## Ordinals in googology

- testitemqlstudop
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### Re: Ordinals in googology

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.

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.

### Re: Ordinals in googology

1) There is no set of all ordinals, just like there is no set of all sets.testitemqlstudop wrote: ↑October 25th, 2019, 7:34 amWhat is the cardinality of the set of:

1) all ordinals;

2) all countable ordinals;

3) all computable ordinals.

2) ℵ_1

3) ℵ_0

- Moosey
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### Re: Ordinals in googology

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

I think that at n = 1 it's already larger than g64, but I'm not sure

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?"

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?"

- Moosey
**Posts:**2788**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

Rule 8 is a draft; do not mention any issues with it until I work it out.testitemqlstudop wrote: ↑October 24th, 2019, 8:45 pmThat does not work with your recursive definition at all.8a. ah_n((0@a)#{b,$_6}) = ah_n((0@(a[n])#{b,$_6}), a a lim ord

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:

EDIT: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

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}
```

(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)

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?"

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?"

- testitemqlstudop
**Posts:**1241**Joined:**July 21st, 2016, 11:45 am**Location:**in catagolue-
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### Re: Ordinals in googology

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.Macbi wrote: ↑October 25th, 2019, 7:45 am3) ℵ_0testitemqlstudop wrote: ↑October 25th, 2019, 7:34 amWhat is the cardinality of the set of:

1) all ordinals;

2) all countable ordinals;

3) all computable ordinals.

- Moosey
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### Re: Ordinals in googology

It makes perfect sense: w_1ck is the size of the set of computable ordinals, and |w_1ck| = aleph0testitemqlstudop wrote: ↑October 25th, 2019, 11:16 pmThat 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.Macbi wrote: ↑October 25th, 2019, 7:45 am3) ℵ_0testitemqlstudop wrote: ↑October 25th, 2019, 7:34 amWhat is the cardinality of the set of:

1) all ordinals;

2) all countable ordinals;

3) all computable ordinals.

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?"

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?"

### Re: Ordinals in googology

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 ℕtestitemqlstudop wrote: ↑October 25th, 2019, 11:16 pmThat 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.Macbi wrote: ↑October 25th, 2019, 7:45 am3) ℵ_0October 25th, 2019, 7:34 amWhat is the cardinality of the set of:

1) all ordinals;

2) all countable ordinals;

3) all computable ordinals.

*for which all but finitely many values are 0*. So ω^ω is much smaller than ℵ_0^ℵ_0.

- Moosey
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### Re: Ordinals in googology

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*}
```

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
```

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(@^_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)

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?"

- Moosey
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### Re: Ordinals in googology

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.

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.

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?"

- Moosey
**Posts:**2788**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

Suppose you defined an OCF thusly:
Does psi_n add the ordinals psi_m gets stuck at to the initial set for all m<n?

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)}
```

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?"

- testitemqlstudop
**Posts:**1241**Joined:**July 21st, 2016, 11:45 am**Location:**in catagolue-
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### Re: Ordinals in googology

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.)

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.)

- Moosey
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### Re: Ordinals in googology

That's not a traditional OCF-- OCFs are defined differently, like this:testitemqlstudop wrote: ↑October 31st, 2019, 2:27 amI 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

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)
```

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?"

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?"

- testitemqlstudop
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### Re: Ordinals in googology

Isnt T0 the result of the diagonalization of e0, z0, n0, etc?Moosey wrote: ↑October 31st, 2019, 6:50 amThat's not a traditional OCF-- OCFs are defined differently, like this:testitemqlstudop wrote: ↑October 31st, 2019, 2:27 amI 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.htmlYou also have a mistake in the original; psi_w(0) = (wait for it) phi_w(0), not gamma_0Code: 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)`

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?"

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

- Moosey
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### Re: Ordinals in googology

T0? Do you mean gamma_0?testitemqlstudop wrote: ↑October 31st, 2019, 9:13 amIsnt 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

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

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.testitemqlstudop wrote:I remember you said somewhere T0 = w^^w, obviously that's a large underestimation

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?"

- toroidalet
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### Re: Ordinals in googology

This is sort of a stupid question, but why isn't ε_1=ε_0^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

-Terry Pratchett

- testitemqlstudop
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### Re: Ordinals in googology

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.

- BlinkerSpawn
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### Re: Ordinals in googology

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

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.

- testitemqlstudop
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### Re: Ordinals in googology

Thank you!Moosey wrote: ↑October 31st, 2019, 4:17 pmT0? Do you mean gamma_0?testitemqlstudop wrote: ↑October 31st, 2019, 9:13 amIsnt 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

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_0No, 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.testitemqlstudop wrote:I remember you said somewhere T0 = w^^w, obviously that's a large underestimation

I have fixed it in

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

- Moosey
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### Re: Ordinals in googology

That's not quite right. e_0^w is not w^e_0, since exponentiation is not commutative.testitemqlstudop wrote: ↑October 31st, 2019, 8:44 pme0 is the first solution to a = w^a, or a transfinite power stack of w's. Adding another w will not change the number.

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

My CA rules can be found here

Also, the tree game

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- gameoflifemaniac
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### Re: Ordinals in googology

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

Also, what is the largest countable ordinal?

Also, what is the largest countable ordinal?

z0 is an index tower, not a tetration tower. And how do you define z1, z2, z3 etc.?Moosey wrote:e0 = w^^^2, z0 = w^^^^2

Last edited by gameoflifemaniac on November 1st, 2019, 5:35 pm, edited 1 time in total.

https://www.youtube.com/watch?v=q6EoRBvdVPQ

One big dirty Oro. Yeeeeeeeeee...

One big dirty Oro. Yeeeeeeeeee...

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### Re: Ordinals in googology

The binary Veblen phi function.gameoflifemaniac wrote: ↑November 1st, 2019, 5:17 pmGamma(0) is the fixed point of the phi function. What is phi function

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.

prove itgameoflifemaniac wrote: ↑November 1st, 2019, 5:17 pmz0 is an index tower, not a tetration tower.Moosey wrote:e0 = w^^^2, z0 = w^^^^2

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

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

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?"

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### Re: Ordinals in googology

Define n%f to be the nth fixed point of the function f.
Does this function work? Or does it become ill-defined?

(This is a more formal version of this:)

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)
```

(This is a more formal version of this:)

Anyways, with the new psi, not the one in quotes, what is, say, psi_5(W)? Does psi(W) exist?Moosey wrote: ↑October 29th, 2019, 7:08 pmSuppose you defined an OCF thusly:Does psi_n add the ordinals psi_m gets stuck at to the initial set for all m<n?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)}`

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?"

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### Re: Ordinals in googology

sup{a<w_1^CK:a}Moosey wrote: ↑November 1st, 2019, 5:35 pmThe binary Veblen phi function.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
**Posts:**843**Joined:**January 22nd, 2017, 11:17 am**Location:**There too

### Re: Ordinals in googology

Can you explain Gamma(0) in terms of a simpler notation? I'm not as advanced in math as you all.Moosey wrote:The binary Veblen phi function.

Yeah, I was too slow to realize it's the same.Me wrote:z0 is an index tower, not a tetration tower.

Sorry, I thought there is one in this case.Moosey wrote: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.Me wrote:Also, what is the largest countable ordinal?

https://www.youtube.com/watch?v=q6EoRBvdVPQ

One big dirty Oro. Yeeeeeeeeee...

One big dirty Oro. Yeeeeeeeeee...

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### Re: Ordinals

this is the

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)

**75th**post in this thread!He said countable, not computable (Also a = w_1ck)

The Veblen function is actually pretty simple--gameoflifemaniac wrote: ↑November 2nd, 2019, 4:13 amCan you explain Gamma(0) in terms of a simpler notation? I'm not as advanced in math as you all.Moosey wrote:The binary Veblen phi function.

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)

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?"