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Circulation and the Fast-Growing Hierarchy
#11
Unless I'm the one confused here, it might be, I suggest to NOT confuse circulation/omegation with FGHs.

The omega in omegation/circulation is NOT an ordinal. It is just notation sugar, it has nothing to do with the ordinal omega. Also the operation \(o(a,b)=\lim a\uparrow^n b\) is not even well defined, let alone taking superfunctions out of it.

Also the point of FGH is that they extend Ackermann-Goodstein \(a\uparrow^n b\) like functions from \(n\in\mathbb N\) to \(n\in {\bf On}\), but not to all the ordinals but only to sufficiently small ones, i.e. transfinite ordinal that must be countable \(|\alpha|\leq\aleph_0\), and are also recursively definable in some technical sense (or you can compute the fundamental sequences).

Also there is not a single way to extend it to transfinite ordinals, but multiple ways to do it, some more natural than others, and all the various ways depend fundamentally on a choice of a system fundamental sequences. A fundamental sequence is a system of choices about how to define it for limit ordinals, and to my limited knowledge, it amounts to an algorithm of diagonalization. In other words \(a\uparrow^\omega b\) has not really something to do with the idea of infinity or limit but it is defined using a trick, something like defining \(a\uparrow^\omega b=a\uparrow^b b\).

MSE MphLee
Mother Law \((\sigma+1)0=\sigma (\sigma+1)\)
S Law \(\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)\)
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#12
Question 
(06/06/2022, 04:00 PM)MphLee Wrote: Unless I'm the one confused here, it might be, I suggest to NOT confuse circulation/omegation with FGHs.

The omega in omegation/circulation is NOT an ordinal. It is just notation sugar, it has nothing to do with the ordinal omega. Also the operation \(o(a,b)=\lim a\uparrow^n b\) is not even well defined, let alone taking superfunctions out of it.

Also the point of FGH is that they extend Ackermann-Goodstein \(a\uparrow^n b\) like functions from \(n\in\mathbb N\) to \(n\in {\bf On}\), but not to all the ordinals but only to sufficiently small ones, i.e. transfinite ordinal that must be countable \(|\alpha|\leq\aleph_0\), and are also recursively definable in some technical sense (or you can compute the fundamental sequences).

Also there is not a single way to extend it to transfinite ordinals, but multiple ways to do it, some more natural than others, and all the various ways depend fundamentally on a choice of a system fundamental sequences. A fundamental sequence is a system of choices about how to define it for limit ordinals, and to my limited knowledge, it amounts to an algorithm of diagonalization. In other words \(a\uparrow^\omega b\) has not really something to do with the idea of infinity or limit but it is defined using a trick, something like defining \(a\uparrow^\omega b=a\uparrow^b b\).
Why is the operation a circulated to the b not well defined? Two circulated to the two still equals four. Two circulated to the five may be some large infinity. Still defined.
Not all of the levels of the Fast-growing hierarchy are recursive. Not all countable ordinals have computable fundamental sequences even if they recursive ordinals. The Church-Kleene ordinal (https://googology.fandom.com/wiki/Church-Kleene_ordinal) for instance, is non recursive and non computable.
It might be interesting to extend the hyper-operation hierarchy to transfinite ordinal ranks.
I wonder how omega minus one would behave in some sort of extension of the Fast-growing hierarchy. I know omega minus one is not an ordinal. But it is a Hyperreal number.
[Image: svg.image?f_%7B\omega-1%7D(x)] would a be a sequence, such that when applied x time to x it results in [Image: svg.image?f_\omega(x)]. What about extending the Fast-growing hierarchy to other Hyperreals? such as omega minus 2?
ฅ(ミ⚈ ﻌ ⚈ミ)ฅ
Please remember to stay hydrated.
Sincerely: Catullus
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#13
(06/07/2022, 01:48 AM)Catullus Wrote: Why is the operation a circulated to the b not well defined?

Because for some \(a,b\) the sequence \(\lim a\uparrow^n b\) diverges. A binary function \(f:\mathbb N\times \mathbb N\to \mathbb N\) defined as  \(f(a,b)\,:=\lim a\uparrow^n b\) DOESN'T EXISTS.
There is a partial function \(f:\,P\to \mathbb N\) where \(P\subseteq \mathbb N\times \mathbb N\) s.t. if \((a,b)\in P\) \(f(a,b)=\lim a\uparrow^n b\).

In that case: "Two circulated to the two still equals four." is correct but
Quote:Two circulated to the five may be some large infinity. Still defined.
Is something we cannot claim. Unless we define it like that. But if we do it, we must decide if we want it to be an ordinal or a cardinal. Let's assume we extend \(f:\,P\to \mathbb N\) to \({\bar f}:\mathbb N\times \mathbb N \to {\bf card}\) then we have a class function.... we still need to allow the argument to be a cardinal in order to take superfunction of that.

I fear there will be many non equivalent ways to do this. Also, the previous argument disqualifies \(f\) to be part of some fast growing hierarchy, by definition.

Quote:\(|\alpha|\leq\aleph_0\) does not hold for all countable ordinals. 
Yes it does, by definition of countable.

Quote:\(|\alpha|\leq\aleph_1\) does hold for all countable ordinals.
Nope it doesn't. By definition of uncountable and of \(\omega_1=\aleph_1\).
Conterproof. let \(\alpha=\omega_1\), then \(|\alpha|\leq\aleph_1\) is true but \(\alpha\) is uncountable.

Quote:Not all of the levels of the Fast-growing hierarchy are recursive. Not all countable ordinals have computable fundamental sequences even if they recursive ordinals.

Sure, it's what I was trying to say in my previous post.

Quote:It might be interesting to extend the hyper-operation hierarchy to transfinite ordinal ranks.
Yea, I believe that fast growing hierarchies are a tool that can be used to achieve this. If I recall correctly, I'd better check the literature, but someone important already tried to extend the lower hyperoperations to ordinals... was it by Doner and Tarski? I'll double check if needed... i have the paper somewhere.


Quote:I wonder how omega minus one would behave in some sort of extension of the Fast-growing hierarchy. I know omega minus one is not an ordinal. But it is a Hyperreal number.
This would be cool. But I'm not sure that this count as an extension of some FGH. The reason is, Hyper-reals do not extend ordinals, are kind orthogonal to them. For example, ordinal addition is a non-commutative monoid. Hyperreals under addition are an abelian group.

Defining families of hyperoperations over other number system (matrix algebras, surreals, hyperreals, tropical fields, cardinals, ordinals... and so on) sounds pretty exciting and something we should explore.

MSE MphLee
Mother Law \((\sigma+1)0=\sigma (\sigma+1)\)
S Law \(\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)\)
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#14
(06/07/2022, 11:53 AM)MphLee Wrote:
Quote:\(|\alpha|\leq\aleph_0\) does not hold for all countable ordinals. 
Yes it does, by definition of countable.

Quote:\(|\alpha|\leq\aleph_1\) does hold for all countable ordinals.
Nope it doesn't. By definition of uncountable and of \(\omega_1=\aleph_1\).
Conterproof. let \(\alpha=\omega_1\), then \(|\alpha|\leq\aleph_1\) is true but \(\alpha\) is uncountable.
Oh, you don't mean absolute value by ||.
ฅ(ミ⚈ ﻌ ⚈ミ)ฅ
Please remember to stay hydrated.
Sincerely: Catullus
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#15
(06/08/2022, 03:50 AM)Catullus Wrote:
(06/07/2022, 11:53 AM)MphLee Wrote:
Quote:\(|\alpha|\leq\aleph_0\) does not hold for all countable ordinals. 
Yes it does, by definition of countable.

Quote:\(|\alpha|\leq\aleph_1\) does hold for all countable ordinals.
Nope it doesn't. By definition of uncountable and of \(\omega_1=\aleph_1\).
Conterproof. let \(\alpha=\omega_1\), then \(|\alpha|\leq\aleph_1\) is true but \(\alpha\) is uncountable.
Oh, you don't mean absolute value by ||.

What do you mean? I thought you meant cardinality, as well.

It would help if you went further into depth in your responses, Catullus. Not trying to sound snobbish or anything, it just helps with legibility.
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#16
(06/08/2022, 04:04 AM)JmsNxn Wrote:
(06/08/2022, 03:50 AM)Catullus Wrote:
(06/07/2022, 11:53 AM)MphLee Wrote:
Quote:\(|\alpha|\leq\aleph_0\) does not hold for all countable ordinals. 
Yes it does, by definition of countable.

Quote:\(|\alpha|\leq\aleph_1\) does hold for all countable ordinals.
Nope it doesn't. By definition of uncountable and of \(\omega_1=\aleph_1\).
Conterproof. let \(\alpha=\omega_1\), then \(|\alpha|\leq\aleph_1\) is true but \(\alpha\) is uncountable.
Oh, you don't mean absolute value by ||.

What do you mean? I thought you meant cardinality, as well.

It would help if you went further into depth in your responses, Catullus. Not trying to sound snobbish or anything, it just helps with legibility.
I wanted to be clear about notation.
ฅ(ミ⚈ ﻌ ⚈ミ)ฅ
Please remember to stay hydrated.
Sincerely: Catullus
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#17
It being absolute values makes no sense to me.
Your claim is false regardless of \(|\cdot |\) being cardinality or absolute value.
Tbh this also feels like a waste of forum space, as it is rn. No offense.

I encourage you to be more detailed. This will help you in getting answers and other users in appreciating your ideas.




ps: back in the forum era, forums used to have rules and mods. Nowadays forums are not as popular as tey used to be, so much of the folklore usage, standards, etiquette went lost.
Here in the tetration forum, for what I know, the need for moderation has always been minimal, and the only rules were the implicit standard of forum usage (no spam, ads, decency, ecc.).
So no rules here about posing here. Anyways this is not a mailing list. I encourage a moderate use of the reply function, overusing the replying function means  that the threads growth is exponential at worst.
If we set \(a\) the length of the average new contribution, if everyone abuse of the reply function the \(n+1\)-th post the thread will have length at worst \(T_{n+1}=2T_n+a\) at best \(T_{n+1}=T_n+2a\). So we have at best \(T_n\sim (2n+1)a\) and at worst \(T_n\sim (2^{n-1}-1)a\) of total text versus \(n\cdot a\) of useful info. A total waste of space and of precious time of readers. Readability of the forum gets horrible.

MSE MphLee
Mother Law \((\sigma+1)0=\sigma (\sigma+1)\)
S Law \(\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)\)
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#18
Question 
Could there be [Image: svg.image?f_%7B-1%7D(x)]? Maybe not, since [Image: svg.image?f_0(0)] equals one. And that would be applying the function [Image: svg.image?f_%7B-1%7D(x)] zero times to zero.
Then again some people think zeration exists, even though there are some supposed inconsistencies with zeration.
ฅ(ミ⚈ ﻌ ⚈ミ)ฅ
Please remember to stay hydrated.
Sincerely: Catullus
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#19
Question 
(06/07/2022, 11:53 AM)MphLee Wrote:
Quote:It might be interesting to extend the hyper-operation hierarchy to transfinite ordinal ranks.
Yea, I believe that fast growing hierarchies are a tool that can be used to achieve this. If I recall correctly, I'd better check the literature, but someone important already tried to extend the lower hyperoperations to ordinals... was it by Doner and Tarski? I'll double check if needed... i have the paper somewhere.
Is it this paper An extended arithmetic of ordinal numbers?
ฅ(ミ⚈ ﻌ ⚈ミ)ฅ
Please remember to stay hydrated.
Sincerely: Catullus
Reply
#20
(07/10/2022, 03:20 AM)Catullus Wrote:
(06/07/2022, 11:53 AM)MphLee Wrote:
Quote:It might be interesting to extend the hyper-operation hierarchy to transfinite ordinal ranks.
Yea, I believe that fast growing hierarchies are a tool that can be used to achieve this. If I recall correctly, I'd better check the literature, but someone important already tried to extend the lower hyperoperations to ordinals... was it by Doner and Tarski? I'll double check if needed... i have the paper somewhere.
Is it this paper An extended arithmetic of ordinal numbers?

Yes It is! Rolleyes

MSE MphLee
Mother Law \((\sigma+1)0=\sigma (\sigma+1)\)
S Law \(\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)\)
Reply


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