# T Sequences - Explicit Enumeration of $\epsilon_0$

#### AplanisTophet

This is fun. Let's figure out where what seems obvious to me is not obvious to others. You must return the favor of doing your best to explain why you don't (or do) understand what I'm saying, or this is obviously pointless. At the end of the day this is just an enumeration of $\omega^2$, so I can't emphasize enough that it isn't too complicated. We can do this. With all the crazy cranks that visit this forum who think what they are saying is obviously correct, and all the locals who repeatedly bash them down, I'd say this has got to be a worthwhile experiment if nothing else. You have someone who is crankish enough to repeat the experiment in me but also cognizant enough to happily admit failure and mistake on my own part (because I don't care if I'm right or wrong, quite frankly), so learn something by all means!

Rules: Where $a, b, c, \dots$ are ordinals, the following rules are used to define function $f$.

$$\text{Rule 1 : }a, a-1, a-2, \dots \implies \{ a-3 \} \text{, where } a \geq 3$$
$$\text{Rule 2 : }0, 1, a, \dots \implies \{a+1\}$$
$$\text{Rule 3 : }a, a+1, a+2, \dots \implies \{ a+3, a + \omega \}$$

Define function $f$:

$$f((a,b,c)) = \bigcup \{ x : a, b, c, \dots \implies x \}, \text{ where } a, b, \text{ and } c \text{ are ordinals and } x \text{ is a set defined by the above rules}$$
Explanation of Rule 1:
Within the class of all ordered triplets of ordinals, there will be one and only one triplet $(a,b,c)$ for each ordinal $\gamma$ in the class of all ordinals where $a,b,c = a,a-1,a-2$ and $\gamma = a â€“ 3$. If the input for function $f$ is any triplet of the form $a,b,c = a, a-1, a-2$, then Rule 1 states that $a-3$ will be an element of $f(a,b,c)$.

Explanation of Rule 2:
Within the class of all ordered triplets of ordinals, there will be one and only one triplet $(a,b,c)$ for each ordinal $\gamma$ in the class of all ordinals that are not limit ordinals where $a,b,c = 0,1,\gamma - 1$ and $\gamma = c + 1$. If the input for function $f$ is any triplet of the form $a,b,c = 0, 1, \gamma - 1$, then Rule 2 states that $c + 1$ will be an element of $f(a,b,c)$.

Explanation of Rule 3:
Rule 3 states that any triplet $(a,b,c)$ in the class of all ordered triplets of ordinals where $(a,b,c) = (a, a+1, a+2)$ will result in both $a+3$ and $a+\omega$ being elements of $f(a,b,c)$.

Now what in the bleep could you possibly not understand about that you eggheads (sorry... just trying to play the part of a crank).

#### Maschke

@Aplanis, I wrote this without reading your most recent post. It's all I can say on the subject. tl;dr: Either define your ellipses or don't. But till you do I can't interact further because Rule 1 is the only thing in your exposition I can grasp.

I want to explain as clearly as I can where I'm coming from. I'm willing to have a mathematical conversation with you about this but you are wildly mistaken about a couple of things.

1. You think you're making sense. You're not. At least not to me. If you are to anyone else, they are free to post and I wish they would. But when you address or interact with me, I would appreciate your assuming that I can't understand a single word you're saying. The burden is on you to be clear and you are a long way from that.

So I started with rule 1. It seemed to be ambiguous or unclear on the meaning of the expression, "0, 1, 2, ..." Sometimes it's used as if it's the FINITE string "0, 1, 2", and other times it's used as ... well I'm not sure what. The expression "0, 1, 2, ..." has a well-established meaning in standard math, namely the sequence of the natural numbers as given by the Peano axioms.

I have no idea what YOU mean by "0, 1, 2, ..." So I asked. Several times. The best I can understand your answer is, that "It does not matter and it should be perfectly obvious."

I say to you that this is by no means obvious. But more than that: It does not seem to be anchored in the world of sanity. I have no idea what you mean when you say that the ellipses don't matter. In standard math they matter a lot. So you're making up a new definition for ellipses. That's perfect legal. But you have to DEFINE it for us.

So here's the deal. Explain your notation or don't. For my part, no further conversation can ensue, simply because without your explaining your notation, I have no hope of making sense of the rest of it.

Define your ellipses notation or retract it. Don't just tell me it "makes no difference" and then go off waving your arms about your function. If it doesn't make any difference, what exactly is it?

That's it. You have to engage with me on this exact point or not. Not because I have to have things my way; but that it's the only thing in your exposition that I can make any sense of at all. Rule 1. That's all I know. I'm operating on the theory that if I can understand rule 1, I might be able to figure out what you're getting at. It's the ONLY thing you wrote I can engage with. So like I say, it's not you, it's me. But you either engage on this point or not. There is nothing else I can do.

Rule 1 and the ellipses notation. Please explain it with sufficient clarity that I can understand it. I can not discuss anything else till this is clear.

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#### Micrm@ss

If you would start by using proper notations that every mathematician in the world agreed upon, that would be a great help. Because as of now, something like

$$a, a-1, a-2,...~\Rightarrow~\{a-3\},~a\geq 3$$
makes absolutely no sense and is not a notation I've ever seen before.

This is fun. Let's figure out where what seems obvious to me is not obvious to others.
A lack of correct notation seems the obvious answer...

Honestly, I'm not very interested in this thread. So if you just keep posting the same explanations without radically changing your presentations, I don't think I'll last very long here.

#### AplanisTophet

If you would start by using proper notations that every mathematician in the world agreed upon, that would be a great help. Because as of now, something like

$$a, a-1, a-2,...~\Rightarrow~\{a-3\},~a\geq 3$$
makes absolutely no sense and is not a notation I've ever seen before.

A lack of correct notation seems the obvious answer...

Honestly, I'm not very interested in this thread. So if you just keep posting the same explanations without radically changing your presentations, I don't think I'll last very long here.
If notation is the issue, then we can pretty much scrap that too so as to radically change the exposition for you and Maschke both.

Start a sequence as $T = 1,2,3$. We want to expand on it. To do so, we'll make a rule.

To make this rule, I am going to consider every possible triplet that can be made from 1,2, and 3: (1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), and (3,2,1). I am then going to assert a rule #1 saying that if I can make the triplet (3,2,1) using the ordinals that comprise the initial segment of my sequence, I can add the element 0 to my sequence. We now have our first rule and our sequence has become $T = 1,2,3,0$.

I now want to make a second rule so as to try to expand on our sequence some more. Naively, let this second rule say that if I can make the triplet (0,1,2) using ordinals that comprise the initial segment of my sequence, I can then assert that 3 is in the sequence. Unfortunately, that rule doesn't add anything, because 3 is already in our sequence, so I'll continue to make a third rule.

Let the third rule be that if I can make the triplet (0,1,3) out of the elements comprising the initial segment of my sequence, I can then assert that 4 is in the sequence. This works, so now my sequence has become $T = 1,2,3,0,4$ after three rules.

I can make a fourth rule that takes the triplet (0,1,4) and asserts 5 is in the sequence, a fifth rule that takes the triplet (0,1,5) and asserts 6 is in the sequence, and so on, until I have $\omega$ rules and my sequence has become $T = 1,2,3,0,4,5,6, \dots$ where $\{ x : x \in T \} = \omega$.

Recall the triplet (0,1,2) now and rule 2, which didn't help us. I can now make rule $\omega$, which states that if the triplet (0,1,2) appears in my sequence, then $\omega$ appears in my sequence.

Taking into account all $\omega+1$ rules from the very beginning changes the sequence. That is, we would start with $T = 1,2,3$ as we did before and, using the six triplets that can be made from 1,2, and 3, we would derive $T = 1,2,3,0$ because rule 1 places 0 in the sequence. We then would have rule 3 place 4 into the sequence as before and we would also have rule $\omega$ place $\omega$ into the sequence so as to derive $T = 1,2,3,0,4,\omega$. We then consider all the triplets that can be made from 1,2,3,0,4, and $\omega$, which would have us add 5 (based on rule 4) to the sequence. We have now $T = 1,2,3,0,4,\omega,5$ and, if we continue, we'll end with $T = 1,2,3,0,4,\omega,5,6,7,8,\dots$ where $\{ x : x \in T \} = \omega + 1$.

I can keep making rules and eventually make it so $T$ enumerates $\omega^2$, or any countable ordinal for that matter.

I trust this is all clear?

In my original post, Rule 1 works the same as it does here.

Rule 2 from my original post is the generalized equivalent of all the rules $x$ where 1 < $x$ < $\omega$ here. It takes any triplet of the form $(0,1,a)$, where $a$ is an ordinal, and implies $a+1$ may be added to $T$. I write it as $\text{ Rule 2 : } 0,1,a \implies \{a+1\}$, where $a$ is any ordinal (yes, I will omit the ellipses based on Maschke's suggestion).

Feeling better with Rule 2 now as it was originally stated in my OP or do I need to change my OP and, if so, can you suggest how?

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#### AplanisTophet

Does this work?

Rules: Where $a, b, c, \dots$ are ordinals that are local variables with respect to each rule, the following rules are used to define function $f$.

$$\text{Rule 1 : }\mu = a, \beta = a-1, \gamma = a-2 \implies \{ a-3 \} \text{, where } a \geq 3$$
$$\text{Rule 2 : }\mu = 0, \beta = 1, \gamma \implies \{\gamma+1\}$$
$$\text{Rule 3 : }\mu = a, \beta = a+1, \gamma = a+2 \implies \{ a+3, a + \omega \}$$

Define function $f$:

$$f((\mu,\beta,\gamma)) = \bigcup \{ x : \mu,\beta,\gamma \implies x \}, \text{ where } \mu,\beta, \text{ and } \gamma \text{ are ordinals and } x \text{ is a set defined by the above rules}$$

#### AplanisTophet

This post makes substantial revisions to the OP thanks to insights provided by Maschke and Micrm@ss (hopefully they let me know if it's still "not even wrong"): 1) The "Brief Explanation of Function $f$" has been shortened and all ellipses have been eliminated from the original paragraph.
2) Each rule has been substantially rewritten for clarity.
3) The definition of function $f$ has been substantially rewritten for clarity.
4) Step 2, part c, of "Define the Sequence $T$" was slightly modified for clarity.
5) Concluding remarks were added that discuss $T^{\alpha}$ for any ordinal $\alpha$, where $\alpha$ represents the number of rules employed by the $T$ sequence.

Introduction: Making a Sequence $T$ based on â€œThe Rule of Threeâ€

The goal of this thread is to develop the basic model for what I refer to as a "$T$ sequence" and then expand on the basic model so as to create an enumeration of the ordinal $\epsilon_0$.

Each $T$ sequence is a listing of ordinals that is iteratively generated based on a list of rules. After developing a basic $T$ sequence model with only three rules that happens to be an enumeration of $\omega^2$, I will then add additional rules so as to (try to) create the explicit enumeration of $\epsilon_0$.

The primary means of generating the sequence $T$ is through the use of a function $f$. In general, function $f$ is going to be a function that takes as input a three-member sequence of ordinal numbers (an ordered triplet) and returns the set of all ordinals that are a continuation of any recognizable pattern or limit formed by the triplet. The term â€˜recognizable pattern or limitâ€™ is defined for this purpose as one that appears in a sequence of rules (see â€œRulesâ€ below).

Brief Explanation of Function $f$:

To get started, consider the ordered triplet $1, 2, 3$ as the initial segment of some pattern of ordinals that could be continued or a limit that could be approached. The number $4$ is a continuation of the ordinalsâ€™ pattern, so we might then say that $1, 2, 3 \implies 4$. The number $4$ is not the only thing that may be implied by $1, 2, 3$, however. We can also say that $1, 2, 3 \implies \omega$. The triplet $1, 2, 3$ is being used as an example because it implies two ordinals, $4$ and $\omega$ (and so $f((1,2,3)) = \{4, \omega \}$), whereas other triplets may imply one or none. The order of the triplet is important too. If we instead consider $3, 2, 1$ we find that $0$ is a continuation of the ordinalsâ€™ pattern (instead of $4$), so we say $3, 2, 1 \implies 0$. To summarize, we are starting to form some basic rules that collectively are going to define the term â€˜recognizable pattern or limitâ€™.

Rules: Where $a, b, c, \dots$ are ordinals that are local variables with respect to each rule, the following rules are used to define function $f$:

$$\text{Rule 1 : }\mu = 3, \beta = 2, \gamma = 1 \implies \{ 0 \}$$
$$\text{Rule 2 : }\mu = 0, \beta = 1, \gamma = a \implies \{a + 1\}$$
$$\text{Rule 3 : }\mu = a, \beta = a+1, \gamma = a+2 \implies \{ a+3, a + \omega \}$$
$$\text{-Note that while we start with these three rules, we will look to add more later so as to enumerate } \epsilon_0.$$

Define function $f$:

$$f((\mu,\beta,\gamma)) = \bigcup \{ x : \mu,\beta,\gamma \implies x \}, \text{ where } \mu,\beta, \text{ and } \gamma \text{ are ordinals and } x \text{ is a set defined by the above rules}$$

Define function $g$:

For any set of ordinals, $A$, let $g(A)$ be the set of all ordered triplets that can be made from the elements of $A$:
$$g(A) = \{ (a,b,c) : a,b,c \in A \}$$

Define $X_{Ord}$:
$$X_{Ord} = X \setminus \{ x \in X : x \text{ is not an ordinal} \} \text{ for any set } X$$

Define the sequence $T$:

Define a sequence $T = t_1, t_2, t_3, \dots$ via iterations where:

Step 1) $t_1 = 1, t_2 = 2,$ and $t_3 = 3$.

Step 2) Each $t_n$, where $n \geq 4$, is defined by the previous elements of the sequence. Starting with $n = 4$:

a) Let $A = \{ t_i \in T : i < n \}$. E.g., $A = \{1, 2, 3 \}$ on the first iteration.

b) Let $B = \{ f((a,b,c))_{Ord} : (a,b,c) \in g(A) \}$. Using the previous elements of the sequence $A$, this step creates a set $B$ of all the new sets of ordinals implied by letting function $f$ range over $g(A)$. The use of the $X_{Ord}$ function in the definition of $B$ may be unnecessary for this particular $T$ sequence.

c) Let $C = (\bigcup B) \setminus A$ and let $c_1, c_2, c_3, \dots$ be an enumeration of $C$ that is also well ordered if $|C| \in \mathbb{N}$. This step removes any redundant elements from $\bigcup B$ before potentially well ordering them so that we can add them to $T$.

d) If $|C| \in \mathbb{N}$, then set $t_n = c_1, t_{n+1} = c_2, t_{n+2} = c_3, \dots, t_{n+j-1} = c_j$.

e) If $|C| = |\mathbb{N}|$ (not applicable for this particular $T$ sequence), then let $Tâ€™ = tâ€™_1, tâ€™_2, tâ€™_3, \dots$, be a subsequence of the remaining undefined elements of $T$ and set $tâ€™_1 = c_1, tâ€™_3 = c_2, tâ€™_5 = c_3, \dots$.

Step 3) Proceed to the next undefined index $j$ in $T$, set $n = j$, and repeat step 2.

The first few elements of $T$ with the above three rules would be:
$$T = 1,2,3,0,4,\omega,5, \omega + 1, 6, \omega + 2,7,\omega+3,\omega \cdot 2,8,\omega+4,\omega \cdot 2 + 1, \dots$$

Where $T$ is generated one iteration at a time, the first iteration takes each ordered triplet that may be comprised from the ordinals $1, 2,$ and $3$ (there are six possible triplets) and tests each triplet to see if its ordering matches one of the rules. We have $f((3,2,1)) = \{0\}$ by Rule 1 and $f((1,2,3)) = \{4, \omega \}$ by Rule 3. The iteration process then takes the union of $\{0\}$ and $\{4, \omega \}$, orders the union to produce a countable sequence that is also well ordered (because it is a finite sequence), and adds the sequence to the initial undefined elements of $T$. We then start the second iteration by taking each ordered triplet that may be comprised from the ordinals $1,2,3,0,4,\omega$ and seeing which new ordinals the rules imply. Here we get the same old $f((3,2,1)) = \{0\}$ and $f((1,2,3)) = \{4, \omega \}$ in addition to $f((0,1,4)) = \{5\}, f((2,3,4)) = \{5\},$ and $f((0,1,\omega)) = \{\omega + 1\}$. Any duplicate ordinals are removed during the iteration process so as to refrain from adding them to $T$ more than once.

If we keep going with these three rules, the $T$ sequence will become an enumeration of $\omega^2$. Note that no rule exists that is capable of taking three ordinals less than $\omega^2$ and implying a set containing an ordinal greater than or equal to $\omega^2$. Rule #1 could produce $\omega^2$ given the triplet $(\omega^2 + 3, \omega^2 + 2, \omega^2 + 1)$, but since there are no ordinals greater than $\omega^2$, Rule 1 is not capable of inserting $\omega^2$ into the sequence either.

Rule 4 will lead to $T$ becoming an enumeration of $\omega^2 \cdot 2$ (assuming my calculations are correct for all of these):

$$\text{Rule 4 : }\mu = a \cdot b, \beta = a \cdot (b+1), \gamma = a \cdot (b+2) \implies \{a \cdot (b + \omega) \}$$

Rule 5 will lead to $T$ becoming an enumeration of $\omega^{\omega}$

$$\text{Rule 5 : }\mu = a+b \cdot c, \beta = a+b \cdot (c+1), \gamma = a+b \cdot (c+2) \implies \{a+b \cdot (c+\omega)\}$$

Rule 6 will lead to $T$ becoming an enumeration of $\omega^{\omega} \cdot 2$:

$$\text{Rule 6 : }\mu = a^b, \beta = a^{b+1}, \gamma = a^{b+2} \implies \{a^{b+\omega}\}$$

Rule 7 will lead to $T$ becoming an enumeration of $\omega^{\omega^2}$:

$$\text{Rule 7 : }\mu = a+b^c, \beta = a+b^{c+1}, \gamma = a+b^{c+2}\implies \{a+b^{c+\omega}\}$$

Rule 8 will lead to $T$ becoming an enumeration of $\omega^{\omega^{\omega}} \cdot 2$:

$$\text{Rule 8 : }\mu = a + b^{c + d \cdot e}, \beta = a + b^{c + d \cdot (e+1)}, \gamma = a + b^{c + d \cdot(e+2)}\implies \{a + b^{c + d \cdot(e+\omega)}\}$$

Rule 9 will lead to $T$ becoming an enumeration of $\epsilon_0 = \omega^{\epsilon_0}$:

$$\text{Rule 9 : }\mu = a + b^{c+d^e}, \beta = a + b^{c+d^{e+1}}, \gamma = a + b^{c+d^{e+2}}\implies \{ a + b^{c+d^{e+w}}\}$$

Concluding Remarks:

Assuming the above calculations are correct, the goal is to discuss a transfinite sequence $(T^{\alpha})_{\alpha \in Ord}$ compiled by adding rules to the above (or some similar) $T$ sequences. Each $T^{\alpha}$ will be a $T$ sequence generated using $\alpha$ rules, where $\alpha$ is some ordinal. An uncountable $\alpha$ would require some clarification and for $T^{\alpha}$ to be a transfinite sequence, presumably. Also, let $T^0 = 1,2,3$.

Each additional rule must meet four specific criteria:

1) Each $T$ sequence may contain only ordinals, but not every $T$ sequence is an enumeration of an ordinal. Equivalently, $\{ x : x \in T \}$ may or may not be an ordinal based on the rules used to generate $T$. I wish to create a transfinite sequence $(T^{\alpha})_{\alpha \in Ord}$ where $\{ x : x \in T^{\alpha} \}$ is an ordinal for each element of the sequence. To do this, it must be ensured that each rule, when added to the previous rules, results in a new $T$ sequence that again enumerates some ordinal.

2) It must hold true that $\bigcup_{\delta < \alpha} \{ x : x \in T^{\delta} \} = \{x : x \in T^{\alpha - 1} \}$ whenever $\alpha$ is not a limit ordinal.

3) A new rule $\alpha$ should imply an ordinal that is equal to $\bigcup_{\delta < \alpha} \{ x : x \in T^{\delta} \}$ so as to extend the sequence:

$$\text{Rule } \alpha \text{ : } \mu = a, \beta = b, \gamma = c \implies \{ \bigcup_{\delta < \alpha} \{x : x \in T^{\delta} \} \}, \text{ where } a,b,c < \bigcup_{\delta < \alpha} \{ x : x \in T^{\delta} \}$$

4) No new rule should should result in $|f((\mu, \beta, \gamma))| \geq \aleph_0$ for any particular triplet $(\mu, \beta, \gamma)$ in the class of all triplets.

Of particular interest is what $T^{\omega_1}$ would equate to after $\omega_1$ rules are added in a fashion that meets the above criteria and assuming no modifications are made to the basic $T$ sequence model to accommodate a transfinite $T$ (note that we could still start with $T = 1,2,3$ and let the iterative process run, so $T^{\omega_1}$ must equate to something despite making no alterations to the $T$ sequence model to accommodate a transfinite $T$, and it may be that $\{x : x \in T^{\omega_1} \} = \omega_1$ ??? ).

#### Maschke

I came to see what was up lately. I haven't the heart for all this. Isn't there a simpler or shorter version just to get me started? I started reading about your idea of permuting {0, 1, 2} and then saying "if you can get 2, 1, 0 out of it" or something like that. It just doesn't make any sense. You have to write less. Try to write one or two paragraphs, short ones, to help the reader get started. I see that you have a great passion for something and I'm motivated to help, but you give me nothing to hold on to.

You keep trying to do ALL of it but you are starting without me. You have to do a LITTLE BIT and give me the chance to ask questions about it.

#### AplanisTophet

I came to see what was up lately. I haven't the heart for all this. Isn't there a simpler or shorter version just to get me started? I started reading about your idea of permuting {0, 1, 2} and then saying "if you can get 2, 1, 0 out of it" or something like that. It just doesn't make any sense. You have to write less. Try to write one or two paragraphs, short ones, to help the reader get started. I see that you have a great passion for something and I'm motivated to help, but you give me nothing to hold on to.

You keep trying to do ALL of it but you are starting without me. You have to do a LITTLE BIT and give me the chance to ask questions about it.
That sounds good to me: one step at a time and you can agree or disagree as we go.

I've got a mountain of papers to grade right now for the tax course I teach. The students often make as much sense to me as I do to you in this thread, but I'm usually able to figure out what they are going for and help them along. I feel your pain. :giggle:

#### AplanisTophet

You keep trying to do ALL of it but you are starting without me. You have to do a LITTLE BIT and give me the chance to ask questions about it.
This version takes out all the confusing notation and I would think anyone could understand it, but maybe I'm wrong. Did it make sense to you? Was it maybe a tl;dr thing and you didn't read it?

If notation is the issue, then we can pretty much scrap that too so as to radically change the exposition for you and Maschke both.

Start a sequence as $T = 1,2,3$. We want to expand on it. To do so, we'll make a rule.

To make this rule, I am going to consider every possible triplet that can be made from 1,2, and 3: (1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), and (3,2,1). I am then going to assert a rule #1 saying that if I can make the triplet (3,2,1) using the ordinals that comprise the initial segment of my sequence, I can add the element 0 to my sequence. We now have our first rule and our sequence has become $T = 1,2,3,0$.

I now want to make a second rule so as to try to expand on our sequence some more. Naively, let this second rule say that if I can make the triplet (0,1,2) using ordinals that comprise the initial segment of my sequence, I can then assert that 3 is in the sequence. Unfortunately, that rule doesn't add anything, because 3 is already in our sequence, so I'll continue to make a third rule.

Let the third rule be that if I can make the triplet (0,1,3) out of the elements comprising the initial segment of my sequence, I can then assert that 4 is in the sequence. This works, so now my sequence has become $T = 1,2,3,0,4$ after three rules.

I can make a fourth rule that takes the triplet (0,1,4) and asserts 5 is in the sequence, a fifth rule that takes the triplet (0,1,5) and asserts 6 is in the sequence, and so on, until I have $\omega$ rules and my sequence has become $T = 1,2,3,0,4,5,6, \dots$ where $\{ x : x \in T \} = \omega$.

Recall the triplet (0,1,2) now and rule 2, which didn't help us. I can now make rule $\omega$, which states that if the triplet (0,1,2) appears in my sequence, then $\omega$ appears in my sequence.

Taking into account all $\omega+1$ rules from the very beginning changes the sequence. That is, we would start with $T = 1,2,3$ as we did before and, using the six triplets that can be made from 1,2, and 3, we would derive $T = 1,2,3,0$ because rule 1 places 0 in the sequence. We then would have rule 3 place 4 into the sequence as before and we would also have rule $\omega$ place $\omega$ into the sequence so as to derive $T = 1,2,3,0,4,\omega$. We then consider all the triplets that can be made from 1,2,3,0,4, and $\omega$, which would have us add 5 (based on rule 4) to the sequence. We have now $T = 1,2,3,0,4,\omega,5$ and, if we continue, we'll end with $T = 1,2,3,0,4,\omega,5,6,7,8,\dots$ where $\{ x : x \in T \} = \omega + 1$.

I can keep making rules and eventually make it so $T$ enumerates $\omega^2$, or any countable ordinal for that matter.

I trust this is all clear?

#### Maschke

This version takes out all the confusing notation
You don't get it and you don't want to listen.

I want you to write one single thing. Like Rule 1. Forget all the metaphysics about natural continuations of sequences or whatever it is you're doing. Give me the raw mathematical formalism, starting with the very first thing there is to know.

Then let me digest that and ask questions.

One idea. A few lines of text or a definition of some formula. Like Rule 1, but now that you've retracted your ellipses notation.

I'll make this concrete for you.

I will not read any post you write that is over 150 words in length.

You do recall I hope that a few posts back I asked you to "clarify or retract" the ellipses. I see you've retracted them. But please remember how often and how vehemently you have thought that it's "obvious what they are," and then later "obvious that there's no difference," then finally abandoned altogether.

So get some self-awareness and some humbleness.

Your lengthy posts are not tethered to rationality in my opinion. I don't seem to be able to communicate to you that your exposition is incoherent and meaningless; and your belief that it's practically obvious is a delusion. Not a mathematical confusion of some sort. A genuine break with reality that should generate in you -- I'll be blunt -- concern.