Consider the recursively defined hyperoperation sequence $\circ_i$

$$\begin{array}{rcrclclcl} x& \small{+}&(y\ {\small+}1)&:=&x& &&{\small+}&1\\ x& \boldsymbol{+}&(y\ {\small+}1)&:=&(x& \boldsymbol{+} &y)&{\small+}&x\\ x& \cdot &(y\ {\small+}1)&:=&(x& \cdot& y)&\boldsymbol{+}&x\\ x&\uparrow &(y\ {\small+}1)&:=&(x&\uparrow& y)&\cdot&x\\ x& \uparrow\uparrow &(y\ {\small+}1)&:=&(x&\uparrow\uparrow& y)&\uparrow&x\\ &&&\vdots\\ x& \uparrow^{n+1} &(y\ {\small+}1)&:=&(x&\uparrow^{n+1}& y)&\uparrow^{n}&x\\ &&&\vdots\\ \end{array}$$

with $x \circ_0 y = x {\small+} 1$ (the successor function), $x \circ_1 y = x + y$, $x \circ_2 y = x \cdot y$, $x \circ_3 y = x \uparrow y = x^y$, and so on. This sequence can be used to define a sequence of successor functions for ordinals:

$$\begin{array}{rcrclclcl} \mathsf{S}_0(x) &:=& x \circ_0 \omega&=&x&\small{+}& \omega& = &x& \small{+}& 1\\ \mathsf{S}_1(x) &:=& x \circ_1 \omega&=&x&\boldsymbol{+}& \omega\\ \mathsf{S}_2(x) &:=& x \circ_2 \omega&=&x&\cdot& \omega\\ \mathsf{S}_3(x) &:=& x \circ_3 \omega&=&x&\uparrow& \omega\\ &\vdots\\ \end{array}$$

With these successor functions and their respective limits one gets the following picture of the countable ordinals (at least the smaller ones):

enter image description here

Note that the small dots on every level correspond to the big dots on the level before.

Is this picture essentially correct and maybe helpful?

Can the limit ordinal labelled ?? (which seems to be countable) be characterized rigorously (not just by dots) and does it have a name or notation, eventually?

[Added] Thanks to Miha's suggestion I learned a little bit about fundamental sequences. So the question is about the limit of the fundamental sequence $\lbrace \omega \uparrow^n \omega\rbrace_{n \in \mathbb{N}}$. Maybe it's enough to characterize this ordinal just like this. But it would be interesting how it could be characterized otherwise.

[Added] Maybe it's trivial, but I had to learn (from here) that the limit ordinal under consideration must be a countable ordinal, because the first uncountable ordinal does'nt have a countable fundamental sequence.

  • 1
    $\begingroup$ The limit, it seems to me (off the bat), is not larger than $\omega_1^{CK}$ the least ordinal which cannot be encoded as a recursive relation on the natural numbers. This ordinal is very much countable. $\endgroup$ – Asaf Karagila Nov 27 '13 at 14:40
  • $\begingroup$ To see that it is countable (assuming the natural formalization of your picture) notice that it is the limit of the $\omega$-sequence of countable ordinals appearing directly above it in the picture: $0,\omega,\omega^2,\ldots$. $\endgroup$ – Trevor Wilson Nov 27 '13 at 16:49
  • $\begingroup$ How are these (limit) ordinals characterized formally: $0, \omega, \omega^2, \omega^\omega$ (to be able to build a formal limit)? $\endgroup$ – Hans-Peter Stricker Nov 27 '13 at 17:09
  • $\begingroup$ .. and does this limit have a name or a notation? $\endgroup$ – Hans-Peter Stricker Nov 27 '13 at 17:33
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    $\begingroup$ I don't know whether your ordinal has a special name, but I would suggest you look at Veblen functions and the Feferman-Schütte ordinal. $\endgroup$ – Miha Habič Nov 28 '13 at 1:45

The limit is $φ_ω(0)$, where φ is the Veblen function.

To formally define hyperoperations, you need to define the case $y=0$, and also (for infinite ordinals) the behavior at limit ordinals. I assume reasonable values at $y=0$, and continuous (or otherwise reasonable) behavior at limit $y$.

Your nesting of parentheses is different from the standard treatment of hyperoperations, which is relevant given non-associativity of exponentiation and higher hyperoperations. For example, in your definition, $x↑↑y = x^{x^y}$ (since $(x^x)^x = x^{x^2}$ and so on).

However, $x ↑^{n+1} (y+1) = (x ↑^{n+1} y) ↑^n x = ((x ↑^{n+1} y) ↑^n (x-1)) ↑^{n-1} (x ↑^{n+1} y)$, so while (for finite arguments) your $↑^n$ is slower (for $n>1$) than the standard $↑^n$, your $↑^{2n}$ is still faster than the standard $↑^n$, and your $↑^n$ has the benefit of working well unmodified for infinite arguments.

$x↑↑↑y$ essentially iterates exponentiation, and if $ω≤x<ε_0$, $x↑↑↑(ω+ωy) = ε_y$.
$(ω+x) ↑^{2n+1} (ω+x)$ is probably between $φ_n(x)$ and $φ_{n+1}(x)$.

However, we do not need precise bounds to get the limit point, which is $φ_ω(0)$. If we start with 0 and iterate $φ_n$ $ωα$-times, adding 1 at limits to escape the fix-points, the result will be $φ_{n+1}(α)+1$. The hyperoperations sequence grows at a similar general rate/recurrence, with finite increments of $n$ corresponding to finite increases of $n$ in $φ_n$, so the limit is $φ_ω(0)$.

To go further, you can continue hyperoperations transfinitely (with continuous or otherwise reasonable values at limit $n$). $↑^{2α}$ (with your nesting of parentheses) will roughly correspond to $φ_α$, and the supremum of ordinals expressible this way (allowing hyperoperations inside $α$) is $Γ_0$.

You can also approach $φ_ω(0)$ with appropriately defined one variable functions/hyperoperations. For example, set $f_0(x)=x$, $f_n(0)=1$ ($n>0$), and $f_{n+1}(x+1) = f_n(f_{n+1}(x)⋅2)$ with $f_n(x)$ continuous in both $n$ and $x$. We have $f_ω(x) = φ_ω(0)$ if $ω<x≤φ_ω(0)$; and $Γ_0$ is the least ordinal $α$ such that $f_α(ω+1)=α$.


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