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

\(\uparrow_{,\uparrow}\) corresponds to \(\Omega\).

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

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

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

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

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

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

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

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

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, because we now have the \(\Omega_\alpha\) in the toollist. 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.

CLARIFICATION: If you are wondering, \(\psi_{\Omega_\omega}(0)\) does not exist. PROOF: Let's assume that it does. Then, by definition, \(\psi_{\Omega_\omega}(0)<\Omega_\omega\). Therefore, we can find a n such that \(\Omega_n\le\psi_{\Omega_\omega}(0)<\Omega_{n+1}\). However, \(\Omega_{n+1}\) is in the toolbox because n+1 is, which contradicts the definition! Therefore, \(\psi_{\Omega_\omega}\) and other similar contradictory \(\psi\) functions do not exist.

\(\uparrow_{,\uparrow\uparrow}\) corresponds to \(\Omega_2\)

\(\uparrow_{,\uparrow\uparrow\uparrow}\) corresponds to \(\Omega_3\)

\(\uparrow_{,\uparrow_\uparrow}\) corresponds to \(\Omega_\omega\)

\(\uparrow_{,\uparrow_{,\uparrow}}\) corresponds to \(\Omega_\Omega\)

\(\uparrow_{,\uparrow_{,\uparrow\uparrow}}\) corresponds to \(\Omega_{\Omega_2}\)

\(\uparrow_{,\uparrow_{,\uparrow_{,\uparrow}}}\) corresponds to \(\Omega_{\Omega_\Omega}\)

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

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

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

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

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

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

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

Part IV: Inaccessibles
Now we introduce inaccessibles. Inaccessibles are cardinals \(I\) such that: then they are inaccessible. For example, \(\Omega\) is not inaccessible because it is a successor, and \(\Omega_\omega\) is not because it is equal to \(\text{sup}\{\omega,\Omega,\Omega_2,\Omega_3,\dots\}\). Now, how do we define the aleph fixed point? We include the \(I_\alpha\) operation in our toollist. So, \(\psi_I(0)\) is obviously the primitive limit less than I which is the aleph fixed point. \(\psi_I(1)\) is the second such fixed point, and so on. We continue with \(\psi_{\Omega_{I+1}}(0)=\varepsilon_{I+1}\), \(I_2\), ...
 * they are not the suprenum of a set of fewer than \(I\) cardinals, and
 * they are limit cardinals: they are not of the form \(\Omega_{\alpha+1}\),

\(\uparrow_{,,\uparrow}\) corresponds to \(I\)

\(\uparrow_{,\uparrow,\uparrow}\) corresponds to \(\Omega_{I+1}\)

\(\uparrow_{,,\uparrow\uparrow}\) corresponds to \(I_2\)

\(\uparrow_{,,\uparrow_\uparrow}\) corresponds to \(I_\omega\)

\(\uparrow_{,,\uparrow_{,,\uparrow}}\) corresponds to \(I_I\)

\(\uparrow_{,,\uparrow_{,,\uparrow_{,,\uparrow}}}\) corresponds to \(I_{I_I}\)

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

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

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

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

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

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

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

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

Part V: \(\alpha\)-Inaccessibles
a-inaccessibles are an extension of normal inaccessibles. For example, 1-inaccessibles are inaccessible and are a limit of inaccessibles. Before you say that \(I_\omega\) is 1-inaccessible, remember that \(I_\omega\), even though it doesn't seem like it, is not inaccessible because it is singular. The first 1-inaccessible is \(I(1,0)\). Similarly, 2-inaccessibles are 1-inaccessible and are a limit of 1-inaccessibles, and the smallest 2-inaccessible is \(I(2,0)\), and so on. \(\alpha\)-inaccessibles, when \(\alpha\) is a limit ordinal, is the limit of \(<\alpha\)-inaccessibles.

So how do we go further?

(1,0)-inaccessibles are of the form \(I(1,0,\alpha)\), \((\alpha_1,\alpha_2,\dots,\alpha_n)\)-inaccessibles are of the form \(I(\alpha_1,\alpha_2,\dots,\alpha_n,\alpha)\)... \((\alpha_1,\alpha_2,\dots,\alpha_n)\)-inaccessibles are \(<(\alpha_1,\alpha_2,\dots,\alpha_n)\)-inaccessible, for short.

People with prior non-formal exposure to OCFs might fall into the trap (or similar traps) that \(I(1,0)=\alpha\to I_\alpha\). NO! What is true is \(\psi_{I(1,0)}(0)=\alpha\to I_\alpha\). So how exactly do we define these functions? Same as before, though we change \(I_\alpha\) to \(I(\alpha_1,\alpha_2,\dots,\alpha_n)\).

\(\uparrow_{,,,\uparrow}\) corresponds to \(I(1,0)\)

\(\uparrow_{,\uparrow,,\uparrow}\) corresponds to \(\Omega_{I(1,0)+1}\)

\(\uparrow_{,,\uparrow,\uparrow}\) corresponds to \(I_{I(1,0)+1}\)

\(\uparrow_{,,,\uparrow\uparrow}\) corresponds to \(I(1,1)\)

\(\uparrow_{,,,\uparrow_\uparrow}\) corresponds to \(I(1,\omega)\)

\(\uparrow_{,,,,\uparrow}\) corresponds to \(I(2,0)\)

\(\uparrow_{,,,,,\uparrow}\) corresponds to \(I(3,0)\)

Dimensional Array Notation
Finally! We arrive at the next stage of R function: Dimensional array notation.

Rules: scan from left to right. Stop at the first arrow with no descendants. Ignore non-comma separators currently.

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. If there is a comma before it, 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 \(\uparrow_{\ddots_{\uparrow_{,,\dots,,\uparrow@_1}}⋰}@_2[n]=\uparrow_{\ddots_{\uparrow_{,,\dots,\uparrow_{\ddots_{\uparrow_{,,\dots,,\uparrow@_1}}⋰}@_2[n-1],@_1}}⋰}@_2\).

5. Otherwise there should be a non-comma before it.


 * If the arrow has no arrows directly after it, then we diagonalize the separator before it: \(\uparrow_{,_{something}\uparrow@}[n]=\uparrow_{,_{something}[n]\uparrow@}\) where @ is a string starting with a separator.


 * If the arrow has 1 or more arrows directly after it, then \(\uparrow_{,_{something}\uparrow\uparrow@}[n]=\uparrow_{\underbrace{,_{something}[n],_{something}[n]\dots,_{something}[n]}_n\uparrow,_{something}\uparrow@}\) where @ is any string.

So how exactly do we diagonalize separators?

Well, of course, \(,_{\uparrow@}[n]=\underbrace{,_@,_@\dots,_@}_n\).

Then, we diagonalize them similarly to Rule 4.

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

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

A counterintuitive property of OCFs: you might think that \(\psi_{I(\omega,0)}(0)\) is contradictory. Another mistake. Think about what is \(I(\omega,0)\). It is the first \(\omega\)-inaccessible. However, the \(\omega\)s here don't behave like regular ones! \(\omega\)-inaccessibles are \(<\omega\)-inaccessible, but \(\text{sup}\{I(n,0)|n<\omega\}\) is singular so it is not \(\omega\)-inaccessible. \(\text{sup}\{I(n,0)|n<\omega\}=\psi_{I(\omega,0)}(0)\). To sum it up, the reason why \(\psi_{I(\omega,0)}(0)\) is not contradictory is because \(I(\omega,0)\ne\text{sup}\{I(n,0)|n<\omega\}\).

Pseudo Nested Array Notation
This is the \(4\frac{1}{2}\)th part of R function.

In Feburary 2018, I started analyzing Googleaarex's EUAN. At the time, I had no formal understanding to OCFs past TFB, and I did not know of R function. Googleaarex didn't write much on his page, so as well as analyzing the notation, I was also inventing it. I fell into the common trap that the limit of PNAN was \(\psi(\varepsilon_{M+1})\), which I expressed as \(\psi(\psi_{\varepsilon_{M+1}}(0))\) at that time. The reason is because I didn't analyze the notation carefully. Thus, I'm doing to analyze it very carefully this time.

Let ; stand for \(,_{,\uparrow}\) here.