Ordinals in googology

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Moosey
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Re: Ordinals in googology

Post by Moosey » November 29th, 2019, 11:17 am

gameoflifemaniac wrote:
November 29th, 2019, 10:38 am
Moosey wrote:
November 26th, 2019, 6:15 pm
they aren't
why?
Because someone tried to diagonalize f_psi_1(1) (3) in a video. How can you diagonalize an uncountable number?
Oh.
I thought you meant my OCF

I don't know why, but there are quite a few possibilities
1) Typos
2) Different psi
3) Anyone can post something false on YouTube. Just because one has heard of OCFs and the fgh doesn't mean one understands them.
4) something else
Last edited by Moosey on November 29th, 2019, 3:32 pm, edited 1 time in total.
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Re: Ordinals in googology

Post by gameoflifemaniac » November 29th, 2019, 3:10 pm

Moosey wrote:
November 29th, 2019, 11:17 am
3) Anyone can post something false on YouTube. Just because you've heard of the fgh doesn't mean you understand them.
f_w(n) = f_n(n)
f_w+1(n) = f_w(f_w(f_w...(n)...)) n times
f_w2(n) = f_w+n(n)
f_w3(n) = f_w2+n(n)
f_w^2(n) = f_wn(n)
f_w^3(n) = f_(w^2)n(n)
f_w^w(n) = f_w^n(n)
f_w^w^w(n) = f_w^w^n(n)
f_e_0 = f_w^^n(n)
Don't underestimate me.
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Re: Ordinals in googology

Post by Moosey » November 29th, 2019, 3:32 pm

gameoflifemaniac wrote:
November 29th, 2019, 3:10 pm
Moosey wrote:
November 29th, 2019, 11:17 am
3) Anyone can post something false on YouTube. Just because you've heard of the fgh doesn't mean you understand them.
f_w(n) = f_n(n)
f_w+1(n) = f_w(f_w(f_w...(n)...)) n times
f_w2(n) = f_w+n(n)
f_w3(n) = f_w2+n(n)
f_w^2(n) = f_wn(n)
f_w^3(n) = f_(w^2)n(n)
f_w^w(n) = f_w^n(n)
f_w^w^w(n) = f_w^w^n(n)
f_e_0 = f_w^^n(n)
Don't underestimate me.
I was referring to the person in YouTube, not you
(Also f_e_0(n), n>1 = f_w^^(n-1)(n) because e_0's fs is {0,1,w,w^w,...} not {1,w,w^w,...})
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Re: Ordinals in googology

Post by gameoflifemaniac » November 29th, 2019, 4:22 pm

Moosey wrote:
November 29th, 2019, 3:32 pm
(Also f_e_0(n), n>1 = f_w^^(n-1)(n) because e_0's fs is {0,1,w,w^w,...} not {1,w,w^w,...})
What? 0 is not in e_0's fundamental sequence because w^0 = 1, and 0 can't be expressed that way.
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Re: Ordinals in googology

Post by Moosey » November 29th, 2019, 4:35 pm

gameoflifemaniac wrote:
November 29th, 2019, 4:22 pm
Moosey wrote:
November 29th, 2019, 3:32 pm
(Also f_e_0(n), n>1 = f_w^^(n-1)(n) because e_0's fs is {0,1,w,w^w,...} not {1,w,w^w,...})
What? 0 is not in e_0's fundamental sequence because w^0 = 1, and 0 can't be expressed that way.
0 is in e_0's fs, as virtually every source or mathematician will tell you:
https://googology.wikia.org/wiki/Ε₀#Hig ... _hierarchy
https://en.wikipedia.org/wiki/Epsilon_n ... _ε_numbers

In fact, what you said is arguably why the FS starts at 0 and not 1: 0 is the "starting point" because nothing can be an earlier starting point
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Re: Ordinals in googology

Post by gameoflifemaniac » November 29th, 2019, 5:00 pm

Moosey wrote:
November 29th, 2019, 4:35 pm
gameoflifemaniac wrote:
November 29th, 2019, 4:22 pm
Moosey wrote:
November 29th, 2019, 3:32 pm
(Also f_e_0(n), n>1 = f_w^^(n-1)(n) because e_0's fs is {0,1,w,w^w,...} not {1,w,w^w,...})
What? 0 is not in e_0's fundamental sequence because w^0 = 1, and 0 can't be expressed that way.
0 is in e_0's fs, as virtually every source or mathematician will tell you:
https://googology.wikia.org/wiki/Ε₀#Hig ... _hierarchy
https://en.wikipedia.org/wiki/Epsilon_n ... _ε_numbers

In fact, what you said is arguably why the FS starts at 0 and not 1: 0 is the "starting point" because nothing can be an earlier starting point
Ok
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Re: Ordinals in googology

Post by BlinkerSpawn » November 29th, 2019, 5:10 pm

gameoflifemaniac wrote:
November 29th, 2019, 2:33 am
BlinkerSpawn wrote:
November 28th, 2019, 5:53 pm
gameoflifemaniac wrote:
November 27th, 2019, 2:31 pm
Can anyone explain me ordinals significantly bigger than the BHO?
In many commonly used ordinal notations, W_1 acts as a fixed point of psi_0.
We can then define psi_1 which creates expressions with W_1, and define W_2 as a fixed point of psi_1.
Many strong notations, such as Pi and Pair Sequence System, make use of this "omega-upgrading" up to its first limit, psi_0(W_w).
Wouldn't that give you uncountable ordinals?
Yeah, but that's fine as long as they appear inside a psi that collapses them into countables.
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Re: Ordinals in googology

Post by gameoflifemaniac » November 30th, 2019, 3:35 am

BlinkerSpawn wrote:
November 29th, 2019, 5:10 pm
gameoflifemaniac wrote:
November 29th, 2019, 2:33 am
BlinkerSpawn wrote:
November 28th, 2019, 5:53 pm

In many commonly used ordinal notations, W_1 acts as a fixed point of psi_0.
We can then define psi_1 which creates expressions with W_1, and define W_2 as a fixed point of psi_1.
Many strong notations, such as Pi and Pair Sequence System, make use of this "omega-upgrading" up to its first limit, psi_0(W_w).
Wouldn't that give you uncountable ordinals?
Yeah, but that's fine as long as they appear inside a psi that collapses them into countables.
I kind of get it, but not quite
Maybe the diagonalization of ordinals larger than the Bachmann-Howard ordinal will help me understand what they are.
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Re: Ordinals in googology

Post by Moosey » December 1st, 2019, 9:40 am

ah v2:

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a serious revamp of ah. Most rules are preserved.

Part one: Starter notations
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

Basic ah rules
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

Enormous arrays
if unspecified, n is the subscript in the first (closest) ah that is in front of the array.

{$_0(@@b,$_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. This includes nesting and incrementing the subscript. 

{$_0@@@(a+1),$_1} = {$_0(@@($_0@@@a,$_1))$_0,$_1} 
{$_0@@@0} = {$_0} 
{$_0@@@a,$_1} = {$_0@@@a[n],$_1}, a a lim ord. 
Apply ah rules otherwise

Dimensional arrays
{$_0[[k]]0} = {$_0} 
{$_0[[k+1]]a+1,$_1} = {$_0[[k]]$_0[[k+1]]a,$_1} 
{$_0[[k]]a+1,$_1} = {$_0[[(k[n])]]$_0[[k]]a,$_1}, k a lim ord 
{$_0[[0]]a+1,$_1} = {$_0@@@$_0[[0]]a,$_1} 
{$_0[[k]]a,$_1} = {$_0,a[[k]]a[n],$_1}, a a lim ord 
{$_0[[k]]0@m,a+1,$_1} = {$_0[[k]](ah_n{$_0[[k]](a+1)@m,a,$_1})@m,a,$_1} 
{$_0[[k]]0@m,a,$_1} = {$_0[[k]]a@m,a[n],$_1}, a a lim ord

The rightmost of the highest-valued (by k) [[k]] is evaluated first.
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Re: Ordinals in googology

Post by gameoflifemaniac » December 3rd, 2019, 10:18 am

Can you diagonalize w_1CK?
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Re: Ordinals in googology

Post by Moosey » December 3rd, 2019, 4:24 pm

gameoflifemaniac wrote:
December 3rd, 2019, 10:18 am
Can you diagonalize w_1CK?
That depends on what you mean by diagonalization in this context
Typically you diagonalize across many things
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Re: Ordinals in googology

Post by gameoflifemaniac » December 4th, 2019, 2:37 am

I mean what the fundamental sequence of it? I can't understand what's in the wiki.
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Re: Ordinals in googology

Post by Moosey » December 4th, 2019, 7:56 am

gameoflifemaniac wrote:
December 4th, 2019, 2:37 am
I mean what the fundamental sequence of it? I can't understand what's in the wiki.
Its fundamental sequence would be using ordinals which are the first not computable using n or fewer symbols, or something like that.
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Re: Ordinals in googology

Post by gameoflifemaniac » December 10th, 2019, 10:28 am

LVO isn't the limit of Veblen notation. There are more fixed points beyond that.
LVO is psi(Ω^Ω^Ω).
For simplicity I'll use a recursive function called v(α) (which I made up):
v(α) = phi(1, α 0's)

v(0) = phi(0) = 1
v(1) = phi(1,0) = ω
v(2) = phi(1,0,0) = Γ_0
v(ω) = SVO (Small Veblen ordinal)
v(v(ω)) = phi(1, SVO 0's)
v(v(v(v(v(...))))) = LVO (Large Veblen ordinal)
But we can go even further.
v(v(v(v(v(...LVO+1...))))) = LVO_1 = psi((Ω^Ω^Ω)2) (next fixed point after LVO)
v(v(v(v(v(...LVO_1+1...))))) = LVO_2 = psi((Ω^Ω^Ω)3)
...

Is this known?
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Re: Ordinals in googology

Post by Moosey » December 11th, 2019, 7:13 am

Presumably it's known that you can keep going farther, but I'm not sure.
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Re: Ordinals in googology

Post by Moosey » December 11th, 2019, 8:37 pm

Just rolled out a new version of ah:
https://www.conwaylife.com/wiki/User:Mo ... nal_arrays
Will update blog tomorrow
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Re: Ordinals in googology

Post by Moosey » January 15th, 2020, 6:52 pm

ah is getting very messy and very powerful with the addition of a new separator, ,,,

basically I index my separator
it was based on an idea for extending beyond ,, to ,,, and so forth but then I realized I could do it in a single separator

So I'm gonna try to analyze it
[[0]] >> gamma_0 in the fgh as determined previously
[[1]] is probably phi(2,0,0)?
[[n]] is probably around phi(w,0,0)
[[$]] is then around gamma_w^2 to phi(w^2,0,0) (?)
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Re: Ordinals in googology

Post by Moosey » February 9th, 2020, 6:59 pm

I'm tired of working on ah for awhile because I ran out of ideas (feel free to suggest how ,,,, should work!)

So instead I'll work on a way to turn trees into ordinals

let $ indicate any tree
let a-b indicate the tree

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b
|
a
let
(a-$)-b
indicate the tree

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$ b
\/
a
so for instance, (0-0-0)-0
is the tree

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0
|
0 0
\/
0
alright so now for evaluation:
let E($) = the value of $
E(0) = 0
E($-0) = $ + 1, if $ is not in parentheses
E((0-a)-0) = sup{a,a-a,a-a-a,a-a-a-a,etc}

That's all for now.

E((0-0)-0) = w already.


Also note that with this representation of trees you could play TREE
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Re: Ordinals in googology

Post by Moosey » February 16th, 2020, 7:11 pm

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Re: Ordinals in googology

Post by BlinkerSpawn » February 21st, 2020, 10:35 pm

Moosey wrote:
February 9th, 2020, 6:59 pm
So instead I'll work on a way to turn trees into ordinals...
This notation, given I'm interpreting it correctly, has limit e_0.
Proof:
From the successorship and expansion rules we get that a-b must have value a+b.
The tree (a+1)-0 is equivalent to a-a-a-a-... with w a's, so the value of (a+1)-0 is w*a.
Therefore, if a-0 has value w^b, then (a+1)-0 has value w^(b+1), and so the limit of these trees is the first limit of additive principals, e_0.
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Re: Infinities in googology

Post by Moosey » February 27th, 2020, 11:12 am

I find it shocking how simple inaccs are
An inaccessible k is not of the form N_(a+1) and has cofinality k
Which is really simple when you know how cofinality works

But can anyone give me a simple understanding of Mahlos like this?
All I need is a better understanding of clubsets
Specifically, the part with the "closed" part
Not the unbounded part

(I also want to learn about everything else on this page, and yes, including Reinhardt cardinals.)

Also, random tangent: a different googologist Reinhardt plays piano well
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Re: Infinities in googology

Post by BlinkerSpawn » February 28th, 2020, 11:31 pm

Moosey wrote:
February 27th, 2020, 11:12 am
But can anyone give me a simple understanding of Mahlos like this?
All I need is a better understanding of clubsets
Specifically, the part with the "closed" part
Not the unbounded part
According to the Wikipedia definition, club sets can be thought of as being indexable by ordinals:
In fact a club set is nothing else but the range of a normal function (i.e. increasing and continuous).
Say you have a function f that takes ordinals and returns elements of k.
k being unbounded is similar to saying that for every f(a) in k there exists f(a+1) > f(a).
k being closed is similar to saying that for every f(a_0), f(a_1), f(a_2), ... in k there exists f(sup(a_0, a_1, a_2, ...)) = sup(f(a_0), f(a_1), f(a_2), ...)) also in k.
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Re: Ordinals in googology

Post by testitemqlstudop » March 2nd, 2020, 6:54 am

E(0) = 0
E($-0) = $ + 1, if $ is not in parentheses
E((0-a)-0) = sup{a,a-a,a-a-a,a-a-a-a,etc}
E(x) = limit of E if all nodes are <= x-1
E($-x) = E(($-(x-1))-($-(x-1))-(x-1)) + 1, if $ not in parentheses
E((x-$)-x) = sup{((x-1)-$)-(x-1), ((x-1)-$)-((x-1)-$)-(x-1), ((x-1)-$)-((x-1)-$)-((x-1)-$)-(x-1), etc}

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Re: Ordinals in googology

Post by PkmnQ » March 2nd, 2020, 8:02 am

Hmm...can you reverse engineer a tree from an ordinal?
∃(w) = (0-0)-0

I won’t try this yet because of sup{}.

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Re: Ordinals in googology

Post by Moosey » March 3rd, 2020, 9:36 am

PkmnQ wrote:
March 2nd, 2020, 8:02 am
Hmm...can you reverse engineer a tree from an ordinal?
∃(w) = (0-0)-0

I won’t try this yet because of sup{}.
sup is just the supremum-- the minimum upper bound

Actually this use of sup is questionable
testitemqlstudop wrote:
March 2nd, 2020, 6:54 am
E(0) = 0
E($-0) = $ + 1, if $ is not in parentheses
E((0-a)-0) = sup{a,a-a,a-a-a,a-a-a-a,etc}
E(x) = limit of E if all nodes are <= x-1
E($-x) = E(($-(x-1))-($-(x-1))-(x-1)) + 1, if $ not in parentheses
E((x-$)-x) = sup{((x-1)-$)-(x-1), ((x-1)-$)-((x-1)-$)-(x-1), ((x-1)-$)-((x-1)-$)-((x-1)-$)-(x-1), etc}
lim would probably be better

Also
Testitem
You're active again

Did you see the explanation of mahlos I found and posted on my discord
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