Veblen to OCF's

Simply Beautiful Art

May 2, 2017 23:55

I'll just explain real quick again:
ε1 = {ε0, ε0^ε0, ε0^ε0^ε0, ...}
ε2 = {ε1, ε1^ε1, ε1^ε1^ε1, ...}
...
εω = {ε1, ε2, ε3, ...}
ε(ω+1) = {εω, εω^εω, εω^εω^εω, ...}
---

φ(1,x) = ε(x)

Veblen function time

Alexander Day

YAY!!

i get dis

for now

Infinite Monkey

On a side note, mirror cubes are solved with the same strategy as the 6 color 3x3 cube but it is way harder. If you really want to get crazy there are also "ghosts cubes"

Simply Beautiful Art

φ(2,0) = {ε0, ε(ε0), ε(ε(ε0)), ...}

Infinite Monkey

Alexander Day

but not dat one!!

@InfiniteMonkey i can solve a mirror cube blindfolded...

@SimplyBeautifulArt k i get it

Simply Beautiful Art

May 2, 2017 23:58

φ(2,1) = {ε(x), ε(ε(x)), ε(ε(ε(x))), ...} where x=φ(2,0)+1
φ(2,2) = {ε(y), ε(ε(y)), ε(ε(ε(y))), ...} where y=φ(2,1)+1

Alexander Day

just add 1?

Simply Beautiful Art

φ(2,ω) = {φ(2,1), φ(2,2), φ(2,3), ...}

Alexander Day

oh wait...

Simply Beautiful Art

Add one and then do lots of ε's

Alexander Day

i get the 2nd line now

eh

???

Simply Beautiful Art

May 3, 2017 00:00

φ(3,0) = {φ(2,0), φ(2,φ(2,0)), φ(2,φ(2,φ(2,0))), ...}

Just like how φ(2,0) was a lot of ε's, we have φ(3,0) as a lot of φ(2,x)'s

Alexander Day

and then φ(w,0)

Simply Beautiful Art

Yup

Alexander Day

and then to the skies!!

Simply Beautiful Art

φ(ω,0) = {φ(1,0), φ(2,0), φ(3,0), ...}

Alexander Day

but then φ(e0, 0)=φ(0, 0, 0)

i think

or is it lgr

Simply Beautiful Art

May 3, 2017 00:02

φ(ω,1) = {φ(1,x), φ(2,x), φ(3,x), ...} where x=φ(ω,0)+1

@AlexanderDay and nope to that

Alexander Day

k

:-<

Simply Beautiful Art

φ(ω+1,0) = {φ(ω,0), φ(ω,φ(ω,0)), φ(ω,φ(ω,φ(ω,0))), ...}

etc.

Alexander Day

then when do we get to φ(0,0,0)?????????????

Simply Beautiful Art

φ(ε0,0) = {φ(ω,0), φ(ω^ω,0), φ(ω^ω^ω,0), ....}

Alexander Day

k

Simply Beautiful Art

May 3, 2017 00:04

φ(1,0,0) = {φ(1,0), φ(φ(1,0),0), φ(φ(φ(1,0),0),0), ...}

Notice that φ(φ(1,0),0) = φ(ε0,0)

Alexander Day

oh...

Simply Beautiful Art

Very very very big

Alexander Day

mind blown

the supermum of that...

BOOOOOOOOOOOOOOOOOMMMMMMMMMMM!!!!!!!!!!!

Simply Beautiful Art

φ(1,0,1) = {φ(x,0), φ(φ(x,0),0), φ(φ(φ(x,0),0),0), ...} where x=φ(1,0,0)+1

φ(1,0,2) = {φ(x,0), φ(φ(x,0),0), φ(φ(φ(x,0),0),0), ...} where x=φ(1,0,1)+1

Alexander Day

k

Simply Beautiful Art

May 3, 2017 00:07

φ(1,1,0) = {φ(1,0,0), φ(1,0,φ(1,0,0)), φ(1,0,φ(1,0,φ(1,0,0))), ...}

Alexander Day

lemme guess...

Simply Beautiful Art

φ(1,1,1) = ?

Alexander Day

@SimplyBeautifulArt repeat these parts...

Simply Beautiful Art

@AlexanderDay Yup!

Alexander Day

@SimplyBeautifulArt with this part...

Simply Beautiful Art

May 3, 2017 00:08

φ(1,1,1) = {φ(x,0), φ(φ(x,0),0), φ(φ(φ(x,0),0),0), ...} where x=φ(1,1,0)+1

Alexander Day

YAY!!

Simply Beautiful Art

φ(2,0,0) = {φ(1,0,0), φ(1,φ(1,0,0),0), φ(1,φ(1,φ(1,0,0),0),0), ....}

Alexander Day

i was typing that...

Simply Beautiful Art

φ(1,0,0,0) = {φ(1,0,0), φ(φ(1,0,0),0,0), φ(φ(φ(1,0,0),0,0),0,0), ...}

Mine!

φ(1,0,0,1) = {φ(x,0,0), φ(φ(x,0,0),0,0), φ(φ(φ(x,0,0),0,0),0,0), ...} where x=φ(1,0,0,0)+1

And it just keeps going like this

Alexander Day

and just keep going...

Simply Beautiful Art

May 3, 2017 00:11

HAHA

Alexander Day

BTW, could you tell me what googolpelxainth is close to in Veblan terms??

Simply Beautiful Art

Small Veblen Ordinal = {φ(1,0), φ(1,0,0), φ(1,0,0,0), ....}

Alexander Day

it is 10^{10^{10^100}}

Simply Beautiful Art

googolplexianth < f_ω²(3)

Nowhere close to Veblen

Alexander Day

:-<

Simply Beautiful Art

May 3, 2017 00:13

We then have things beyond all of Veblen ordinals

Mainly, we have the ordinal collapsing function (OCF)

and I have to wash dishes

Alexander Day

K

what about Meameamealokkapoowa oompa the number i mentioned a few days ago to get you concerned?

Alexander Day

May 3, 2017 00:35

U back yet?

Simply Beautiful Art

Back

@AlexanderDay you beat it with super extended ordinal collapsing functions

Alexander Day

going...

going....

...AND IT"S OUTTA HERE!!

Simply Beautiful Art

lol

Alexander Day

it's way outtta Veblan function!!

Simply Beautiful Art

yup

Alexander Day

May 3, 2017 00:43

how big is it?

Simply Beautiful Art

Very big

Alexander Day

i heard the inventor of Bower's operators came up with it.

Simply Beautiful Art

That is false

Alexander Day

what's next?

wait---WHAT!!

it was in my book 1001 mathemathics!!

who did??????????????

Simply Beautiful Art

C(α,0) = {0,1,ω,Ω}
C(α,n+1) = C(α,n) U {δ+Γ, δΓ, δ^Γ, ψ(µ) | δ,Γ,µ in C(α,n) and µ<α}
ψ(α) = {δn | δn in C(α,n) and δn<Ω, n in N}
Ω > ψ(α)

Alexander Day

May 3, 2017 00:47

?????

Simply Beautiful Art

This is the scary looking ordinal collapsing function

Alexander Day

i see gamma, delta,and in. i never got in.

Simply Beautiful Art

don't worry, we'll go through it step by step

Alexander Day

could u give me a crash course in the in??

??

?

Simply Beautiful Art

First of all, there is no such ψ(-1). We never have negative numbers

Alexander Day

May 3, 2017 00:49

???

Simply Beautiful Art

ψ(0) is the first number

Alexander Day

k

Simply Beautiful Art

C(0,0) = {0,1,ω,Ω}

To get to C(0,1), just add, multiply, and exponentiate all of this

Alexander Day

???

Simply Beautiful Art

C(0,1) = {0, 1, 2, ω, ω+1, ω2, ω^2, ω^ω, Ω, Ω+1, Ω+ω, Ω2, Ωω, ...}

Alexander Day

May 3, 2017 00:51

??

still confused.

talk about scary-looking

Simply Beautiful Art

2 = 1+1
ω2 = ω+ω
ω^2 = ω*ω

All I did was add, multiply, exponentiate. We started with {0,1,ω,Ω} and now we have this

Alexander Day

BUT that is in Veblan function??

Simply Beautiful Art

It looks like Veblen function... but it will stop looking like Veblen function later

Alexander Day

but is it momentarily?

Simply Beautiful Art

Yes

max(C(0,1) and less than Ω) = ω^ω

Alexander Day

May 3, 2017 00:53

k

Simply Beautiful Art

You agree with this so far?

Alexander Day

very big

all i have to say

Simply Beautiful Art

Mhm

C(0,2) = add, multiply, exponentiate all the previous stuff

C(0,2) = {0, 1, 2, 3, 4, ω, ω+1, ω+2, ω+3, ω2, ω2+1, ω2+2, ω3, ω4, ω^2, ω^2+1, ω^2+2, ω^2+ω, ω^2+ω2, ω^ω, ...., Ω, Ω+1, Ω+2, ....}

Alexander Day

like this:{0, 1, 2, 3, ω, ω+1, w+2, ω2, w3ω^2, w^3ω^ω, w^w^w, Ω, Ω+1, Ω+2, Ω+ω, Ω2, Ωω, ...}

Simply Beautiful Art

Yeah, pretty much what I got, you get the general idea

Alexander Day

May 3, 2017 00:56

beat me to it. AURRRRUUUUUUUUUUGGGGGGGHHHHHHHHHHH!!!!

Simply Beautiful Art

max(C(0,2) and less than Ω) = ω^ω^ω

etc.

ψ(0) = {ω^ω, ω^ω^ω, ω^ω^ω^ω, ....}

= ε0

C(1,0) = {0,1,ω,Ω}

Alexander Day

the same!!

or is it??

Simply Beautiful Art

The thing different about C(1,1) is that not only do we add, multiply and exponentiate, but we also do ψ(µ) for µ in {0,1,ω,Ω} and µ<1

Alexander Day

& is still veblan....

Simply Beautiful Art

So C(1,1) is just like C(0,1) except we also have ψ(0) thrown into the mix, since 0 is in {0,1,ω,Ω} and 0<1

Alexander Day

May 3, 2017 00:59

???

Simply Beautiful Art

C(α,0) = {0,1,ω,Ω}

C(α,n+1) = C(α,n) U {δ+Γ, δΓ, δ^Γ, ψ(µ) | δ,Γ,µ in C(α,n) and µ<α}

ψ(α) = {δn | δn in C(α,n) and δn<Ω, n in N}
Ω > ψ(α)

Alexander Day

?????

Simply Beautiful Art

C(1,1) = {0, 1, 2, ω, ω+1, ω2, ω^2, ω^ω, ψ(0), Ω, Ω+1, Ω+ω, Ω2, Ωω, ...}

Alexander Day

head explodes

oh...

Simply Beautiful Art

So it turns out that...
ψ(1) = {ψ(0), ψ(0)^ψ(0), ψ(0)^ψ(0)^ψ(0), ....}
= ε1

Alexander Day

May 3, 2017 01:03

so then C(2,0) is where ψ(2) is the t/z thingie??

Simply Beautiful Art

No

Alexander Day

or e(e0)

Simply Beautiful Art

ψ(2) = ε2

Much smaller

ψ(3) = ε3

Alexander Day

:(

Simply Beautiful Art

etc.

Infinite Monkey

May 3, 2017 01:04

@SimplyBeautifulArt so that is where you where going!

Simply Beautiful Art

@InfiniteMonkey Just you wait

And then... we have something weird

Alexander Day

@InfiniteMonkey hi

Simply Beautiful Art

ψ(φ(2,0)) = ε(φ(2,0)) = φ(2,0)

Since φ(2,0) = ε(ε(ε(...))) is the t/z thingie

I'll just call it z0

Alexander Day

@SimplyBeautifulArt oh...

Simply Beautiful Art

ψ(z0) = z0

Alexander Day

May 3, 2017 01:05

k

Simply Beautiful Art

Now, we have magic!

Alexander Day

alacadabra, alacazam, make this easier, no we cant!

Simply Beautiful Art

ψ(z0+1) = z0
ψ(z0+2) = z0
ψ(z0*2) = z0
...
ψ(Ω) = z0

Alexander Day

:(

Simply Beautiful Art

Alaca...what?!

xD

Remember, Ω is very big. It is bigger than ψ(x) for any x.

Alexander Day

May 3, 2017 01:07

alacadabra!

um...

Infinite Monkey

@SimplyBeautifulArt looks like a weird "wall"

Alexander Day

be back soon!

i gotta go.

Simply Beautiful Art

Indeed, it is a wall

But this is what makes the ordinal collapsing function... powerful?

@AlexanderDay Bye!

Alexander Day

like tomorrow....

sorry...

Simply Beautiful Art

notice that no matter what you do, you only start with {0,1,ω,Ω}

@AlexanderDay No problem!

From there, you can only add, multiply, exponentiate, and take ψ(µ)

The problem is that we can't take ψ(z0), since to do this, we need to somehow reach z0 first.

Infinite Monkey

May 3, 2017 01:10

I see...

Simply Beautiful Art

But you can't reach z0 using addition, multiplication, exponentiation, and ψ(µ) for µ values we've reached. If we reached µ=z0, we wouldn't be needing to take ψ(z0) to begin with...

So we're stuck. But not forever.

C(Ω+1,0) = {0,1,ω,Ω}

Infinite Monkey

So ordinal collapsing functions let us "break through" the z0 wall

Simply Beautiful Art

C(Ω+1,1) = {0, 1, 2, ...., ω^ω, ψ(0), ψ(1), ψ(ω), ψ(Ω), Ω, Ω+1, ...}

Notice the last of these ψ's

We can break through!

Until we hit ψ(Ω+z1) = z1

But we can break through! ψ(Ω+Ω) = ψ(Ω2) = z1

and then we get stuck, and then we break through

eventually ψ(Ωx) = x, so we get stuck.
and then we break through! ψ(Ω²) = x

You get the idea

ψ(Ω^Ω^ω) is the small Veblen ordinal.

Infinite Monkey

Yes

Simply Beautiful Art

ψ(Ω^Ω^Ω) is the large Veblen ordinal

ψ(Ω^Ω^Ω^Ω^...) is the Bachmann-Howard ordinal

and then we get stuck

Infinite Monkey

May 3, 2017 01:15

but we can break through?

is there anything special to be said about the points where we get stuck or they're not interesting?

Simply Beautiful Art

Why? Well note that you can never exceed an infinite tower of Ω's due to the restriction of addition, multiplication, and exponentiation, and since Ω>ψ(x), we can't make more Ω's that way

@InfiniteMonkey Yes, at the very low levels, we get some fairly obvious Veblen equivalents

But there is a way to go beyond....

Say hello to ψ1(0) = Ω^Ω^Ω^Ω^...

Just like ψ(0) = ω^ω^ω^ω^...

Infinite Monkey

And the game starts allover again!

Simply Beautiful Art

Mhm

To break through ψ1(x) = x, we use Ω_2

The second Ω

Again, we get stuck at infinitely many Ω_2's powered to each other, so we use ψ2(0)

then ψ3(0)

etc.

then ψω(0)

We can even reach ψΩ(0)

^^^ Just imagine that!

We can even have Ω_Ω_Ω

and the normal extended OCF ends at infinitely many Ω's tied to each other

Which is where we bring in ψI(0)

Infinite Monkey

That gets quite scary 0.0

Simply Beautiful Art

the inaccessible cardinals...

Oh yeah. It goes pretty darn far

I's will get stuck, so use I+1, then I+2, etc. even Ω_(I+1), etc. eventually I(1), the second inaccessible

You can even go up to things like I(Ω), or I(I(I(...)))

I(1,0) is beyond all this...

$Mathematics$ This is the Realm of Simply Beautiful…

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