## FANDOM

10,821 Pages

EDIT: I abandoned this OCF (and it's analysis with standard ordinal notation) for a stronger one, that you can find here.

Sometimes, there are notations that are so strong that norrmal ordinal notation can't handle it. So I will make a new ordinal OCF, Ψa(b), and two functions to generate the uncountable ordinals used in the OCF, ΨLa(b) and L(x).

WIP!

## What's a limit?

For a set of ordinals X, the limit of X is the smallest ordinal that is bigger than all of the elements of X. For example, the limit of the set of non-negative integers is ω, and the limit of the set of countable ordinals is Ω.

## Defining L(x), attempt 1

Now, let's try to make a function L(x) to assign uncountable ordinals to ordinals of the form Ωx. Also, we don't want to have any gaps (if the limit of the function is X, we want all uncountable ordinals < X of the form Ωx to be able to be expressed using our system), but we don't care about fixed points (they can be defined later in the OCF). We can count the values of this function L(x) when x increases, and we start at x = 0.

• L(0) = Ω.

We can assign 1 to the smallest ordinal with greater cardinality than L(0), which is Ω2. And we can assign x+1 to the smallest ordinal with greater cardinality than L(x).

• L(1) = Ω2
• L(2) = Ω3
• L(3) = Ω4
• ...

Now, what's L(ω)? Is it the limit of the set of ordinals L(n), where n is an ordinal in the set of non-negative integers?

If we use the definition of ''L(x) is the limit of the set of ordinals L(n) where n is an ordinal in the set that x is a limit of'', then we get:

• L(x) (x < ω) = Ωx+1
• Else, L(x) = Ωx

Not very strong. Can we do better?

## Defining L(x), attempt 2

Let's call the previous L() function as J(x) instead of L(x). Now, we can redefine the L() function.

Let's start with the rule in J(): L(x+1) is the smallest ordinal with greater cardinality than L(x), which is the same as:

• L(0) = Ω
• L(x+1) = the limit of a set that is equal to L(0), but with all ω's in each term replaced with L(x)'s

Using this definition, we get:

• L(0) = Ω
• L(1) = Ω2
• L(2) = Ω3
• ..

Now, what is L(ω)? We can't define it as Ωω, because that would make L(x) equal to J(x), but we can't something greater than Ωω either, because if we did that, we would be leaving out some ordinals of the form Ωx either. However...

...we can define Ωω with another function.

Let's call this function ΨLL(ω)(x), because it makes L(ω) stronger. To calculate any function ΨLL(x)(y) (where x is a limit ordinal), first look at L(x[n]) for any positive integer n, and find the value that changing L(x[n]) to L(x[n+1]) adds 1 to it. This value is called the key value. Then, ΨLL(x)(y) would be the value of key value that L(x[n]) would have if that value was changed to y. Finally, define L(x) (if x is a limit ordinal) to be the smallest ordinal that cannot be made using 1, addition, and ΨLL(x)().

However, we are still leaving some ordinals out, like the omega fixed point. But the ordinals we are leaving out are fixed points, which we can define later in the OCF.

Using this new L(x), we get:

• L(ω) = I
• L(ω+1) = ΩI+1
• L(ω+2) = ΩI+2
• L(ω2) = I2
• L(ω3) = I3

But at L(ω2), we run into a problem. What's ΩIω+1 in this system?

To solve this, we need to allow some ΨLa(b) to behave like L(x): if the a in ΨLa(b) is of the form ΨLc(d), find a function F(x) such that ΨLc(d) is the smallest ordinal that cannot be made using 1, addition, and F(). Then, F(b) = ΨLa(b).

We can continue calculating the values of L(x) now:

• L(ω2) = I(1,0)
• L(ω22) = I(1,1)
• L(ω3) = I(2,0)
• L(ω4) = I(3,0)
• L(ωω) = I(1,0,0)
• L(ωω2) = I(2,0,0)
• L(ωω2) = I(1,0,0,0)
• ...

Now, what is L(L(0))? Well, it's just the limit of the set L(x) for all x that are elements of the set L(0). This gives us L(L(0)) = M.

If @ is the rest of the ordinal inside of L(), and if x doesn't have a fundamental sequence, then L(@x) is the limit of the set L(@y) for all y that are elements of the set x.

This way, we get an extremely strong ordinal function. You will see some examples of the strength of this L(x) later.

## Defining L(x), attempt 3

L(x) is really strong, but we can do better. Define K(x) as our second L(x) (the first one was J(x)) and ΨKa(b) as our previous ΨLa(b), so we can define a third L(x).

First, define L(x) in the same way as we defined J(x), up to L(L(0)). Then, we can do something similar to what we did when defining J(x)

If x doesn't have a fundamental sequence, to get ΨLL(x)(y), first order the elements of the set that x is a limit of by increasing size, assign each member of that set after it's ordered (the first member of the set is assigned 1, the second one is assigned 2, the third one is assigned 3, and so on), and define a function F(n) that when an ordinal of that set is the input, the output is the ordinal that was assigned to the input. Next, find the key value. This time, it's a value in the value of F(n) for all ordinals n smaller than x and greaterequal (greater than or equal to) to ω that is equal to n. Then, the value of ΨLL(x)(y) is the value that F(n) (where n is an ordinal greater than ω) would have if the key value was replaced with y.

If you thought K(x) was strong... it's only ΨLL(L(0))(x).

## Defining Ψ_a(b)

Finally, we need to define the OCF. These are the rules:

• @ is the rest of the ordinal inside of Ψ() (the ends of parentheses don't count as part of the ordinal), and \$ is a copy of @ (if @ is empty, then the last \$ is 0).
• X is 0 or an ordinal that doesn't have a fundamental sequence and that is the smallest ordinal with cardinality Y.
• If X doesn't have a fundamental sequence and it is the smallest ordinal with cardinality Y, then X+ is the smallest ordinal with the sucessor cardinal of Y as it's cardinality. If X is 0, then X+ = J(0) = Ω.
• Ψ(0) = ε0
• Ψ(x+1) = sup(Ψ(x),Ψ(x)Ψ(x),Ψ(x)Ψ(x)Ψ(x),...)
• For limit ordinal x, Ψ(@x) = sup(Ψ(@x[1]),Ψ(@x[2]),Ψ(@x[3]),...)
• Ψ0(x) = Ψ(x)
• Ψ(@ΨX(0)) = sup(Ψ(@X),Ψ(@XX),Ψ(@XXX),...)
• Ψ(@ΨX(x+1)) = sup(Ψ(@ΨX(x)),Ψ(@ΨX(x)ΨX(x)),Ψ(@ΨX(x)ΨX(x)ΨX(x)),...)
• Ψ(@X+) = sup(Ψ(@),Ψ(@ΨX(\$)),Ψ(@ΨX(\$ΨX(\$))),...)
• If X is the smallest ordinal that can't be made using 1, addition and a function F(), Ψ(@X) = sup(Ψ(@),Ψ(@F(\$)),Ψ(@F(\$F(\$))),...)

We can plug in the ordinals that we already defined, but when we reach K(J(0)), we run into a problem. What is the function F(x) such that K(J(0)) is the smallest ordinal that can't be made using F() or functions that are eventually overgrowed by F()?

Here is a solution for ordinals that lack this F(x) (I will call the ordinal that lacks the F(x) as X):

• First, assign each member of the set that X is a limit of an ordinal (the first element is assigned 1, the second element is assigned 2, the third element is assigned 3, etc).
• Make a function G(x) such that when an x < X is the input, the output is the element of the set that X is a limit of that x was assigned to.
• If G(x) appears once, to represent it, it's the same as x, but change all ω's into X's.
• If G(x) appears more than once, to represent the first one, it's the same as x, but change all ω's into X's, and to represent the other ones, it's the same as the first one, but inside of FX() (where F is ΨK if X < L(L(0)) and ΨL otherwise)

Now, how strong is this OCF?

## Some more analysis

You can find the analysis of this OCF with standard ordinal notation here. The short version is this:

• J(x) = Ωx
• K(ω) = I
• K(ω+1) = ΩI+1
• K(ω2) = I2
• K(ω2) = I(1,0)
• K(ω3) = I(2,0)
• K(ωω) = I(1,0,0)
• K(J(0)) = M
• K(J(0)2) = M2
• K(J(0)2) = M(1,0)
• K(J(1)) = Ξ(2,0)
• K(J(2)) = Ξ(3,0)
• K(J(J(0))) = Ξ(K,0)
• K(J(J(1))) = K
• K(J(J(1)2)) = Ξ2(0,0)
• K(J(J(2))) = ΞU(0,0)
• K(J(J(ω)2)) = S(TTT)
• K(J(J(J(0)))) = T
• K(J(J(J(0)+3))) = 2-superstage cardinal
• K(J(J(J(0)2))) = 1-duperstage cardinal
• K(J(J(J(0)3))) = 1-truperstage cardinal
• K(J(J(J(0)ω))) = A stage cardinal over x-perstage cardinals