Ordinals in googology

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testitemqlstudop
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Re: Ordinals in googology

Post by testitemqlstudop » October 25th, 2019, 7:34 am

My argument for Latex is with such a complicated definition, with no formatting it's difficult to understand.

What is the cardinality of the set of:

1) all ordinals;
2) all countable ordinals;
3) all computable ordinals.

1 is at least beth_2.

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Re: Ordinals in googology

Post by Macbi » October 25th, 2019, 7:45 am

testitemqlstudop wrote:
October 25th, 2019, 7:34 am
What is the cardinality of the set of:

1) all ordinals;
2) all countable ordinals;
3) all computable ordinals.
1) There is no set of all ordinals, just like there is no set of all sets.
2) ℵ_1
3) ℵ_0

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Re: Ordinals in googology

Post by Moosey » October 25th, 2019, 3:32 pm

So, how powerful is ah_n((0@n)#{w+1})?
I think that at n = 1 it's already larger than g64, but I'm not sure
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Re: Ordinals in googology

Post by Moosey » October 25th, 2019, 5:05 pm

testitemqlstudop wrote:
October 24th, 2019, 8:45 pm
8a. ah_n((0@a)#{b,$_6}) = ah_n((0@(a[n])#{b,$_6}), a a lim ord
That does not work with your recursive definition at all.
Rule 8 is a draft; do not mention any issues with it until I work it out.
But feel free to suggest ways to improve it (and remove issues). Just don't point out issues without suggestions of how to improve it.

Currently, just treat it as though the rules are:
Moosey wrote:Rules:
I define a#b to be the concatenation of the arrays a and b, and n@m to be an array of m n’s — specifically,
n@m =
n, m = 1
{n}#(n@(m-1))
Additionally, I 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

1. If ah is iterated: ah^a_n {$_0} is ah_ah_ah_ah…_n {$_0} {$_0} {$_0} {$_0}… with a ah’s and {$_0}’s
Formally, define g(a,n,B) =
ah_g(a-1,n,B) {B}, a > 1;
n, a = 0.
ah^a_n {$_0} = g(a,n,$_0)
2. ah_n{$_1,z} = ah_n{$_1}, z = 0
alternatively, ah_n ({$_1}#{z}) = ah_n {$_1}, z = 0
3. ah_n{} = n+1
4. ah_n{a+1,$_2} = ah^n_n{a,$_2}
5. ah_n{a,$_3} = ah_n{a[n],$_3}, a a lim ord
6. ah_n ((0@b)#{a+1,$_4}) = ah_(n+1) (((a+1)@b)#{a,$_4}), b > 0
7. ah_n ((0@b)#{a,$_5}) = ah_n ((a@b)#{a[n],$_5}), a a lim ord & b > 0
EDIT:
even ah_1{0,w+1} is huge: during my world history class I worked on evaluating it and got the following:
ah_1{0,w+1} =

Code: Select all

ah_2{w+1,w}
ah_ah_2{w,w}{w,w}
ah_ah_2{2,w}{w,w}
ah_ah_ah_2{1,w}{1,w}{w,w}

ah_2{1,w}=
ah_ah_2{0,w}{0,w}
ah_ah_2{w,2}{0,w}
ah_ah_2{2,2}{0,w}

ah_2{2,2} =
ah_ah_2{1,2}{1,2}
ah_ah_ah_2{0,2}{0,2}{1,2}

ah_2{0,2} =
ah_3{2,1}
ah_ah_ah_3{1,1}{1,1}{1,1}
ah_ah_ah_ah_ah_3{0,1}{0,1}{0,1}{1,1}{1,1}

ah_3{0,1} = 8

ah_ah_3{0,1}{0,1} =
ah_8{0,1}
18

ah_ah_ah_3{0,1}{0,1}{0,1} =
ah_18{0,1}
38

Therefore
ah_1{0,w+1} =
ah_ah_ah_ah_ah_ah_ah_38{1,1}{1,1}{0,2}{1,2}{0,w}{1,w}{w,w}
I'd imagine ah_ah_ah_ah_ah_ah_ah_38{1,1}{1,1}{0,2}{1,2}{0,w}{1,w}{w,w} >> g64
(and thus ah_1{0,w+1} >> g64)
but I don't know how to prove it

Regardless, ah_2{0,0,w+1} is much much much larger. ah_64 ((0@64)#{w+1}) is of course >> g64

In fact, one can prove that ah_63 ((0@63)#{w+1}) is also >> g64, and one can probably extend this to at least, say, something like ah_32 ((0@32)#{w+1})

Note: ah_n ((0@n)#{w+1}) is by definition ah_(n+1) (((w+1)@n)#{w})

also note that ah_n{a} = f_a(n)

Thus, it is easy to show that if ah_n{anything} ever has something of the form ah_m{w+1,$_0} for m >= 64, it is provably larger than g64 (even if $_0 is no entries)
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My CA rules can be found here

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Re: Ordinals in googology

Post by testitemqlstudop » October 25th, 2019, 11:16 pm

Macbi wrote:
October 25th, 2019, 7:45 am
testitemqlstudop wrote:
October 25th, 2019, 7:34 am
What is the cardinality of the set of:

1) all ordinals;
2) all countable ordinals;
3) all computable ordinals.
3) ℵ_0
That does not make sense; you can infinitely exponentiate w and add them together (this is further enlargened by the branching possibilities of each exponent and coefficient), so under e0 it already makes sense to have a set larger than aleph1.

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Re: Ordinals in googology

Post by Moosey » October 26th, 2019, 6:18 am

testitemqlstudop wrote:
October 25th, 2019, 11:16 pm
Macbi wrote:
October 25th, 2019, 7:45 am
testitemqlstudop wrote:
October 25th, 2019, 7:34 am
What is the cardinality of the set of:

1) all ordinals;
2) all countable ordinals;
3) all computable ordinals.
3) ℵ_0
That does not make sense; you can infinitely exponentiate w and add them together (this is further enlargened by the branching possibilities of each exponent and coefficient), so under e0 it already makes sense to have a set larger than aleph1.
It makes perfect sense: w_1ck is the size of the set of computable ordinals, and |w_1ck| = aleph0
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Re: Ordinals in googology

Post by Macbi » October 26th, 2019, 7:14 am

testitemqlstudop wrote:
October 25th, 2019, 11:16 pm
Macbi wrote:
October 25th, 2019, 7:45 am
testitemqlstudop wrote:
October 25th, 2019, 7:34 am
What is the cardinality of the set of:

1) all ordinals;
2) all countable ordinals;
3) all computable ordinals.
3) ℵ_0
That does not make sense; you can infinitely exponentiate w and add them together (this is further enlargened by the branching possibilities of each exponent and coefficient), so under e0 it already makes sense to have a set larger than aleph1.
Cardinal exponentiation and ordinal exponentiation have different rules. The cardinal ℵ_0^ℵ_0 is the cardinality of the set of all functions from ℕ to ℕ, whereas the ordinal ω^ω corresponds to the lexicographic ordering on the set of all functions from ℕ to ℕ for which all but finitely many values are 0. So ω^ω is much smaller than ℵ_0^ℵ_0.

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Re: Ordinals in googology

Post by Moosey » October 27th, 2019, 10:03 am

testitemqlstudop wrote:
October 24th, 2019, 8:45 pm
PLEASE USE LATEX
Fine.

Code: Select all

\begin{eqnarray*}g(a,n,B)=\\
ah_{g(a-1,n,B)} \{B\}, a > 1;\\
n, a = 0\\
rule\ 1: ah^a_n\{\$_0\} = g(a,n,\$_0)\\
rule\ 2: ah_n\{\$_1,\beta\} = ah_n\{\$_1\}, \beta = 0\\
rule\ 3: ah_n\{\} = n+1\\
rule\ 4: ah_n\{\alpha+1,\$_2\} = ah^n_n\{\alpha,\$_2\}\\
rule\ 5: ah_n\{\alpha,\$_3\} = ah_n\{\alpha[n],\$_3\}, \alpha\ a\ lim\ ord\\
rule\ 6: ah_n ((0@b)\#\{\alpha+1,\$_4\}) = ah_{n+1} (((\alpha+1)@b)\#\{\alpha,\$_4\}),\ b > 0\\
rule\ 7: ah_n ((0@b)\#\{\alpha,\$_5\}) = ah_n ((\alpha@b)\#\{\alpha[n],\$_5\}), \alpha\ a\ lim\ ord, b > 0
\end{eqnarray*}
See here

Possible extension:
I'm considering using a sort of equivalent to legion bars from BEAF

EDIT:
{a,$_0//b} = {a@a@a@a@a w/b a's,$_0} doesn't quite work.
But
(Note: this is an old definition. A much definition is found further down this post)
ah_n{a,$_0//b} = ah_n(((a@(ah_n{a,$_0//(b-1)}))#{$_0})#{//(b-1)}), b>0
ah_n{a,$_0//b} = ah_n{a,$_0}, b=0
does.
That sounds really really really confusing, but it's just this:
ah_3{w,1,2//1} = ah_3{w,w,w,w,w,w,w,w,w,w,w,w,w...,1,2} with ah_3{w,1,2} w's

(btw, don't scream "LATEX!!! Where is the LaTeX?" Because this is hardly any clearer.

Another demo:
ah_1{1//1} = ah_1{1@(ah_1{1})} = ah_1{1,1} = ah_2{1} = ah_ah_2{0}{0} = 4
ah_1{1//2} = ah_1{1@(ah_1{1//1})//1} = ah_1{1,1,1,1//1} = ah_1{1@(ah_1{1,1,1,1}),1,1,1}
ah_1{1,1,1,1} =

Code: Select all

ah_1{0,1,1,1}
ah_2{1,0,1,1}
ah_ah_2{0,0,1,1}{0,0,1,1}
ah_ah_3{1,1,0,1}{0,0,1,1}
... probably rather large
Therefore:
ah_1{1//2} is rather large

Really, an even better definition is this:
ah_n{$_0//(b+1)} = ah_n((($_0)(@^_n)ah_n{$_0//b})#{//b})
ah_n{$_0//0} = ah_n{$_0}
and of course, the dreaded limit ordinal case:
ah_n{$_0//b} = ah_n{$_0//(b[n]} for lim ord b
Where ($_1)(@^_n)a =
($_1)@(ah_n(($_1)(@^_n)(a-1))), a > 1,
ah_n{$_1}, a=1

Using this definition, which is closer to legion bars anyways, one gets functions like ah_n{2//w} or even ah_n{(gamma_0)//gamma_0}
(And of course, something like ah_n{(gamma_0)//gamma_0}, if I wrote all this as I wanted to and it's well defined, is by far the most powerful function that I have coined.)

Even ah_n{1//w} is huge very fast:
n = 1:
ah_1{1//1} = ah_1(1(@^_1)(ah_1{1})) =
(ah_1{1} = 2)
ah_1(1(@^_1)2) =
ah_1(1@(ah_1{1})) =
ah_1{1,1} = 4
n = 2:
ah_2{1//2} = ah_2((1(@^_2)(ah_2{1//1}))//1)
Already, this must be > ah_2((1,1,1,1)//1)

Code: Select all

ah_2{1//1} =
ah_2(1(@^_2)4) =
ah_2(1@(ah_2(1@(ah_2(1@(ah_2(1@(ah_2{1})))) =
ah_2(1@(ah_2(1@(ah_2(1@(ah_2{1,1,1,1}))
...
ah_2{1//w} =
ah_2((1(@^_2)(ah_2(1@(ah_2(1@(ah_2(1@(ah_2{1,1,1,1}))))//1)
Yikes, this grows fast
Can anyone give reasonable bounds in terms of other functions?

(Also, it looks like ah_n{gamma_0//gamma_0} is defined, making it my fastest growing function, since of course it trumps f_gamma_0(n) = ~dco(gamma_0+n))


And now, the @@ operator, essentially the legion array of operator:
ah_n(a@@b) expands to
ah_n(a//(a@@(b-1))), b>0, b not a lim ord
ah_n(a), b = 0
ah_n(a@@(b[n])), b a lim ord
(Note: legion arrays can currently only associate rightwards, I.e. a//b//c is (a//b)//c which is, in an ah array, this mess:
ah_n{(a//b)//c} = ah_n(((a//b)(@^_n)ah_n{(a//b)//(c-1)})#{//(c-1)})

I think ah_n(n@@n) is very fast growing but I have a hunch that it's somewhat illdefined. So let's just stick to legions (//) for now.

Y'know, I'll add an extra rule for a@b so that it does the whole lim ord mess thing too:
ah_n({$_0}#a@b) = ah_n({$_0}#a@(b[n]))

EDIT:
Better @@:
ah_n($_0@@b) expands to
ah_n($_0//(ah_n($_0@@(b-1)))), b>0, b not a lim ord
ah_n($_0), b = 0
ah_n($_0@@(b[n])), b a lim ord
This time they associate more powefully:
ah_n(gamma_0@@gamma_0) is much more powerful than ah_n(gamma_0//gamma_0)

And now, two more:
ah_n{$_0/m+1/(b+1)} = ah_n((($_0)(@@^m_n)ah_n{$_0//b})#{//b})
ah_n{$_0/m+1/0} = ah_n{$_0}
ah_n{$_0/m/b} = ah_n{$_0/m/(b[n]} for lim ord b
ah_n{$_0@@b}, m = 0
Where ($_1)(@@^m_n)a =
($_1)@@(ah_n(($_1)(@@^m_n)(a-1)), a > 1, m=0
($_1)@@(ah_n(($_1)(@@^m_n)(a-1)#{/m/$_0)), a > 1, m>0
ah_n{$_1}, a=1
These new ones don't exactly match my original stuff and they may be illdefined, but they're pretty crazy.

Anyways, how powerful is ah_n(gamma_0@@gamma_0)
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My CA rules can be found here

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Re: Ordinals in googology

Post by Moosey » October 28th, 2019, 10:35 am

Bump
I added some new rules to ah.
What is ah_1{gamma_0@@gamma_0}?
How large is it compared to, say, g64?

The forums have been so inactive recently.
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Also, the tree game
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Re: Ordinals in googology

Post by Moosey » October 29th, 2019, 7:08 pm

Suppose you defined an OCF thusly:

Code: Select all

C_0,0(a) = {0,1}
C_(n+1),0(a) = C_n,0(a) U min{a|Forall b, psi_n(b) <= a}
C_n,m+1(a) = C_n,m(a) U {c+d,cd,c^d,epsilon_c,psi_n(h),forall c,d,h in C_n,m(a),h<a}
C_n(a) = U(m<w) C_n,m
psi_n(a) = sup{all b<w_1 in C_n(a)} 
Does psi_n add the ordinals psi_m gets stuck at to the initial set for all m<n?
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Re: Ordinals in googology

Post by testitemqlstudop » October 31st, 2019, 2:27 am

I have defined an OCF, a fat version of it, and an accompanying fundamental sequence system.
I have took the limit of the fat OCF function and defined the fundamental sequence for it as well.
(Note that the fat OCF system isn't exactly "more powerful", but instead "simpler".)
You can see where this is going...
... and I fed this into the FGH.

I'm pretty sure the ordinal T is at least the SVO. Possibly even the BHO? (in which case this function would obliterate the lower bound for tree)

See my blog post here:

https://testitem.github.io/colg/fpt.html

(Before p-bot says "there's no such thing as the w-th fixed point" I'm going to point out that the w-th fixed point of a |-> w^a is e_w.)

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Re: Ordinals in googology

Post by Moosey » October 31st, 2019, 6:50 am

testitemqlstudop wrote:
October 31st, 2019, 2:27 am
I have defined an OCF, a fat version of it, and an accompanying fundamental sequence system.
...
See my blog post here:
https://testitem.github.io/colg/fpt.html
That's not a traditional OCF-- OCFs are defined differently, like this:

Code: Select all

C_0(x) = set
C_n+1(x) = C_n(x) U {operations with a and b in C_n(x)} U {psi(d), d in C_n(x), d<x}
C(x) = U (n<w) C_n(x)
psi(x) = min x not in C(x)
You also have a mistake in the original; psi_w(0) = (wait for it) phi_w(0), not gamma_0
I'd imagine that psi_n(m) = phi_n(m), so fatphi(n) for n < w is probably along the lines of gamma_n.
It's quite possible that, in that case, T is only a|->gamma_a (which makes it much less that the Ackerman ordinal, and I think it's phi(1,0,1))

Just, please, don't curse posterity by deleting the whole post or something if you get frustrated by how fatphi let you down.

Speaking of phi_w(0), it really needs a name. How about we refer to it as "Joe the ordinal?"
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Re: Ordinals in googology

Post by testitemqlstudop » October 31st, 2019, 9:13 am

Moosey wrote:
October 31st, 2019, 6:50 am
testitemqlstudop wrote:
October 31st, 2019, 2:27 am
I have defined an OCF, a fat version of it, and an accompanying fundamental sequence system.
...
See my blog post here:
https://testitem.github.io/colg/fpt.html
That's not a traditional OCF-- OCFs are defined differently, like this:

Code: Select all

C_0(x) = set
C_n+1(x) = C_n(x) U {operations with a and b in C_n(x)} U {psi(d), d in C_n(x), d<x}
C(x) = U (n<w) C_n(x)
psi(x) = min x not in C(x)
You also have a mistake in the original; psi_w(0) = (wait for it) phi_w(0), not gamma_0
I'd imagine that psi_n(m) = phi_n(m), so fatphi(n) for n < w is probably along the lines of gamma_n.
It's quite possible that, in that case, T is only a|->gamma_a (which makes it much less that the Ackerman ordinal, and I think it's phi(1,0,1))

Just, please, don't curse posterity by deleting the whole post or something if you get frustrated by how fatphi let you down.

Speaking of phi_w(0), it really needs a name. How about we refer to it as "Joe the ordinal?"
Isnt T0 the result of the diagonalization of e0, z0, n0, etc?
i.e. e0 = w^^2, z0 = w^^^2, n0 = w^^^^2
I remember you said somewhere T0 = w^^w, obviously that's a large underestimation

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Re: Ordinals in googology

Post by Moosey » October 31st, 2019, 4:17 pm

testitemqlstudop wrote:
October 31st, 2019, 9:13 am
Isnt T0 the result of the diagonalization of e0, z0, n0, etc?
i.e. e0 = w^^2 [sic], z0 = w^^^2, n0 = w^^^^2
I remember you said somewhere T0 = w^^w, obviously that's a large underestimation
T0? Do you mean gamma_0?
gamma_0 is certainly not the supremum (which is what you mean-- I won't argue with P-bot saying that it the diagonalisation of the sequence using some more formal definition of diagonalisation) of e0, z0, h0, and so forth-- the supremum of phi_1(0) (e0), phi_2(0) (z0), phi_3(0) (h0), phi_4(0), etc, is of course phi_w(0), not gamma_0
testitemqlstudop wrote:I remember you said somewhere T0 = w^^w, obviously that's a large underestimation
No, I was referring to saibian and bower's estimation. Also, all of your things are missing an arrow: e0 = w^^^2, z0 = w^^^^2 (unless you're saibian or bowers), etc.
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Re: Ordinals in googology

Post by toroidalet » October 31st, 2019, 8:32 pm

This is sort of a stupid question, but why isn't ε_1=ε_0^ω?
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Re: Ordinals in googology

Post by testitemqlstudop » October 31st, 2019, 8:44 pm

e0 is the first solution to a = w^a, or a transfinite power stack of w's. Adding another w will not change the number.

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Re: Ordinals in googology

Post by BlinkerSpawn » October 31st, 2019, 9:49 pm

toroidalet wrote:
October 31st, 2019, 8:32 pm
This is sort of a stupid question, but why isn't ε_1=ε_0^ω?
Epsilon numbers are the fixed points of w^, so e_a = w^(e_a) for any a.

w^(e_0^w) = w^(e_0+e_0^w) = e_0^e_0^w > e_0^w, so e_1 cannot be e_0^w.
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Re: Ordinals in googology

Post by testitemqlstudop » October 31st, 2019, 11:14 pm

Moosey wrote:
October 31st, 2019, 4:17 pm
testitemqlstudop wrote:
October 31st, 2019, 9:13 am
Isnt T0 the result of the diagonalization of e0, z0, n0, etc?
i.e. e0 = w^^2 [sic], z0 = w^^^2, n0 = w^^^^2
I remember you said somewhere T0 = w^^w, obviously that's a large underestimation
T0? Do you mean gamma_0?
gamma_0 is certainly not the supremum (which is what you mean-- I won't argue with P-bot saying that it the diagonalisation of the sequence using some more formal definition of diagonalisation) of e0, z0, h0, and so forth-- the supremum of phi_1(0) (e0), phi_2(0) (z0), phi_3(0) (h0), phi_4(0), etc, is of course phi_w(0), not gamma_0
testitemqlstudop wrote:I remember you said somewhere T0 = w^^w, obviously that's a large underestimation
No, I was referring to saibian and bower's estimation. Also, all of your things are missing an arrow: e0 = w^^^2, z0 = w^^^^2 (unless you're saibian or bowers), etc.
Thank you!

I have fixed it in

https://testitem.github.io/colg/fpt_new.html

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Re: Ordinals in googology

Post by Moosey » November 1st, 2019, 6:23 am

testitemqlstudop wrote:
October 31st, 2019, 8:44 pm
e0 is the first solution to a = w^a, or a transfinite power stack of w's. Adding another w will not change the number.
That's not quite right. e_0^w is not w^e_0, since exponentiation is not commutative.
e_0^w is, rather, using the definition of e_0 and exponentiation rules, (w^e_0)^w = w^(e_0*w) = w^((w^e_0)*w) = w^w^(e_0 +1)
Compare e_0^w with e_1, the supremum of {e_0 +1, w^(e_0 +1), w^w^(e_0 +1), w^w^w^(e_0 +1) ...}
Yes, e_0^w happens to be in e_1's fundamental sequence
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Re: Ordinals in googology

Post by gameoflifemaniac » November 1st, 2019, 5:17 pm

Gamma(0) is the fixed point of the phi function. What is phi function

Also, what is the largest countable ordinal?
Moosey wrote:e0 = w^^^2, z0 = w^^^^2
z0 is an index tower, not a tetration tower. And how do you define z1, z2, z3 etc.?
Last edited by gameoflifemaniac on November 1st, 2019, 5:35 pm, edited 1 time in total.
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Re: Ordinals in googology

Post by Moosey » November 1st, 2019, 5:35 pm

gameoflifemaniac wrote:
November 1st, 2019, 5:17 pm
Gamma(0) is the fixed point of the phi function. What is phi function
The binary Veblen phi function.
gameoflifemaniac wrote:
November 1st, 2019, 5:17 pm
Also, what is the largest countable ordinal?
There is no such thing. If you say that a is the largest countable ordinal, then I can point out that a+1 is larger.
gameoflifemaniac wrote:
November 1st, 2019, 5:17 pm
Moosey wrote:e0 = w^^^2, z0 = w^^^^2
z0 is an index tower, not a tetration tower.
prove it
It's a very informal metaphor. Calling e_0 w^^w is already informal
Additionally, w^^w = e_0, e_0^^w = e_1, so perhaps we should say that w^^^w = e_w
It seems that w^^^(2+n) = e_n, which would mean that z_0 is the first fixed point of w^^^n
gameoflifemaniac wrote:
November 1st, 2019, 5:17 pm
And how do you define z1, z2, z3 etc.?
In terms of tetration/pentation/whatever of w? Or in normal terms? If the latter, up to h_0 you can define it thusly:
z_0 = sup{0,e_0,e_e_0,e_e_e_0,...}
z_(n+1) = sup{(z_n)+1,e_((z_n)+1),e_e_((z_n)+1),e_e_e_((z_n)+1},...}
z_n = sup{z_(n[0]),z_(n[1]),z_(n[2]),z_(n[3]),z_(n[4])...}, n a lim ord
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Re: Ordinals in googology

Post by Moosey » November 1st, 2019, 5:59 pm

Define n%f to be the nth fixed point of the function f.

Code: Select all

C_0,0(a) = {0,1,w,W}
C_m+1,0(a) = C_m,0(a) U {0%psi_m}
C_m,n+1(a) = C_m,n(a) U {b+c,bc,b^c,psi_m(d); b,c,d in C_m,n(a), d<a}
C_m(a) = U (n<w) C_m,n(a)
psi_m(a) = min b|b not in C_m(a)
C(a) = U (m<w) C_m(a)
psi(a) = min b|b not in C(a)
Does this function work? Or does it become ill-defined?
(This is a more formal version of this:)
Moosey wrote:
October 29th, 2019, 7:08 pm
Suppose you defined an OCF thusly:

Code: Select all

C_0,0(a) = {0,1}
C_(n+1),0(a) = C_n,0(a) U min{a|Forall b, psi_n(b) <= a}
C_n,m+1(a) = C_n,m(a) U {c+d,cd,c^d,epsilon_c,psi_n(h),forall c,d,h in C_n,m(a),h<a}
C_n(a) = U(m<w) C_n,m
psi_n(a) = sup{all b<w_1 in C_n(a)} 
Does psi_n add the ordinals psi_m gets stuck at to the initial set for all m<n?
Anyways, with the new psi, not the one in quotes, what is, say, psi_5(W)? Does psi(W) exist?
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Re: Ordinals in googology

Post by testitemqlstudop » November 2nd, 2019, 3:11 am

Moosey wrote:
November 1st, 2019, 5:35 pm
The binary Veblen phi function.
gameoflifemaniac wrote:
November 1st, 2019, 5:17 pm
Also, what is the largest countable ordinal?
There is no such thing. If you say that a is the largest countable ordinal, then I can point out that a+1 is larger.
sup{a<w_1^CK:a}

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Re: Ordinals in googology

Post by gameoflifemaniac » November 2nd, 2019, 4:13 am

Moosey wrote:The binary Veblen phi function.
Can you explain Gamma(0) in terms of a simpler notation? I'm not as advanced in math as you all.
Me wrote:z0 is an index tower, not a tetration tower.
Yeah, I was too slow to realize it's the same.
Moosey wrote:
Me wrote:Also, what is the largest countable ordinal?
There is no such thing. If you say that a is the largest countable ordinal, then I can point out that a+1 is larger.
Sorry, I thought there is one in this case.
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Re: Ordinals

Post by Moosey » November 2nd, 2019, 7:16 am

this is the 75th post in this thread!
testitemqlstudop wrote:
November 2nd, 2019, 3:11 am
sup{a<w_1^CK:a}
He said countable, not computable (Also a = w_1ck)
gameoflifemaniac wrote:
November 2nd, 2019, 4:13 am
Moosey wrote:The binary Veblen phi function.
Can you explain Gamma(0) in terms of a simpler notation? I'm not as advanced in math as you all.
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)
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