A Sequence T based on the Rule of Three

[AplanisTophet]

TL;DR Summary

Compiling a sequence of ordinals and analyzing it

Introduction: Making a Sequence $T$ based on “The Rule of Three”
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 ‘recognizable pattern or limit’ is defined for this purpose as one meeting a sequence of rules (see “Rules” 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: $1, 2, 3, \dots$. What would come next? The number $4$ is a continuation of the ordinals’ pattern, so we might then say that $4$ is implied by $1, 2, 3, \dots$ or, similarly, $1, 2, 3, \dots \implies 4$. The number $4$ is not the only thing that may be implied by $1, 2, 3, \dots$, however. We can also say that $1, 2, 3, \dots \implies \omega$. If there are any other ordinal numbers that happen to be universally agreeable continuations or limits for the three-member sequence $1, 2, 3, …$, we could consider them too. In this case, I am unaware of any. 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 ordinals’ 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 ‘recognizable pattern or limit’.
Rules: Let $a, b, c, \dots$ be ordinals.

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

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

$$\text{Rule 3 : }a \cdot b, a \cdot (b+1), a \cdot (b+2), \dots \implies \{ a \cdot (b+3), a \cdot (b + \omega) \}$$

$$\text{Rule 4 : }0, 1, a, \dots \implies \{a+1\}$$

$$\text{Rule 5 : }0, 2, a, \dots \implies \{a+2\}$$

$$\text{Rule 6 : }a^{b+c}, a^{b+(c+1)}, a^{b+(c+2)}, \dots \implies \{a^{b+(c+3)}, a^{b + (c + \omega)}\}$$

$$\text{Rule 7 : }a_{b+c}, a_{b+(c+1)}, a_{b+(c+2)}, \dots \implies \{a_{b+(c+3)}, a_{b+ (c+ \omega)}\}$$

$$\text{Rule 8 : }a^{b \cdot c}, a^{b \cdot (c+1)}, a^{b \cdot (c+2)}, \dots \implies \{a^{b \cdot (c+3)}, a^{b \cdot (c + \omega)}\}$$

$$\text{Rule 9 : }a_{b \cdot c}, a_{b \cdot (c+1)}, a_{b \cdot (c+2)}, \dots \implies \{ a_{b \cdot (c+3)}, a_{b \cdot (c + \omega)}\}$$

$$\text{Rule 10 : }a, s(a), s(s(a)), \dots \implies \{ s(s(s(a))), a^{b^{b^{b^{\vdots}}}}\} \text{, where } s(a) = a^b$$

$$\text{Rule 11 : }a, s(a), s(s(a)), \dots \implies \{ s(s(s(a))), a_{b_{b_{b_{\vdots}}}}\} \text{, where } s(a) = a_b$$

$$\text{Rule 12 : }0, 1, \omega, \dots \implies \{\omega_1\}$$

$$\text{Rule 13 : }1, \omega, \omega_1, \dots \implies \{\omega_2\}$$

$$\text{Rule 14 : }a_{0}, a_{1}, a_{b}, \dots \implies \{a_{b+1}\}$$

$$\text{Rule 15 : }a_{0}, a_{2}, a_{b}, \dots \implies \{a_{b+2}\}$$

$$\text{Rule 16 : }a^{0}, a^{1}, a^{b}, \dots \implies \{a^{b+1}\}$$

$$\text{Rule 17 : }a^{0}, a^{2}, a^{b}, \dots \implies \{a^{b+2}\}$$

$$\text{Rule 18 : }a^{0}, a^{1}, a^{\omega}, \dots \implies \{a^{\omega_1}\}$$

$$\text{Rule 19 : }a^{1}, a^{\omega}, a^{\omega_1}, \dots \implies \{a^{\omega_2}\}$$

$$\text{Rule 20 : }a^{b_c}, a^{b_{c+1}}, a^{b_{c+2}}, \dots \implies \{a^{b_{c+3}}, a^{b_{c + \omega}}\}$$

$$\text{Rule 21 : }a_{0}, a_{1}, a_{\omega}, \dots \implies \{a_{\omega_1}\}$$

$$\text{Rule 22 : }a_{1}, a_{\omega}, a_{\omega_1}, \dots \implies \{a_{\omega_2}\}$$

$$\text{Rule 23 : }a_{b_c}, a_{b_{c+1}}, a_{b_{c+2}}, \dots \implies \{a_{b_{c+3}}, a_{b_{c+\omega}}\}$$
We can now define function $f$:

$$f((a,b,c)) = \{ x : a, b, c, \dots \implies x \}, \text{ where } a, b, \text{ and } c \text{ are ordinals and } x \text{ is a set containing ordinals}$$
For any set of ordinals, $A$, let $g(A)$ be the set of all ordered triplets that can be made from the elements of $A$:

$$g(A) = \{ (a,b,c) : a,b,c \in A \}, \text{ where } A \text{ is a set containing only ordinals (or is empty).}$$Define the sequence $T$:
Define a sequence $T = t_1, t_2, t_3, \dots$, where:
1) $t_1 = 1, t_2 = 2,$ and $t_3 = 3$.
2) Each $t_n$, where $n \geq 4$, is defined by the previous elements of the sequence:

a. Let $A = \{ t_i \in T : i < n \}$.

b. Let $B = \{ f((a,b,c)) : (a,b,c) \in g(A) \}$.

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 $|C| \in \mathbb{N}$.

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 $T’ = t’_1, t’_2, t’_3, \dots$, be a subsequence of the remaining undefined elements of $T$ and then set $t’_n = c_1, t’_{n+2} = c_2, t’_{n+4} = c_3, \dots$.
3) After completing step 2 for a given element of $T$, proceed to the next undefined element of $T$, set it equal to $t_n$, and repeat step 2.Assuming my calculations are correct (it's tricky ), the first few elements of $T$ would be:

$$T = 1,2,3,0,4, \omega, 5, \omega + 1, \omega +2, \omega_1, 6, \omega + 3, \omega \cdot 2, \omega_1 + 1, \omega_1 + 2, \omega_2, 7, \dots$$
The question arises: Is there a subsequence of $T$ that is unbounded in $\omega_1$ and/or is $T$ itself unbounded across the class of all ordinals?
The above definition for this particular $T$ sequence leaves options and is still considered a working draft. I may update a few of the rules that will add even more items to $T$ if possible. It’s going to be a regular pandora’s box of ordinals. I soon won’t be able to tell what ordinals, if any, aren’t in $T$ and then will ask others here if they can find any:
A second question arises: What ordinals do not appear in $T$?
And the last question too: What additional rules might we add?
I can also see models for $T$ sequences where there are an infinite number of ordinals implied by the rules at each iteration instead of just a finite number, but I’m not sure it would end up adding more elements to $\bigcup T$ overall at this point. I also see versions of this where the rules themselves are enumerated by the sequence instead of being plainly stated so as to produce a potentially infinite number of rules that could make some of the same ordered triplets imply (potentially infinitely many) more ordinals as the sequence progresses.
Finally, the model could accept any countably infinite set of implications for a given sequence of three with some minor tweaks. Think of all the things that could follow $0, 1, \omega, \dots$, for example. We can add any countable number of them.
If you find this, I hope it’s interesting!

 

[SSequence]

As a first question, does the above enumeration mean that:
$t_1=1$
$t_2=2$
$t_3=3$
$t_4=0$
$t_5=4$
$t_6=\omega$
$t_7=5$
$t_8=\omega+1$
$t_9=\omega+2$
$t_{10}=\omega_1$
...

or something else?

 

[AplanisTophet]

Rules 4 and 5 for the triplet 1,2,3 on the first iteration would imply both 4 and 5 get added (instead of just 4), so:
$$T = 1,2,3 \text{ to start, then}$$
$$T = 1,2,3,0,4,5, \omega \text{ after the first iteration}$$
$$T = 1,2,3,0,4,5, \omega, 6, 7, \omega +1, \omega +2, \omega_1, \text{ after the second iteration, again, if my quick calc is correct}$$

I am sure there are other ways to define $g$. The whole $f((a,b,c,))$ looks funny perhaps too, but it works.

It was tricky and I just finished my working draft to post, sorry.

I hope that helps. Thank you!

 

[SSequence]

Yes, it is very long post so I haven't read it.

But it seems that with $f(a,b,c)=S$ where $a,b,c \in Ord$ and $S \subseteq Ord$ ... you essentially want to generate a subset of ordinals that depends on the parameters $a,b,c$. So perhaps you could write something like:
$X(a,b,c) \subseteq Ord$ where $a,b,c \in Ord$

So this would indicate that the "set" X depends on the parameters in the input. However another question:
(Q) Is $X(a,b,c) \subseteq Ord$ always supposed to be a set in the strict sense for any parameters $a,b,c \in Ord$. Meaning can $X(a,b,c) \subseteq Ord$ be (potentially) unbounded in $Ord$ for some parameters $a,b,c \in Ord$ or will there always be a cut-off value $\alpha \in Ord$ such that $X(a,b,c) \subseteq \alpha$?Edit:
It seems that you used the symbol $A$ already in OP. I have replaced "$A$" with "$X$" in my post to avoid ambiguity.

 

[AplanisTophet]

For each iteration we are given a particular $t_n$ and the set $A$ gets defined as $A = \{ t_i \in T : i < n \}$. Outside of its specific definition within a given iteration, $A$ is just any set of ordinals (unbounded).

 

[SSequence]

OK thanks.

One of the reasons I was asking was that the number of rules that you have are a lot. So actually, I think it might be very relevant to determine that whether for all $\alpha \in Ord$ we can show that the following "set" is unbounded in $Ord$ or not :
$\{\cup X(a,b,c) : a,b,c \in \alpha \}$
And also getting a good handle on the upper-bound in the case that the above is bounded (and hence a set in strict sense always).

Above I have used the symbol $X(a,b,c)$ in the same sense as in post#4.

So before worrying about the formation of sequence $t_i$ it is an idea to look at what the transition rules yield.

===============

As an example, can you find any ordinal $\geq \omega_{\omega^2}$ using your transition rules in the "set":
$\{\cup X(a,b,c) : a,b,c \in \omega_1 \}$

 

[AplanisTophet]

I am not sure what you mean by $X(a,b,c)$ entirely, but I assume it's any set of three ordinals. By definition, $\{\bigcup X (a, b, c) : a, b, c \in \alpha \}$, where $\alpha$ is any ordinal, would be the class of all ordinals. I'm not sure I follow...

Given my list of rules, does there exist at least some triplet that, according to one of the rules, would imply $\alpha$ for any given $\alpha$? Good question. Where the list of possible triplets is restricted to those that can be formed from initial segments of $T$, we are not drawing on all possible triplets. If we were drawing on all possible triplets as opposed to just those that appear as initial segments of $T$, then would there be a rule for any given $\alpha$? I speculate yes. What is absolutely true (I believe) is that there exists a new rule that could be added to ensure that at least some triplet implies $\alpha$ for any given $\alpha$, however, whether or not that triplet may be formed out of some initial segment of $T$ then becomes the primary concern.

 

[SSequence]

I am not sure what you mean by $X(a,b,c)$ entirely, but I assume it's any set of three ordinals. By definition, $\{\bigcup X (a, b, c) : a, b, c \in \alpha \}$, where $\alpha$ is any ordinal, would be the class of all ordinals. I'm not sure I follow...

By $\alpha$, I mean any "fixed" ordinal (just like we set a value to a constant $C$ for natural or real numbers). For example, we may set it to:
$\alpha=\omega_1$
$\alpha=\omega_1 \cdot \omega$
$\alpha=\omega_2$
etc.

But once its fixed we can't change it.

 

[AplanisTophet]

Yes, so choose any ordinal $\alpha$. I believe your question is: Is there some triplet $\beta$ in the class of all triplets that would, given my finite list of rules, lead to $f(\beta) = \{ \alpha \}$? I speculate yes.

Can $\beta$ be made out of some initial segment of $T$? If we know anything about cardinality, I would suspect not.

Can a new rule be added that would make it so $\alpha$ is actually implied by a triplet made out of some initial segment of $T$? Well yes, always. This is vacuously true because we can make any rule we want. E.g., just update rule #1 to assert that $a, a+1, a+2, \dots \implies \{a+3, a+\omega, \alpha \}$.

 

[SSequence]

Yes, so choose any ordinal $\alpha$. I believe your question is: Is there some triplet $\beta$ in the class of all triplets that would, given my finite list of rules, lead to $f(\beta) = \{ \alpha \}$? I speculate yes.

Can $\beta$ be made out of some initial segment of $T$? If we know anything about cardinality, I would suspect not.

Can a new rule be added that would make it so $\alpha$ is actually implied by a triplet made out of some initial segment of $T$? Well yes, always. This is vacuously true because we can make any rule we want. E.g., just update rule #1 to assert that $a, a+1, a+2, \dots \implies \{a+3, a+\omega, \alpha \}$.

Hmmm ... what I meant was something really simple. Choose any three values you want:
$\alpha_1,\alpha_2,\alpha_3 <\omega_1$

Now can you think of any choice of the values $\alpha_1,\alpha_2,\alpha_3$ such that $f(\alpha_1,\alpha_2,\alpha_3)$ contains ordinals greater than say:
$\omega_{\omega^2}$
$\omega_{\omega^\omega}$
$\omega_{\omega_1}$
and so on...Edit:
Also, have you thought about the way the terms of your sequence are defined on the limit? Maybe you have thought about it already, but you also need to define $t_i$ for limit values $i \in Ord$.

One possible way is:
$t_i=sup\{t_x: x<i\}$ ----for limit values $i \in Ord$
But maybe you might have something else in mind.

Second Edit:
Though in general it might be helpful to consider that if you have a rule such that from one "iteration" to "next" you can always generate $\omega_{i+1}$ from $\omega_i$, then having a "limit rule" such as one I described above ... the terms of the sequence $t_i$ ($i \in Ord$) would be unbounded in $Ord$.

 

[AplanisTophet]

Currently, the only rule that takes a triplet of ordinals all less than $\omega_1$ and returns a set containing at least one ordinal that is equal to or greater than $\omega_1$ happens to be Rule #12. I didn't build in any others, but once $\omega_1$ finds its way into an initial segment of $T$ (just like plain old $\omega$ did at some point), we start finding there will be an infinite number of ordinals with cardinality greater than or equal to $\omega_1$.

How about this: If $\alpha$ is countable, are we not guaranteed its existence at some point within an initial segment of $T$? I don't think so... ??

 

[SSequence]

As I mentioned in previous post, if you defining $t_i$ such that $i$ can only take values in $\mathbb{N}$ then your sequence will be bounded by a given ordinal of course. So if you are thinking that it goes unbounded in $Ord$ the only way it might be possible is to define $t_i$ for all $i \in Ord$. And for that you need to define a "limit rule" for your sequence (as I mentioned in the "Edits" in the previous post#10).

My point is that even if you are producing $\geq \omega$ number of terms of $t_i$ in your iterations, even then you need a limit rule as the number of iterations approach $\omega$. Otherwise, you will again have your sequence bounded by some given ordinal.

[Of course some other questions still remain relevant in the above two cases even without a limit rule]The last alternative is that you just ask the question of the type like I asked in post#10.

 

[AplanisTophet]

As I mentioned in previous post, if you defining $t_i$ such that $i$ can only take values in $\mathbb{N}$ then your sequence will be bounded by a given ordinal of course.

Yes, and in particular, which ordinal less than $\omega_1$ bounds the subsequence $T^{\omega_1}$ of $T$ that contains only members of $T$ less than $\omega_1$ (assuming the regular ordinal alphabet, $\Sigma = \{0\dots9, \omega, <, +, \cdot \}$, if that matters).

So if you are thinking that it goes unbounded in $Ord$ the only way it might be possible is to define $t_i$ for all $i \in Ord$. And for that you need to define a "limit rule" for your sequence (as I mentioned in the "Edits" in the previous post#10).

It might be possible?
Chewing through the limit ordinals is exactly what $T$ sequences are designed to do. They also enumerate, for lack of a better way to put it, an awful lot of ordinals.

My point is that even if you are producing $\geq \omega$ number of terms of $t_i$ in your iterations, even then you need a limit rule as the number of iterations approach $\omega$. Otherwise, you will again have your sequence bounded by some given ordinal.

So add a Rule #24 that enumerates as a set the supremums of $T$ sequences with only the first 23 Rules and each successive supremum? E.g.:
$$Rule 24 : \exists E = e_1, e_2, e_3, \dots, \text{ where each }e_i \text{ such that:}$$

$$e_i = \text{ the supremum of a } T \text{ sequence defined by:}$$

$$\text{1) Rules 1-23, and,}$$

$$\text{2) where Rule #1 is rewritten as : } a, a + 1, a + 2, \dots \implies \{a + 3, a + \omega, e_1, e_2, e_3, \dots, e_j \} \text{, where } j < i$$

$$\text{Then Rule 24 becomes : } 1, 2, 3, \dots \implies \{e_i : e_i \in E \}$$

[Of course some other questions still remain relevant in the above two cases even without a limit rule]

Yes, we can easily show a proof for any number of given ordinals that they appear in the sequence. How about the Church-Kleene ordinal? Let me give an example of some easy proofs that might help show my line of thought:
We know $\omega$ appears in the sequence by $1,2,3$ and Rule 1.

We then know $\omega + 1$ and $\omega + 2$ appear in the sequence by Rules 4 and 5.

We then know $\omega \cdot 2$ appears by Rule 1.

We then know $\omega \cdot 2 + 1$ and $\omega \cdot 2 + 2$ appear by Rules 4 and 5.

We then know $\omega \cdot 3$ appears by Rule 1.

We then know $\omega^2$ appears by Rule 3.

We then know that $\omega^2 + 1$ and $\omega^2 + 2$ appear by Rules 4 and 5.

We then know that $\omega^2 + \omega$ appears by Rule 1.

… if we keep going, we’ll easily hit ordinals like $$\omega^{\omega^{\omega^{\vdots}}}$$

And beyond! Well, way, way, way beyond, actually. Or maybe?
I am really curious what ordinal it is after only Rules 1-23 that is the supremum of $T^{\omega_1}$ defined above. What are we doing with my hypothetical Rule 24? We could keep going…

 

[SSequence]

Chewing through the limit ordinals is exactly what $T$ sequences are designed to do. They also enumerate, for lack of a better way to put it, an awful lot of ordinals.

One can just define a function $f:Ord \rightarrow Ord$ such that:
$f(0)=0$
$f(x+1)=f(x)+1$
$f(x)=sup\{ f(i) : i<x\}$ ---- for all limit value $x \in Ord$

But this is just another way of writing the function $x \mapsto x$ , that is $f(x)=x$. If we wanted a function enumerating limits then we change the first and second rules above to:
$f(0)=\omega$
$f(x+1)=f(x)+\omega$

But this is also just another way of writing $f(x)=\omega+\omega \cdot x$.

So I don't quite understand what you are trying to do. Since once we add some kind of limit rule, we have to be careful in delineating the formation of function/"sequence" if the domain of our function/"sequence" is $\omega$ (essentially $\mathbb{N}$). Adding a limit rule in naive way as above examples simply means that we aren't talking about the domain of the function as $\omega$ anymore.

 

[AplanisTophet]

One can just define a function $f:Ord \rightarrow Ord$ such that:
$f(0)=0$
$f(x+1)=f(x)+1$
$f(x)=sup\{ f(i) : i<x\}$ ---- for all limit value $x \in Ord$

But this is just another way of writing the function $x \mapsto x$ , that is $f(x)=x$. If we wanted a function enumerating limits then we change the first and second rules above to:
$f(0)=\omega$
$f(x+1)=f(x)+\omega$

But this is also just another way of writing $f(x)=\omega+\omega \cdot x$.

So I don't quite understand what you are trying to do. Since once we add some kind of limit rule, we have to be careful in delineating the formation of function/"sequence" if the domain of our function/"sequence" is $\omega$ (essentially $\mathbb{N}$). Adding a limit rule in naive way as above examples simply means that we aren't talking about the domain of the function as $\omega$ anymore.

Your enumeration of the limits is then $\omega, \omega \cdot 2, \omega \cdot 3, \dots, \omega ^2, \omega^2 + \omega, \omega^2 + \omega \cdot 2, \dots, \omega^2 \cdot 2, \dots$. Yes, all of these would appear in a $T$ sequence with only the first 23 Rules given above. Of course, your enumeration has a bunch of "$\dots$" along the way whereas a $T$ sequence does not, so I'm still focused on what the least upper bound of $T^{\omega_1}$ is (where $T^{\omega_1}$ is a subsequence of $T$ containing only elements in $\omega_1$).

I'm not as focused on the ordinals greater than $\omega_1$ at this point. I could update a few more of the rules, where the more rules we add, the more ordinals we enumerate. At this point, enumerating ordinals less than $\omega_1$ has become very easy (think of Kleene's O and other efforts to name and classify the ordinals), so I want to know where we sit after these first 23 Rules.

After that, what rules could we add?

PS - I see a few clerical clean-ups that could be made to the OP too, but I think we're on the same page.

 

[AplanisTophet]

This post makes substantial revisions to the OP after gaining numerous insights since the thread was first posted based on guidance received from some mathematicians much more capable than I. Thank you!
Introduction: Making a Sequence $T$ based on “The Rule of Three”
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 $\epsilon_0$. I also offer a concluding paragraph that discusses $T^{\alpha}$ for any ordinal $\alpha$, where $\alpha$ represents the number of rules employed by the $T$ sequence, and pose a concluding question that perhaps someone here would be willing to address.
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 ‘recognizable pattern or limit’ is defined for this purpose as one that appears in a sequence of rules (see “Rules” 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 ordinals’ 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 ordinals’ 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 ‘recognizable pattern or limit’.
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 $A$:

$$g(A) = \{ (a,b,c) : a,b,c \in A, a \neq b, a \neq c, \text{ and } b \neq c \}$$
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 $T’ = t’_1, t’_2, t’_3, \dots$, be a subsequence of the remaining undefined elements of $T$ and set $t’_1 = c_1, t’_3 = c_2, t’_5 = c_3, \dots$.
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 $f((0,1,4)) = \{5\}$ by rule 2, $f((2,3,4)) = \{5, \omega\}$ by rule 3, and $f((0,1,\omega)) = \{\omega + 1\}$ by rule 2. 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, if put into a general form (similar to rules 3 and above where $\mu = a + 3, \beta = a + 2, \gamma = a + 1 \implies \{ a\}$), 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.
Adding Rules
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 except $T^0$. 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 $\{x : x \in T^{\omega_1} \} = \omega_1$ ? ).