,\Gamma_0,\varphi(1,\Gamma_0,0),\varphi(1,\varphi(1,\Gamma_0,0),0),...\}=\varphi(2,0,0)$

$\varphi(2,0,0)$ is the first fixed point of $\alpha\mapsto\omega\uparrow\uparrow\uparrow\alpha$.

$\omega\uparrow\uparrow\uparrow\uparrow\alpha$ could be defined in a similar way to how $\omega\uparrow\uparrow\alpha$ and $\omega\uparrow\uparrow\uparrow\alpha$ are defined, making $\omega\uparrow\uparrow\uparrow\uparrow(\omega+1)$ equal $\varphi(2,0,\omega)$.

And then, the ordinal $\omega\uparrow\uparrow\uparrow\uparrow(\omega+2)$ is equal to $\varphi(2,0,\omega\uparrow\uparrow\uparrow\omega)=\varphi(2,0,\Gamma_0)$.

After that, comes the ordinal $\omega\uparrow\uparrow\uparrow\uparrow(\omega+3)$, which is equal to $\varphi(2,0,\omega\uparrow\uparrow\uparrow\omega\uparrow\uparrow\uparrow\omega)=\varphi(2,0,\varphi(1,\Gamma_0,0))$.

You could keep going, until you reach $\omega\uparrow\uparrow\uparrow\uparrow(\omega*2)=\varphi(2,0,\varphi(2,0,0))$.

And then, eventually, you reach $\omega\uparrow\uparrow\uparrow\uparrow(\omega^2)=\varphi(2,1,0)$.

After that, you eventually reach $\omega\uparrow\uparrow\uparrow\uparrow(\omega^3)=\varphi(2,2,0)$.

Please notice that $\omega\uparrow\uparrow\uparrow\uparrow(\omega^{1+\alpha})=\varphi(2,\alpha,0)\forall\alpha\in\textrm{Ord}$.

$\omega\uparrow\uparrow\uparrow\uparrow\uparrow\omega=\sup\{\omega\uparrow\uparrow\uparrow\uparrow\uparrow0,\omega\uparrow\uparrow\uparrow\uparrow\uparrow1,\omega\uparrow\uparrow\uparrow\uparrow\uparrow2,\omega\uparrow\uparrow\uparrow\uparrow\uparrow3,...\}=\sup\{1,\omega,\varphi(2,0,0),\varphi(2,\varphi(2,0,0),0),...\}=\varphi(3,0,0)$

$\varphi(3,0,0)$ is the first fixed point of $\alpha\to\omega\uparrow\uparrow\uparrow\uparrow\alpha$.

UNCOUNTABLY MANY UP ARROWS

In a way that could be considered an ordinal collapsing function, up arrow notation can be extended to uncountably many up arrows.

Please let $\Omega$ stand for $\omega_1$.

For countable transfinite ordinals, $\alpha\uparrow^\Omega\beta=\alpha\uparrow^\beta\alpha$, just like how for finite numbers a and b $a\uparrow^\omega b=a\uparrow^ba$, under Denis Maksudov's transfinite arrow notation.

This means that $\omega\uparrow^\Omega\omega=\omega\uparrow^\omega\omega$.

For a countable ordinal $\alpha$, $\omega\uparrow^{2+\alpha}\omega=\varphi(\alpha,0,0)$.

This means that $\omega\uparrow^\omega\omega=\varphi(\omega,0,0)$.

$\omega\uparrow^{\Omega+1}\omega=\omega\uparrow^{\omega\uparrow^{\omega\uparrow^\vdots\omega}\omega}\omega=\omega$$\{\{1\}\}$$\omega=\varphi(1,0,0,0)$.

$\varphi(1,0,0,0)$ is the first fixed point of $\alpha\mapsto\omega\uparrow^\Omega\alpha$.