User:12AbBa/"Normal" OCFs vs a different version of R function

In this article, we will be rigorously but informally (so that everyone can understand) defining OCFs, and comparing them to R Function. The real thing is too cumbersome for me to understand, so I rewrite the expressions in arrows. It turns out that Extended Up-Arrow Notation is curiously similar to R function, so I can do that.

Part I: Below \(\varepsilon_0\)
In this part we have no OCFs. All notations are evaluated starting from the right, but arrows start from the left.

Basic Notation
\(\uparrow_0\) is short for \(\uparrow\).

Rules:

1. Base Case 1: \(a\uparrow b=a^b\)

2. Base Case 2: \(a@1=a\)

3. Recursive case 1: \(a\uparrow@b=a@a\uparrow@(b-1)\)

4. Recursive case 2: \(a\uparrow_n@b=a\underbrace{\uparrow_{n-1}\uparrow_{n-1}\dots\uparrow_{n-1}}_b@a\)

Now, let's evaluate \(2LIMIT2\). I KNOW it is =4, but lets just do this.

\(2\uparrow_{\uparrow_\uparrow}2=2\uparrow_22=2\uparrow_1\uparrow_12=2\uparrow\uparrow\uparrow_12=2\uparrow_12=2\uparrow\uparrow2=4\)

Nested Notation
We change \(\uparrow_n\) into \(\uparrow_{\underbrace{\uparrow\uparrow\dots\uparrow}_n}\).

The b in a@b is the diagonalizer.

Rules: scan from left to right. Stop at the first arrow with no descendants. a@1=a.

1. If that arrow is the only one then \(a\uparrow b=a^b\)

2. Otherwise, if it is the first one then \(a\uparrow@b=a@a\uparrow@(b-1)\)

3. Otherwise, look to the upper left. \(\uparrow_{\uparrow@}[n]=\underbrace{\uparrow_@\uparrow_@\dots\uparrow_@}_n\)

\(2\uparrow_{,\uparrow}2=2\uparrow_{\uparrow_\uparrow}2=4\)

Part II: Bachmann's \(\psi\)
Start with a set \(\{0,1,\omega,\Omega\}\) where \(\Omega=\aleph_1\) and apply these operations: +, ^, and \(\psi\) if it is already defined. \(\psi(n)\) is defined by the smallest countable ordinal that "covers" the final set.

I will be expressing absolutely all ordinals in \(\omega\)s and OCFs.

Linear Array Notation
Now to extend the notation, we introduce the comma.

Rules: scan from left to right. Stop at the first arrow with no descendants.

1. If that arrow is the only one then \(a\uparrow b=a^b\)

2. If it is the first one then \(a\uparrow@b=a@a\uparrow@(b-1)\)

3. If it is the first thing in its level, look to the upper left. \(\uparrow_{\uparrow@}[n]=\underbrace{\uparrow_@\uparrow_@\dots\uparrow_@}_n\)

4. Otherwise, you should have some thing like this: \(\uparrow_{,,\dots,,\uparrow@}\). Look to the upper left until you find a level that has rank less than or equal to \(\uparrow_{,,\dots,,@}\). Now, we diagonalize inside that level. If there is no such level, we diagonalize the whole thing. Now, we have \(\uparrow_{\ddots_{\uparrow_{,,\dots,,\uparrow@_1}}⋰}@_2[n]=\uparrow_{\ddots_{\uparrow_{,,\dots,\uparrow_{\ddots_{\uparrow_{,,\dots,,\uparrow@_1}}⋰}@_2[n-1],@_1}}⋰}@_2\). WHOA!

Subrule1: \(\ddots_{\uparrow_{,\dots,,\uparrow@}}⋰[1]=\ddots_{\uparrow_{,\dots,\uparrow,@}}⋰\)

Subrule2: \(\uparrow_{@,}=\uparrow_@\)

\(2\uparrow_{,\uparrow\uparrow}2\)

\(=2\uparrow_{\uparrow_{\uparrow,\uparrow},\uparrow}2\)

\(=2\uparrow_{\uparrow_{,\uparrow}\uparrow_{,\uparrow},\uparrow}2\)

\(=2\uparrow_{\uparrow_{\uparrow_{\uparrow_\uparrow\uparrow_{,\uparrow},\uparrow}}\uparrow_{,\uparrow},\uparrow}2\)

\(=2\uparrow_{\uparrow_{\uparrow_{\uparrow\uparrow\uparrow_{,\uparrow},\uparrow}}\uparrow_{,\uparrow},\uparrow}2\)

\(=2\uparrow_{\uparrow_{\uparrow_{\uparrow\uparrow_{,\uparrow},\uparrow}\uparrow_{\uparrow\uparrow_{,\uparrow},\uparrow}}\uparrow_{,\uparrow},\uparrow}2\)

\(=2\uparrow_{\uparrow_{\uparrow_{\uparrow_{,\uparrow},\uparrow}\uparrow_{\uparrow_{,\uparrow},\uparrow}\uparrow_{\uparrow\uparrow_{,\uparrow},\uparrow}}\uparrow_{,\uparrow},\uparrow}2\)

\(=2\uparrow_{\uparrow_{\uparrow_{\uparrow_{\uparrow_{\uparrow_\uparrow,\uparrow}\uparrow_{\uparrow_{,\uparrow},\uparrow}\uparrow_{\uparrow\uparrow_{,\uparrow},\uparrow}},\uparrow}\uparrow_{\uparrow_{,\uparrow},\uparrow}\uparrow_{\uparrow\uparrow_{,\uparrow},\uparrow}}\uparrow_{,\uparrow},\uparrow}2\dots\)

This is NEVER going to end, even though we can prove that the final answer is 4!

Why is \(\psi(\varepsilon_{\Omega+1})\) the limit? Well, notice that we only add countable ordinals to our set, so after this point the OCF gets stuck forever, even in an infinite sense.

Part III: More cardinals
In order to make the OCF unstuck, we introduce more cardinals and more \(\psi\) functions. \(\Omega_n=\aleph_n\), by the way, so \(\Omega_0=\omega\) and \(\Omega_1=\Omega\). We will be going as far as the aleph fixed point, which is \(\Omega_{\Omega_\ddots}=\aleph_{\aleph_\ddots}\). We can remove \(\omega\) and \(\Omega\) from our toolbox. Now, we add more \(\psi\) functions. Define \(\psi_\alpha(\beta)\) is the smallest cardinal such that the set of ordinals less than \(\psi_\alpha(\beta)\) contains all of the elements of the union of the toolbox at stage \(\beta\) and the set of ordinals less than \(\alpha\). The function gets unstuck because \(\psi_{\Omega_2}(0)=\varepsilon_{\Omega+1}\) is in the toolbox.