## Ordinals in googology

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gameoflifemaniac
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### Re: Ordinals in googology

Moosey wrote:The Veblen function is actually pretty simple--
phi_0(n) = w^n,
phi_n+1(0) = the 0th fixed point of phi_n(a),
phi_n(0) for limit ordinal n = the limit of {phi_(n[0])(0), phi_(n[1]), (0) phi_(n[2])(0),...},
phi_n(m+1) = the next fixed point like phi_n(m),
phi_n(m) for limit ordinal m = the limit of {phi_n(m[0]),phi_n(m[1]),phi_n(m[2]),...}.

The multivariable Veblen function is basically the same, but it's organized like so: phi(mth entry, m-1th entry,...3rd entry,2nd entry, 1st entry)
The (n+1)th entry indicates fixed points in the nth entry, so gamma_0 = phi(1,0,0) because it's a|->phi_a(0)
Can you give some examples? I still don't really get it.
I was so socially awkward in the past and it will haunt me for the rest of my life.

Code: Select all

x = 21, y = 5, rule = B3/S23
b2o2bo2b2o2b2o2bobobo$o3bobobobobobobobobo$bo2bobob2o2b2o3bo2bo$2bobob obobobobo2bo$2o3bo2bobobobo2bo2bo!


Moosey
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### Re: Ordinals in googology

gameoflifemaniac wrote:
November 2nd, 2019, 10:15 am
Moosey wrote:The Veblen function is actually pretty simple--
phi_0(n) = w^n,
phi_n+1(0) = the 0th fixed point of phi_n(a),
phi_n(0) for limit ordinal n = the limit of {phi_(n[0])(0), phi_(n[1]), (0) phi_(n[2])(0),...},
phi_n(m+1) = the next fixed point like phi_n(m),
phi_n(m) for limit ordinal m = the limit of {phi_n(m[0]),phi_n(m[1]),phi_n(m[2]),...}.

The multivariable Veblen function is basically the same, but it's organized like so: phi(mth entry, m-1th entry,...3rd entry,2nd entry, 1st entry)
The (n+1)th entry indicates fixed points in the nth entry, so gamma_0 = phi(1,0,0) because it's a|->phi_a(0)
Can you give some examples? I still don't really get it.
phi_1(0) = e_0
phi_2(5) = z_5
phi_4(0) = a|->h_a
phi_w(0) = sup{e_0,z_0,h_0,...}
phi_w+1(0) = a|->phi_w(a)
I am a prolific creator of many rather pathetic googological functions

My CA rules can be found here

Also, the tree game
Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"

Nanho walåt derwo esaato?

gameoflifemaniac
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### Re: Ordinals in googology

Moosey wrote:
November 2nd, 2019, 10:34 am
gameoflifemaniac wrote:
November 2nd, 2019, 10:15 am
Moosey wrote:The Veblen function is actually pretty simple--
phi_0(n) = w^n,
phi_n+1(0) = the 0th fixed point of phi_n(a),
phi_n(0) for limit ordinal n = the limit of {phi_(n[0])(0), phi_(n[1]), (0) phi_(n[2])(0),...},
phi_n(m+1) = the next fixed point like phi_n(m),
phi_n(m) for limit ordinal m = the limit of {phi_n(m[0]),phi_n(m[1]),phi_n(m[2]),...}.

The multivariable Veblen function is basically the same, but it's organized like so: phi(mth entry, m-1th entry,...3rd entry,2nd entry, 1st entry)
The (n+1)th entry indicates fixed points in the nth entry, so gamma_0 = phi(1,0,0) because it's a|->phi_a(0)
Can you give some examples? I still don't really get it.
phi_1(0) = e_0
phi_2(5) = z_5
phi_4(0) = a|->h_a
phi_w(0) = sup{e_0,z_0,h_0,...}
phi_w+1(0) = a|->phi_w(a)
So
phi_0(0) = w
phi_0(1) = w^w
phi_0(2) = w^w^w
phi_1(0) = e0
phi_1(1) = e1
phi_1(2) = e2
phi_2(0) = z0
phi_2(1) = z1
phi_2(2) = z2
.
.
.
and
phi_w(0) = Gamma(0), kind of a limit of new kinds of fixed points?

Also, is
z1 = z0^z0^z0^z0^z0^...
z2 = z1^z1^z1^z1^z1^...
.
.
.
?

And how do you write
phi_3(n) = ?
phi_4(n) = ?
.
.
.
I was so socially awkward in the past and it will haunt me for the rest of my life.

Code: Select all

x = 21, y = 5, rule = B3/S23
b2o2bo2b2o2b2o2bobobo$o3bobobobobobobobobo$bo2bobob2o2b2o3bo2bo$2bobob obobobobo2bo$2o3bo2bobobobo2bo2bo!


Moosey
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### Re: Ordinals in googology

gameoflifemaniac wrote:
November 2nd, 2019, 11:19 am
Moosey wrote:
November 2nd, 2019, 10:34 am
gameoflifemaniac wrote:
November 2nd, 2019, 10:15 am

Can you give some examples? I still don't really get it.
phi_1(0) = e_0
phi_2(5) = z_5
phi_4(0) = a|->h_a
phi_w(0) = sup{e_0,z_0,h_0,...}
phi_w+1(0) = a|->phi_w(a)
So
phi_0(0) = w
phi_0(1) = w^w
phi_0(2) = w^w^w
No, phi_0(0) = 1, phi_0(1) = w, phi_0(2) = w^2
gameoflifemaniac wrote:
November 2nd, 2019, 10:15 am
phi_1(0) = e0
phi_1(1) = e1
phi_1(2) = e2
phi_2(0) = z0
phi_2(1) = z1
phi_2(2) = z2
Yes
gameoflifemaniac wrote:
November 2nd, 2019, 10:15 am
.
.
.
and
phi_w(0) = Gamma(0), kind of a limit of new kinds of fixed points?
No. gamma_0 = phi_gamma_0(0). phi_w(0) << gamma_0. Actually I don't know why phi_w(0), which is a rather important ordinal being the supremum of {1,e_0, z_0, ...} does not have a name besides phi_w(0), but that's a real problem because everyone (myself included) that knows about ordinals has (or is) mistakenly thought that gamma_0 was the supremum of {1,e_0,z_0}...
gameoflifemaniac wrote:
November 2nd, 2019, 10:15 am
Also, is
z1 = z0^z0^z0^z0^z0^...
z2 = z1^z1^z1^z1^z1^...
.
.
.
?
I think those are e_(z_0 +1) and e_(z_1 +1), not the much larger z_1 and z_2.
I think that for limit ordinal n > some point between e_0 and z_0, e_(n+1) = the 0th a|->n^a, or n^^w, which is why e_(W+1) = the 0th a|->W^a
gameoflifemaniac wrote:
November 2nd, 2019, 10:15 am
And
phi_3(n) = ?
phi_4(n) = ?
.
.
.
the former is η_n, also known as the nth a|->z_a. η_0 can be though of as z_z_z_z_z_z_z...
The latter is the nth a|->η_a.
Generally we call things beyond eta-naught by their names in the Veblen function anyways, so there isn't a much better (or any other canonical) name for phi_4(n) than phi_4(n).
Last edited by Moosey on November 2nd, 2019, 11:39 am, edited 1 time in total.
I am a prolific creator of many rather pathetic googological functions

My CA rules can be found here

Also, the tree game
Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"

Nanho walåt derwo esaato?

gameoflifemaniac
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### Re: Ordinals in googology

Ok, now I understand everything, but this
how do you construct e1, e2, e3... from e0; z1, z2, z3... from z0; and η1, η2, η3... from η0?
And how do you define Gamma(1), Gamma(2), Gamma(3)...?
What's the ϑ function, θ function, ψ function and Ω? And can you explain how does the multivariable phi function work?
I was so socially awkward in the past and it will haunt me for the rest of my life.

Code: Select all

x = 21, y = 5, rule = B3/S23
b2o2bo2b2o2b2o2bobobo$o3bobobobobobobobobo$bo2bobob2o2b2o3bo2bo$2bobob obobobobo2bo$2o3bo2bobobobo2bo2bo!


testitemqlstudop
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### Re: Ordinals in googology

Is the well ordering of w_1 provable?

Moosey
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### Re: Ordinals in googology

gameoflifemaniac wrote:
November 2nd, 2019, 11:38 am
Ok, now I understand everything, but this
how do you construct e1, e2, e3... from e0; z1, z2, z3... from z0; and η1, η2, η3... from η0?
e_1 = sup{e_0 +1,w^(e_0 +1), w^w^(e_0 +1),...}
e_2 = sup{e_1 +1,w^(e_1 +1), w^w^(e_1 +1),...}
...
z_1 = sup{z_0 +1,e_(z_0 +1),e_e_(z_0 +1)...}
...
h_1 = sup{h_0 +1,z_(h_0 +1),z_z_(h_0 +1)...}
For the epsilons, you could alternatively say that, very loosely,
e_(n+1) = (e_n)^^w
gameoflifemaniac wrote:
November 2nd, 2019, 11:38 am
And how do you define Gamma(1), Gamma(2), Gamma(3)...?
The next fixed points of phi_a(0) after gamma_0
gameoflifemaniac wrote:
November 2nd, 2019, 11:38 am
What's the ϑ function, θ function, ψ function and Ω?
Those are OCFs, which have gotten a bit of discussion here. Ω is w_1 = the first uncountable ordinal
gameoflifemaniac wrote:
November 2nd, 2019, 11:38 am
And can you explain how does the multivariable phi function work?
Moosey wrote:
November 2nd, 2019, 7:16 am
The multivariable Veblen function is basically the same, but it's organized like so: phi(mth entry, m-1th entry,...3rd entry,2nd entry, 1st entry)
The (n+1)th entry indicates fixed points in the nth entry, so gamma_0 = phi(1,0,0) because it's a|->phi_a(0)
So the 3rd entry does to the second entry what the second entry does to the first--series of fixed points for the previous entry
I am a prolific creator of many rather pathetic googological functions

My CA rules can be found here

Also, the tree game
Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"

Nanho walåt derwo esaato?

gameoflifemaniac
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### Re: Ordinals in googology

Thanks now I understand everything. But what about w-1? Does it exist?
I was so socially awkward in the past and it will haunt me for the rest of my life.

Code: Select all

x = 21, y = 5, rule = B3/S23
b2o2bo2b2o2b2o2bobobo$o3bobobobobobobobobo$bo2bobob2o2b2o3bo2bo$2bobob obobobobo2bo$2o3bo2bobobobo2bo2bo!


toroidalet
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### Re: Ordinals in googology

testitemqlstudop wrote:
November 2nd, 2019, 11:47 pm
Is the well ordering of w_1 provable?
Isn't well-orderedness part of the definition of an ordinal?

If I'm missing something, then see this proof (set X to be the set of natural numbers to get the conclusion that ω_1 exists and is well-ordered)
gameoflifemaniac wrote:
November 3rd, 2019, 12:33 pm
Thanks now I understand everything. But what about w-1? Does it exist?
If such an ordinal existed, it would be an infinite ordinal less than ω, which contradicts its definition as the smallest infinite ordinal. Limit ordinals are not successors and so subtraction is undefined in general.
However, surreal numbers include all ordinals and real numbers and things like ω-1, 1/ω, and √ω.
"Build a man a fire and he'll be warm for a day. Set a man on fire and he'll be warm for the rest of his life."

-Terry Pratchett

gameoflifemaniac
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### Re: Ordinals in googology

Interesting. If Gamma(0) is phi_phi_phi...(0)(0)(0), is Gamma(1) phi_phi_phi...(1)(1)(1), Gamma(2) phi_phi_phi...(2)(2)(2) etc.?
Also, I heard somewhere that w+2 and 2+w aren't the same. Is that true?
I was so socially awkward in the past and it will haunt me for the rest of my life.

Code: Select all

x = 21, y = 5, rule = B3/S23
b2o2bo2b2o2b2o2bobobo$o3bobobobobobobobobo$bo2bobob2o2b2o3bo2bo$2bobob obobobobo2bo$2o3bo2bobobobo2bo2bo!


Moosey
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### Re: Ordinals in googology

gameoflifemaniac wrote:
November 4th, 2019, 9:11 am
Interesting. If Gamma(0) is phi_phi_phi...(0)(0)(0), is Gamma(1) phi_phi_phi...(1)(1)(1), Gamma(2) phi_phi_phi...(2)(2)(2) etc.?
no. Probably gamma_1 = sup{gamma_0+1,phi_gamma_0+1(0),phi_phi_gamma_0+1(0)(0),etc.}
gameoflifemaniac wrote:
November 4th, 2019, 9:11 am
Also, I heard somewhere that w+2 and 2+w aren't the same. Is that true?
HOW DID YOU MAKE IT THIS FAR WITHOUT KNOWING THAT WAS TRUE?
Yes, w+2 = {0,1,2,3...w,w+1}, which is > 2+w = {0,0,0,1,2,3,...}
The same goes for w+1 and 1+w, or 2w and w2 (the former is = w, and the latter is larger)

On an unrelated note, I should update the definition of // from ah:

Code: Select all

ah_n{$_0//(b+1),$_1} = ah_n((($_0)(@^_n)ah_n{$_0//b})#{//b,$_1}) ah_n{$_0//0} = ah_n{$_0} ah_n{$_0//b,$_1} = ah_n{$_0//(b[n]),$_1} for lim ord b Else: apply ah's rules. (e.g. ah_{$_0//,0,0,w+1} = ah_{$_0//,w+1,w+1,w}) Where ($_1)(@^_n)a =
($_1)@(ah_n(($_1)(@^_n)(a-1))), a > 1,
ah_n{$_1}, a=1 This makes expressions such as ah_5{w+5//7,w+100223} valid. Additionally, it allows for a better @@: Code: Select all ah_n{$_0@@0} = ah_n{$_0} ah_n{$_0@@a,$_1} = ah_n{$_0@@a[n],$_1},a a lim ord ah_n{$_0@@(a+1),$_1} = ah_n({$_0//}#{$_0@@(a),$_1}
Apply basic ah rules otherwise

This, of course, can be extended indefinitely:

Code: Select all

ah_n{$_0(@@b)0} = ah_n{$_0}
ah_n{$_0((@@b)a,$_1} = ah_n{$_0(@@b)a[n],$_1},a a lim ord
ah_n{$_0(@@0)(a+1),$_1} = ah_n({$_0//}#{$_0@@(a),$_1} ah_n{$_0(@@b+1)(a+1),$_1} = ah_n({$_0(@@b)}#{$_0@@(a),$_1}
ah_n{$_0(@@b)$_2} = ah_n{$_0(@@b[n])$_2}, if b is a lim ord
Apply basic ah rules otherwise
Which of course leads to my new fastest-growing-function (by Moosey):
ah_n{gamma_0(@@gamma_0)gamma_0}
Last edited by Moosey on November 5th, 2019, 7:26 am, edited 1 time in total.
I am a prolific creator of many rather pathetic googological functions

My CA rules can be found here

Also, the tree game
Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"

Nanho walåt derwo esaato?

gameoflifemaniac
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### Re: Ordinals in googology

Moosey wrote:
November 4th, 2019, 9:54 am
HOW DID YOU MAKE IT THIS FAR WITHOUT KNOWING THAT WAS TRUE?
Umm what do you mean? I knew that, but I wanted to be 100% sure.
Moosey wrote:
November 4th, 2019, 9:54 am
no. Probably gamma_1 = sup{gamma_0+1,phi_gamma_0+1(0),phi_phi_gamma_0+1(0)(0),etc.}
Probably?
I was so socially awkward in the past and it will haunt me for the rest of my life.

Code: Select all

x = 21, y = 5, rule = B3/S23
b2o2bo2b2o2b2o2bobobo$o3bobobobobobobobobo$bo2bobob2o2b2o3bo2bo$2bobob obobobobo2bo$2o3bo2bobobobo2bo2bo!


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### Re: Ordinals in googology

gameoflifemaniac wrote:
November 4th, 2019, 10:19 am
Moosey wrote:
November 4th, 2019, 9:54 am
no. Probably gamma_1 = sup{gamma_0+1,phi_gamma_0+1(0),phi_phi_gamma_0+1(0)(0),etc.}
Probably?
Yes. Remember, phi(1,0,a) enumerates fixed points of phi(0,a,0).
Then phi(1,1,a) enumerates fixed points of phi(1,0,a), phi(1,2,a) the fixed points of phi(1,1,a), and so on until
phi(2,0,a) enumerates the fixed points of (1,a,0), phi(3,0,a) enumerates the fixed points of phi(2,a,0), and finally
phi(1,0,0,a) enumerates the fixed points of phi(a,0,0).
This continues until you reach the Small Veblen Ordinal, psi(W^W^w).
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testitemqlstudop
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### Re: Ordinals in googology

toroidalet wrote:
November 3rd, 2019, 3:43 pm
testitemqlstudop wrote:
November 2nd, 2019, 11:47 pm
Is the well ordering of w_1 provable?
Isn't well-orderedness part of the definition of an ordinal?

If I'm missing something, then see this proof (set X to be the set of natural numbers to get the conclusion that ω_1 exists and is well-ordered)
Several things:

The well ordering of w_1 implies the well ordering of all countable ordinals.

Am I dumb or does that immediately prove the infinite-time computability of w_1^CK?

All proof systems have PTO < w_1, so wouldn't a proof be part of the most powerful proof system?

gameoflifemaniac
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### Re: Ordinals in googology

Can you give me some examples of values of Weiermann's ϑ? Maybe I'll understand it this way
I was so socially awkward in the past and it will haunt me for the rest of my life.

Code: Select all

x = 21, y = 5, rule = B3/S23
b2o2bo2b2o2b2o2bobobo$o3bobobobobobobobobo$bo2bobob2o2b2o3bo2bo$2bobob obobobobo2bo$2o3bo2bobobobo2bo2bo!


Moosey
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### Re: Ordinals in googology

gameoflifemaniac wrote:
November 5th, 2019, 1:56 pm
Can you give me some examples of values of Weiermann's ϑ? Maybe I'll understand it this way
Why that one? I've never even heard of that one! Maybe I can explain some simple OCF like Madore's ψ or some simpler one.

Here's a simple example:

Code: Select all

C_0(α)={0,1,Ω}
C_n+1(α)={γ+δ,ψ(η)|γ,δ,η∈C_n(α);η<α}
C(α)=⋃(n<ω)C_n(α)
ψ(α)=min{β∈Ω|β∉C(α)}
ψ(0) = w (the minimum value you can't get by adding finitely many finite numbers together)
ψ(1) = w^2 (The minimum value you cannot get by adding finitely many ws together)
ψ(n) = w^(n+1) for finite n
ψ(w) = w^w
etc.
ψ(e_0) = e_0
ψ(e_0 + 1) = e_0
... epsilon_0 for a long time
ψ(W) = e_0
ψ(W+1) = (e_0)w (I think)
The limit of ψ = e_w

Suppose we introduce multiplication:

Code: Select all

C_0(α)={0,1,Ω}
C_n+1(α)={γ+δ,γδ,ψ(η)|γ,δ,η∈C_n(α);η<α}
C(α)=⋃(n<ω)C_n(α)
ψ(α)=min{β∈Ω|β∉C(α)}
Now, it starts off at ψ(0):
Still w
ψ(1) = the smallest ordinal you can't get by multiplying 0,1, or w finitely many times:
w^w
ψ(n) for finite n = w^^(n+1)
ψ(w) = e_0
ψ(w2) = I think e_1
ψ(wn) for finite n = I think e_(n-1)
ψ(w^2) = e_w in that case
...

Code: Select all

That
C_0(α)={0,1,Ω}
C_n+1(α)={γ+δ,γδ,γ^δ,ψ(η)|γ,δ,η∈C_n(α);η<α}
C(α)=⋃(n<ω)C_n(α)
ψ(α)=min{β∈Ω|β∉C(α)}

Code: Select all

Madore's psi
C_0(α)={0,1,w,Ω}
C_n+1(α)={γ+δ,γδ,γ^δ,ψ(η)|γ,δ,η∈C_n(α);η<α}
C(α)=⋃(n<ω)C_n(α)
ψ(α)=min{β∈Ω|β∉C(α)}
It would start off more slowly but would be the same for all practical purposes.

If you want to see someone analyse it go to that link on the words Madore's psi.

Unfortunately you can't just extend it thru the hyper operators:

Code: Select all

C_0(α)={0,1,w,Ω}
C_n+1(α)={γ+δ,γδ,γ^δ,γ{ξ}δ,ψ(η)|γ,δ,ξ,η∈C_n(α);η<α}
C(α)=⋃(n<ω)C_n(α)
ψ(α)=min{β∈Ω|β∉C(α)}
Unless you define a system for ordinal hyperoperators. If you do, awesome. (Technically since BEAF arrays of {a,b,c} = a{c}b you could use ordinal BEAF, mentioned here, as a system for this. Don't ask me how ordinal beaf works because, I hate to tell you this, but I hardly understand normal BEAF.)

But you can define this extremely simple one:

Code: Select all

C_0(α)={0,Ω}
C_n+1(α)={γ+δ,γδ,γ^δ,φ_γ(δ),ψ(η)|γ,δ,η∈C_n(α);η<α}
C(α)=⋃(n<ω)C_n(α)
ψ(α)=min{β∈Ω|β∉C(α)}
Which would get you to gamma_0 at ψ(0). Yes, ψ of zero. Without so much as a 1 in the starter set.

With the finitely-many-entry Veblen function instead of the Veblen function, you could get to the SVO at 0, and with the any-amount-of-entries-in-the-function-including-infinitely-many you could get to the LVO.

Probably another good system would be

Code: Select all

C_0(α)={0,Ω}
C_n+1(α)={γ+δ,γδ,γ^δ,φ_γ(δ),ψ_0(η),w_γ|γ,δ,η∈C_n(α);η<α}
C(α)=⋃(n<ω)C_n(α)
ψ_0(α)=min{β∈Ω|β∉C(α)}
ψ_1(α)=sup(C(α))
In this case you could have two functions in one (and also the first one, a modification of the one in the code box above the one I'm talking about right now, would be more powerful than the other one in the other code box), and the uncountable-generating one would reach a|->w_a which I think is about inaccessible.
In LaTeX:

Code: Select all

\begin{eqnarray*}
C_0(\alpha) &=& \{0, \Omega\}\\
C_{n+1}(\alpha) &=& \{\gamma + \delta, \gamma\delta,\gamma^\delta,\phi_{\gamma}(\delta),\omega_\gamma,\psi_0(\eta) | \gamma, \delta, \eta \in C_n (\alpha); \eta < \alpha\} \\
C(\alpha) &=& \bigcup_{n < \omega} C_n (\alpha) \\
\psi_0(\alpha) &=& \min\{\beta \in \Omega|\beta \notin C(\alpha)\} \\
\psi_1(\alpha) &=& \sup(C(\alpha)) \\
\end{eqnarray*}
I am a prolific creator of many rather pathetic googological functions

My CA rules can be found here

Also, the tree game
Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"

Nanho walåt derwo esaato?

gameoflifemaniac
Posts: 1191
Joined: January 22nd, 2017, 11:17 am
Location: There too

### Re: Ordinals in googology

Moosey wrote:
November 5th, 2019, 5:38 pm
Why that one? I've never even heard of that one!
Because it's used to write any countale ordinal larger than Gamma(0) and smaller than w_1CK in the googology wiki.
Ackermann's ordinal is Weiermann's ϑ(Ω^3), for example, but what does that mean?
I was so socially awkward in the past and it will haunt me for the rest of my life.

Code: Select all

x = 21, y = 5, rule = B3/S23
b2o2bo2b2o2b2o2bobobo$o3bobobobobobobobobo$bo2bobob2o2b2o3bo2bo$2bobob obobobobo2bo$2o3bo2bobobobo2bo2bo!


Moosey
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### Re: Ordinals in googology

gameoflifemaniac wrote:
November 6th, 2019, 2:57 am
Moosey wrote:
November 5th, 2019, 5:38 pm
Why that one? I've never even heard of that one!
Because it's used to write any countale ordinal larger than Gamma(0) and smaller than w_1CK in the googology wiki.
Ackermann's ordinal is Weiermann's ϑ(Ω^3), for example, but what does that mean?
If you need that one specifically...
... Then they have a nice explanation on the wiki:
Essentially, it wrote: C(a,b) is the set of all ordinals constructible using only the following:
0, all ordinals less than b, and Ω.
Finite applications of c+d, κ↦ωκ [or w^c in other terms], and ϑ(n) for n < a, assuming n, c, and d are produceable using those methods or are 0 or W.

Then, ϑ(a) is the smallest ordinal b such that α∈C(a,b), and b is greater than all the countable ordinals in C(a,b)
I am a prolific creator of many rather pathetic googological functions

My CA rules can be found here

Also, the tree game
Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"

Nanho walåt derwo esaato?

gameoflifemaniac
Posts: 1191
Joined: January 22nd, 2017, 11:17 am
Location: There too

### Re: Ordinals in googology

Moosey wrote:
November 6th, 2019, 7:39 am
gameoflifemaniac wrote:
November 6th, 2019, 2:57 am
Moosey wrote:
November 5th, 2019, 5:38 pm
Why that one? I've never even heard of that one!
Because it's used to write any countale ordinal larger than Gamma(0) and smaller than w_1CK in the googology wiki.
Ackermann's ordinal is Weiermann's ϑ(Ω^3), for example, but what does that mean?
If you need that one specifically...
... Then they have a nice explanation on the wiki:
Essentially, it wrote: C(a,b) is the set of all ordinals constructible using only the following:
0, all ordinals less than b, and Ω.
Finite applications of c+d, κ↦ωκ [or w^c in other terms], and ϑ(n) for n < a, assuming n, c, and d are produceable using those methods or are 0 or W.

Then, ϑ(a) is the smallest ordinal b such that α∈C(a,b), and b is greater than all the countable ordinals in C(a,b)
I don't really understand the definition. I would get it if you'd gave me some examples.

Also, am I bothering you in this topic? You all are asking much more complicated things than me and I don't want to disturb you.
I was so socially awkward in the past and it will haunt me for the rest of my life.

Code: Select all

x = 21, y = 5, rule = B3/S23
b2o2bo2b2o2b2o2bobobo$o3bobobobobobobobobo$bo2bobob2o2b2o3bo2bo$2bobob obobobobo2bo$2o3bo2bobobobo2bo2bo!


Moosey
Posts: 3555
Joined: January 27th, 2019, 5:54 pm
Location: A house, or perhaps the OCA board. Or [click to not expand]
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### Re: Ordinals in googology

gameoflifemaniac wrote:
November 6th, 2019, 1:16 pm
I don't really understand the definition. I would get it if you'd gave me some examples.
I'm not that good with more complicated OCFs, so I can't answer that question.
gameoflifemaniac wrote:
November 6th, 2019, 1:16 pm
Also, am I bothering you in this topic? You all are asking much more complicated things than me and I don't want to disturb you.
No. Personally I like the activity. It keeps this thread up and I desperately want a response to my ah stuff (like this):
Moosey wrote:
November 4th, 2019, 9:54 am
On an unrelated note, I should update the definition of // from ah:

Code: Select all

ah_n{$_0//(b+1),$_1} = ah_n((($_0)(@^_n)ah_n{$_0//b})#{//b,$_1}) ah_n{$_0//0} = ah_n{$_0} ah_n{$_0//b,$_1} = ah_n{$_0//(b[n]),$_1} for lim ord b Else: apply ah's rules. (e.g. ah_{$_0//,0,0,w+1} = ah_{$_0//,w+1,w+1,w}) Where ($_1)(@^_n)a =
($_1)@(ah_n(($_1)(@^_n)(a-1))), a > 1,
ah_n{$_1}, a=1 This makes expressions such as ah_5{w+5//7,w+100223} valid. Additionally, it allows for a better @@: Code: Select all ah_n{$_0@@0} = ah_n{$_0} ah_n{$_0@@a,$_1} = ah_n{$_0@@a[n],$_1},a a lim ord ah_n{$_0@@(a+1),$_1} = ah_n({$_0//}#{$_0@@(a),$_1}
Apply basic ah rules otherwise

This, of course, can be extended indefinitely:

Code: Select all

ah_n{$_0(@@b)0} = ah_n{$_0}
ah_n{$_0((@@b)a,$_1} = ah_n{$_0(@@b)a[n],$_1},a a lim ord
ah_n{$_0(@@0)(a+1),$_1} = ah_n({$_0//}#{$_0@@(a),$_1} ah_n{$_0(@@b+1)(a+1),$_1} = ah_n({$_0(@@b)}#{$_0@@(a),$_1}
ah_n{$_0(@@b)$_2} = ah_n{$_0(@@b[n])$_2}, if b is a lim ord
Apply basic ah rules otherwise
Which of course leads to my new fastest-growing-function (by Moosey):
ah_n{gamma_0(@@gamma_0)gamma_0}
I am a prolific creator of many rather pathetic googological functions

My CA rules can be found here

Also, the tree game
Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"

Nanho walåt derwo esaato?

gameoflifemaniac
Posts: 1191
Joined: January 22nd, 2017, 11:17 am
Location: There too

### Re: Ordinals in googology

What is the large Veblen ordinal? I know it's defined by α↦φ_Ω^α(0), but can you even talk about phi of an uncountable cardinal?
What is phi(1,1,0), phi(2,0,0)?
And is the Ackermann ordinal equal to Γ_Γ_Γ_Γ_Γ_Γ_Γ_Γ_Γ_Γ...?
Last edited by gameoflifemaniac on November 8th, 2019, 4:23 pm, edited 1 time in total.
I was so socially awkward in the past and it will haunt me for the rest of my life.

Code: Select all

x = 21, y = 5, rule = B3/S23
b2o2bo2b2o2b2o2bobobo$o3bobobobobobobobobo$bo2bobob2o2b2o3bo2bo$2bobob obobobobo2bo$2o3bo2bobobobo2bo2bo!


Moosey
Posts: 3555
Joined: January 27th, 2019, 5:54 pm
Location: A house, or perhaps the OCA board. Or [click to not expand]
Contact:

### Re: Ordinals in googology

gameoflifemaniac wrote:
November 8th, 2019, 11:42 am
What is the large Veblen ordinal? I know it's defined by α↦φ_Ω^α(0), but can you even talk about phi of an uncountable cardinal?
No, itś defined like this:

Code: Select all

1. w
2. phi(1,0,0,0,0,0,0,0,...) w/ w 0´s (SVO)
3. phi(1,0,0,0,0,0,...) w/ that many 0s
4. phi(1,0,0,0,...) w/ that many 0s
...

The LVO is the supremum of all of these
I am a prolific creator of many rather pathetic googological functions

My CA rules can be found here

Also, the tree game
Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"

Nanho walåt derwo esaato?

gameoflifemaniac
Posts: 1191
Joined: January 22nd, 2017, 11:17 am
Location: There too

### Re: Ordinals in googology

gameoflifemaniac wrote:
November 8th, 2019, 11:42 am
What is phi(1,1,0), phi(2,0,0)?
And is the Ackermann ordinal equal to Γ_Γ_Γ_Γ_Γ_Γ_Γ_Γ_Γ_Γ...?
This
Also, I can't appreciate your ah function cuz small brain
I was so socially awkward in the past and it will haunt me for the rest of my life.

Code: Select all

x = 21, y = 5, rule = B3/S23
b2o2bo2b2o2b2o2bobobo$o3bobobobobobobobobo$bo2bobob2o2b2o3bo2bo$2bobob obobobobo2bo$2o3bo2bobobobo2bo2bo!


Moosey
Posts: 3555
Joined: January 27th, 2019, 5:54 pm
Location: A house, or perhaps the OCA board. Or [click to not expand]
Contact:

### Re: Ordinals in googology

gameoflifemaniac wrote:
November 8th, 2019, 11:42 am
What is phi(1,1,0), phi(2,0,0)?
And is the Ackermann ordinal equal to Γ_Γ_Γ_Γ_Γ_Γ_Γ_Γ_Γ_Γ...?
And is the Ackermann ordinal equal to Γ_Γ_Γ_Γ_Γ_Γ_Γ_Γ_Γ_Γ...?
No.
I think min a|->gamma_a = phi(1,1,0) but I could be wrong.
phi(2,0,0) = min a|->phi(1,a,0) much like how gamma_0 = phi(1,0,0) = min a|->phi(a,0)
gameoflifemaniac wrote:
November 8th, 2019, 3:57 pm
Also, I can't appreciate your ah function cuz small brain
Simply put, ah is my extension of something by pkmnQ

Since Ę_0,a(n) is hardly more powerful than f_a(n), I extended it much farther-- for instance, ah_n(a@w) is at least f_a*n-ish, and ah_n(a(@@a)a) is probably much more powerful for large enough a. Using the extension of ah rules I will write now.
(Can someone check how powerful it is? Is it e_(a+1)-ish?)

ah rules:

Code: Select all

Part one:

define $_n as any entries (including no entries) in an array. It’s my symbol for we-don’t-care entries. In any one use of any one rule, if n is the same,$_n is the same

a#b = concatenation of a and b

n@m =
n, m = 1
{n}#(n@(m-1)), m > 1, m not a lim ord
n@(m[n]) if m is a lim ord

g(a,n,B) =
ah_g(a-1,n,B) {B}, a > 1;
n, a = 0

ah definition:
Rule 1.
ah^a_n {$_0} = g(a,n,$_0)
Rule 2.
ah_n ({$_1}#{z}) = ah_n {$_1}, z = 0
Rule 3.
ah_n{} = n+1
Rule 4.
ah_n{a+1,$_2} = ah^n_n{a,$_2}
Rule 5.
ah_n{a,$_3} = ah_n{a[n],$_3}, a a lim ord
Rule 6.
ah_n ((0@b)#{a+1,$_4}) = ah_(n+1) (((a+1)@b)#{a,$_4}), b > 0
Rule 7.
ah_n ((0@b)#{a,$_5}) = ah_n ((a@b)#{a[n],$_5}), a a lim ord & b > 0

The rest of the array rules:
ah_n{$_0//(b+1),$_1} = ah_n((($_0)(@^_n)ah_n{$_0//b})#{//b,$_1}) ah_n{$_0//0} = ah_n{$_0} ah_n{$_0//b,$_1} = ah_n{$_0//(b[n]),$_1} for lim ord b Else: apply ah's 7 rules, starting after legion bar. (e.g. ah_{$_0//,0,0,w+1} = ah_{$_0//,w+1,w+1,w}) ($_1)(@^_n)a =
($_1)@(ah_n(($_1)(@^_n)(a-1))), a > 1,
ah_n{$_1}, a=1 ah_n{$_0@@0} = ah_n{$_0} ah_n{$_0@@a,$_1} = ah_n{$_0@@a[n],$_1},a a lim ord ah_n{$_0@@(a+1),$_1} = ah_n({$_0//}#{$_0@@(a),$_1}
Apply basic ah rules otherwise to everything after the @@

ah_n{$_0(@@b,$_2)0} = ah_n{$_0} ah_n{$_0((@@b,$_2)a,$_1} = ah_n{$_0(@@b)a[n],$_1},a a lim ord
ah_n{$_0(@@0)(a+1),$_1} = ah_n({$_0//}#{$_0(@@0)(a),$_1} ah_n{$_0(@@b+1,$_2)(a+1),$_1} = ah_n({$_0(@@b,$_2)}#{$_0@@(a),$_1}
ah_n{$_0(@@b,$_2)$_3} = ah_n{$_0(@@b[n],$_2)$_3}, if b is a lim ord
Apply basic ah rules otherwise to everything after the (@@$) or everything inside the (@@$) depending on what is necessary

These array rules can be applied without the ah_n of the array, but any n refers to the n in the ah subscript.

Note that we can now nest things as powerful as (@@$) to make extremely powerful functions. What is the power of ah_n{gamma_0(@@gamma_0)gamma_0} ? EDIT: Redefine legion bars and the like in a way that formalizes the array rules without the ah: Code: Select all  if unspecified, n is the subscript in the first ah that is in front of the array. Now we can redefine the rest of the array rules: The rest of the array rules: {$_0//(b+1),$_1} = ((($_0)(@^_n)ah_n{$_0//b})#{//b,$_1})
{$_0//0} = {$_0}
{$_0//b,$_1} = {$_0//(b[n]),$_1} for lim ord b
Else: apply ah's 7 rules, starting after legion bar. This includes nesting and incrementing the subscript (e.g. ah_n{$_0//,0,0,w+1} = ah_(n+1){$_0//,w+1,w+1,w})

($_1)(@^_n)a = ($_1)@(ah_n(($_1)(@^_n)(a-1))), a &gt; 1, ah_n{$_1}, a=1

{$_0@@0} = {$_0}
{$_0@@a,$_1} = {$_0@@a[n],$_1},a a lim ord
{$_0@@(a+1),$_1} = ({$_0//}#{$_0@@(a),$_1} Apply basic ah rules otherwise to everything after the @@. See legion bar notes for more details {$_0(@@b,$_2)0} = {$_0}
{$_0((@@b,$_2)a,$_1} = {$_0(@@b)a[n],$_1},a a lim ord {$_0(@@0)(a+1),$_1} = ({$_0//}#{$_0(@@0)(a),$_1}
{$_0(@@b+1,$_2)(a+1),$_1} = ({$_0(@@b,$_2)}#{$_0@@(a),$_1} {$_0(@@b,$_2)$_3} = {$_0(@@b[n],$_2)$_3}, if b is a lim ord Apply basic ah rules otherwise to everything after the (@@$) or everything inside the (@@$) depending on what is necessary. See legion bar notes for more details. (Now things such as ah_n{a(@@(a(@@a)a))a} are well defined.) {$_0@@@(a+1),$_1} = {$_0(@@($_0@@@a,$_1))$_0,$_1}
{$_0@@@0} = {$_0}
{$_0@@@a,$_1} = {$_0@@@a[n],$_1}, a a lim ord.
Last edited by Moosey on November 16th, 2019, 8:28 am, edited 2 times in total.
I am a prolific creator of many rather pathetic googological functions

My CA rules can be found here

Also, the tree game
Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"

Nanho walåt derwo esaato?

gameoflifemaniac
Posts: 1191
Joined: January 22nd, 2017, 11:17 am
Location: There too

### Re: Ordinals in googology

Wow.

What is the Bachmann-Howard ordinal?
I was so socially awkward in the past and it will haunt me for the rest of my life.

Code: Select all

x = 21, y = 5, rule = B3/S23
b2o2bo2b2o2b2o2bobobo$o3bobobobobobobobobo$bo2bobob2o2b2o3bo2bo$2bobob obobobobo2bo$2o3bo2bobobobo2bo2bo!