very slow growing ordinal function analysis

here is a new ordinal definition which has very little growth for your definition here I only give the outline, different from ordinary ordinals such as ω, ε, ζ, and η, in this case I will make a new revolution, for this revolution I will only use addition on omega and not use exponentiation,

If you have ever heard about the set C(x) in the Bhuccholz function, the set C(x) itself contains {1,0,ω,Ω} but remember that we can only use multiplication or exponentiation in a finite amount of time. ,

here I will not use AME, I will only use addition which must be repeated in a finite amount of time or {Finite Addition, 1,0,ω,Ω}, so only addition, this will make the function work slowly but I I'm sure this will definitely reach the tetrathoth at some point, you'll see, I'm just giving the analysis now

It can be said that the fixed points of some of the ordinal ordinals that I have made are only at the multiplication level, not at the tetration level as faced by regular ordinals that you usually encounter, this is the difference so the growth is slow,

The hierarchy here is called mole hierarchy because of its slow growth, the hierarchy and ordinal here are different from the regular ordinals you usually encounter, we start from omega

we can add omega to 1:

$\omega+1$

we can add omega to 1 infinitely because this is not fixed point:

$\omega +1+1+1+1+1+1+...$

This is called omega plus omega

$\omega + \omega$

and we can add omega by omega BUT ONLY FINITE AMOUNT OF TIME

$\omega +\omega +\omega +\omega +\omega +...$

and this is the fixed point for omega in the mole hierarchy

This is what differentiates it from the ordinary ordinals that you usually see, for ordinary ordinals like omega in the regular ordinal system omega itself can reach its fixed point when it is in the tetration stage to produce epsilon nought but here it is not like that, for this system the ordinal must reach fixed point in the position of repeated addition above or multiplication which is different from usual, this is called the mole hierarchy

for the fixed point omega in the mole hierarchy that I made myself, it will produce a self made ordinal, This ordinal is the regular letter E but the difference is that it is written in the old English font \mathfrak{} which is also available on mathjax, this is to differentiate it from the epsilon nought ordinal because the definition and fiex point are different, the definition is

The counting sequence for ${\mathfrak {E}}_{0}$is $1,\omega ,\omega +\omega ,\omega +\omega +\omega ,\omega +\omega +\omega +\omega$and so on infinitely, so ${\mathfrak {E}}_{0}$ approximately $\omega^2$ in regular hierarchy

for example $g_{{\mathfrak {E}}_{0}}(5)=5^{2}=25$using slow growing hierarchy

$g_{{\mathfrak {E}}_{0}}(3)=9$

$g_{{\mathfrak {E}}_{0}}(4)=16$

$g_{{\mathfrak {E}}_{0}}(6)=36$

so $g_{{\mathfrak {E}}_{0}}(n)=n^{2}$

Then what about ${\mathfrak {E}}_{1}$, the counting sequence for ${\mathfrak {E}}_{1}$ is $\{({\mathfrak {E}}_{0}+1),\omega +({\mathfrak {E}}_{0}+1),\ \omega +\omega +({\mathfrak {E}}_{0}+1),\ \omega +\omega +\omega +({\mathfrak {E}}_{0}+1)\}$and so on ad infinitum

for example:

$g_{{\mathfrak {E}}_{1}}(5)=5+5+5+5+(5^{2})+1=5\times 4+(5^{2}+1)=46$

$g_{{\mathfrak {E}}_{1}}(3)=3+3+(3^{2}+1)=16$

Approximately: $g_{{\mathfrak {E}}_{1}}(n)=n\times (n-1)+(n^{2}+1)$remember using PEMDAS

Then what about ${\mathfrak {E}}_{2}$, remember the counting sequence for ${\mathfrak {E}}_{2}$ is ${\mathfrak {E}}_{1}+1$, $\omega +{\mathfrak {E}}_{1}+1,\omega +\omega +{\mathfrak {E}}_{1}+1...$and so on ad infinitum,

we know that in slow growing hierarchy ${\mathfrak {E}}_{1}$ is approximately $g_{{\mathfrak {E}}_{1}}(n)=n\times (n-1)+(n\times 2+1)$

very slow right, I don't know if this well defined or not, but this is using standard ordinal pattern,

so in general:

1. ${\mathfrak {E}}_{0}$ is smallest ordinal that not expressible using omega doing addition in finite number of times (or mole number form)
2. Informal visualizations: ${\mathfrak {E}}_{0}=\omega +\omega +\omega +\omega ...=\omega 2$
3. ${\mathfrak {E}}_{0}=\min\{a|a=\omega +\alpha \}=\sup\{0,1,\omega ,\omega +\omega ,\omega +\omega +\omega ...\}$
4. ${\mathfrak {E}}_{\alpha +1}=\min\{\beta |\beta =\omega +\beta \land \beta >{\mathfrak {E}}_{\alpha }=\sup\{{\mathfrak {E}}_{\alpha }+1,\omega +{\mathfrak {E}}_{\alpha }+1,\omega +\omega +{\mathfrak {E}}_{\alpha }+1...\}$

Then we can continue to

${\mathfrak {E}}_{{\mathfrak {E}}_{0}},{\mathfrak {E}}_{{\mathfrak {E}}_{{\mathfrak {E}}_{0}}},{\mathfrak {E}}_{{\mathfrak {E}}_{{\mathfrak {E}}_{{\mathfrak {E}}_{0}}}},{\mathfrak {E}}_{{\mathfrak {E}}_{{\mathfrak {E}}_{{\mathfrak {E}}_{{\mathfrak {E}}_{0}}}}}$

and we have arrived to ${\mathfrak {E}}_{\alpha }$fixed point, then we need to create new ordinal that is:

${\mathfrak {Z}}_{0}$(\mathfrak{Z}₀)

${\mathfrak {Z}}_{0}$ ordinal can be defined as larger fixed points of the map $\alpha \rightarrowtail {\mathfrak {E}}_{0}$for example ${\mathfrak {Z}}_{1}$is a fixed point that is greater than ${\mathfrak {Z}}_{0}$

Formally

• \(\mathfrak{Z}_0=\textrm{min}(\{\gamma:\gamma=\mathfrak{E}_{\gamma\}})\)
• \(\mathfrak{Z}_0_\alpha=\textrm{min}(\{\gamma:\gamma=\mathfrak{E}_\gamma\land(\forall(\beta<\alpha)(\gamma>\mathfrak{Z}_\beta))\})\)
• One fundamental sequence for \(\mathfrak{Z}_1\) is \(\mathfrak{Z}_1[0]\)=\(\mathfrak{Z}_0+1\) and \(\mathfrak{Z}_1[n+1]=\mathfrak{E}_{\mathfrak{Z}_1[n]}\), and, in general, \(\mathfrak{Z}_{\alpha+1}[0]\)=\(\mathfrak{Z}_{\alpha}+1\) and \(\mathfrak{Z}_{\alpha+1}[n+1]=\mathfrak{E}{\mathfrak{Z}_{\alpha+1}[n]}\).

In general, the counting sequence for ${\mathfrak {Z}}_{0}$ is ${\mathfrak {E}}_{0},{\mathfrak {E}}_{{\mathfrak {E}}_{0}},{\mathfrak {E}}_{{\mathfrak {E}}_{{\mathfrak {E}}_{0}}},{\mathfrak {E}}_{{\mathfrak {E}}_{{\mathfrak {E}}_{{\mathfrak {E}}_{0}}}}$and so on ad infinitum

and, in general:

${\mathfrak {Z}}_{\alpha +1}[0]={\mathfrak {Z}}_{\alpha },$and ${\mathfrak {Z}}_{\alpha +1}[n+1]={\mathfrak {E}}_{{\mathfrak {Z}}_{\alpha }+1[n]}$

then we can do this

${\mathfrak {Z}}_{{\mathfrak {Z}}_{0}},{\mathfrak {Z}}_{{\mathfrak {Z}}_{{\mathfrak {Z}}_{0}}},{\mathfrak {Z}}_{{\mathfrak {Z}}_{{\mathfrak {Z}}_{{\mathfrak {Z}}_{0}}}},$and we have arrive to ${\mathfrak {Z}}_{0}$ fixed point

then after this we will not create another ordinal, to save time we only need to create a generalization function, for this function it is similar to the bhucholz function but this is only used in mole hierarchical systems and nothing else so the definition will be different, I am sure the numbers here I don't know if it has reached tetratoth level or less (in SGH) please just calculate it yourself

Of course I will only show the analysis and outline, I can't write long here because it's a waste of time, I'm also working on a school project right now and there are lots of other things that need to be done, so please

for this function I will call it the mole function because I only think of the word mole, the symbol is the letter λ in this case, don't confuse it with the other λ please

The definition of the mole function is similar to the Bhuccholz function but only includes addition, which is different from the Bhuccholz function which uses AME. There are several sets and terms used here

S is a set containing {0,1, ω,Ω}

Z(0) is the set containing {S}

Z(x) is a set containing 0,1,ω,Ω and everything up to λ(x-1) depends on the value but only using addition in finite number of times

then we finally have λ(x) itself which I have given previously, the mechanism is similar to the bhuccholz function but can only use finite addition, cannot use multiplication or exponentiation

so the function λ(x) is defined as the smallest ordinal and is not part of the set Z(x)

Firstly, we begin from calculate λ(0), λ(0) is smallest ordinal not a member from set {0,1,ω,Ω}, we can use 1 and can add it to a finite amount of time, but fortunately we have ω, we can only add omega to a finite amount of time.

$\omega +\omega +\omega +\omega +\omega +\omega ...$

so that $\lambda (0)=\omega ^{2}$

Then Z(1) is equal to $\{Z(0)\cup \lambda (0)\}$

so that Z(1) = $\{Finite,Addition,0,1,\omega ,\Omega \cup \lambda (0)\}$

We know that $\lambda (0)=\omega ^{2}$so the set Z(1) is include $\omega^2$, we can perform addition to omega but only finite number of time, we cant use multiplication or exponentiation, only addition allowed

$\omega ^{2}+\omega ^{2}+\omega ^{2}+\omega ^{2}+...$

This is same as $n\omega ^{2}$remember we only allowed to used addition in finite, so in can be conclude that

$\lambda (1)=\omega \omega ^{2}=\omega ^{3}$very slow right, this is because we only can use addition only

so in general $\lambda (x)=\omega \omega ^{x+2}=\omega ^{x+2}$

This is very slow because we in this time not deal with ordinal

For example: $g_{\lambda (1)}(3)=9$

$g_{\lambda (2)}(3)=24$

$g_{\lambda (3)}(3)=g_{\lambda (\omega )}(3)=243$

then if we use omega in the lambda or mole function it will cause the function to be diagonalized, and will include all members up to λ(finite), an example of its use is as follows

$g_{\lambda (\omega )}(4)=256$

$g_{\lambda (\omega )}(5)=3127$

$g_{\lambda (\omega )}(6)=46658$

$g_{\lambda (\omega )}(10)=10^{12}$

and so on like that

If we use $\lambda (\omega +1)$then the set Z will contain everything until $\lambda(\omega)$member so that $\lambda (\omega +1)$is smallest ordinal that cant be express using addition operator in finite amount of time of everything until $\lambda(\omega)$when omega can be diagonalized

and this is my estimation for bigger ordinal

$\lambda (x)=\omega ^{x+2}$

$\lambda (\lambda (0))=\lambda (0)^{2}=(\omega ^{2})^{2}=\omega ^{4}$

$\lambda (\lambda (\lambda (0)))=\lambda (1)^{2}=((\omega ^{2})^{2})^{2}=\omega ^{8}$

then i dont know what is $\lambda (\Omega )$or something like that, i stop here, that for today

Complete definition for \(\lambda\) function, it is similar to madore function but only can use addition in finite amount of time,

\(Z_0(\alpha) = \{0, 1, \omega, \Omega\}\)

\(Z_{n+1}(\alpha) = \{\gamma + \delta\mid \gamma, \delta \in Z_n (\alpha)\} \cup \{\lambda(\eta) \mid \eta \in Z_n (\alpha) \land \eta < \alpha\} \)

\(Z(\alpha) = \bigcup_{n < \omega} Z_n (\alpha) \)

\(\lambda(\alpha) = \min\{\beta < \Omega|\beta \notin C(\alpha)\} \)

Let set Z₀ (x) is including \(\{0,1,\omega,\Omega\}\) then,

Let set Z_a (x) is including anything in union with \(\lambda(a-1)\) using only addition operator in finite amount of time

Let \(\lambda(x)\) is smallest ordinal that not member of Z_a (x) and smaller than \(\Omega\) by using only addition operator in finite amount of time