# Another Opus Here

#### AplanisTophet

This should help a lot if anyone revisits:

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

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 meeting 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: $1, 2, 3, \dots$. What would come next? The number $4$ is a continuation of the ordinalsâ€™ 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 $1, 2, 3, â€¦$, 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 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: Let $a, b, c, \dots$ be ordinals.
$$\text{Rule 1 : }a, a+1, a+2, \dots \implies \{ a+3, a + \omega \}$$
$$\text{Rule 2 : }a, a-1, a-2, \dots \implies \{ a-3 \} \text{ where } a \geq 3$$
$$\text{Rule 3 : }a \cdot b, a \cdot (b+1), a \cdot (b+2), \dots \implies \{ a \cdot (b+3), a \cdot (b + \omega) \}$$
$$\text{Rule 4 : }0, 1, a, \dots \implies \{a+1\}$$
$$\text{Rule 5 : }0, 2, a, \dots \implies \{a+2\}$$
$$\text{Rule 6 : }a^{b}, a^{b+1}, a^{b+2}, \dots \implies \{a^{b + \omega}\}$$
$$\text{Rule 7 : }a_{b}, a_{b+1}, a_{b+2}, \dots \implies \{a_{b+\omega}\}$$
$$\text{Rule 8 : }a^{b}, a^{b \cdot 1}, a^{b \cdot 2}, \dots \implies \{a^{b \cdot \omega}\}$$
$$\text{Rule 9 : }a_{b}, a_{b \cdot 1}, a_{b \cdot 2}, \dots \implies \{a_{b \cdot \omega}\}$$
$$\text{Rule 10 : }a^b, a^{b^b}, a^{b^{b^b}}, \dots \implies \{a^{b^{b^{b^{\vdots}}}}\}$$
$$\text{Rule 11 : }a_b, a_{b_b}, a_{b_{b_b}}, \dots \implies \{a_{b_{b_{b_{\vdots}}}}\}$$
$$\text{Rule 12 : }0, 1, \omega, \dots \implies \{\omega_1\}$$
$$\text{Rule 13 : }1, \omega, \omega_1, \dots \implies \{\omega_2\}$$
$$\text{Rule 14 : }a_{0}, a_{1}, a_{b}, \dots \implies \{a_{b+1}\}$$
$$\text{Rule 15 : }a_{0}, a_{2}, a_{b}, \dots \implies \{a_{b+2}\}$$
$$\text{Rule 16 : }a^{0}, a^{1}, a^{b}, \dots \implies \{a^{b+1}\}$$
$$\text{Rule 17 : }a^{0}, a^{2}, a^{b}, \dots \implies \{a^{b+2}\}$$
$$\text{Rule 18 : }a^{0}, a^{1}, a^{\omega}, \dots \implies \{a^{\omega_1}\}$$
$$\text{Rule 19 : }a^{1}, a^{\omega}, a^{\omega_1}, \dots \implies \{a^{\omega_2}\}$$
$$\text{Rule 20 : }a^{\omega}, a^{\omega_1}, a^{\omega_2}, \dots \implies \{a^{\omega_{\omega}}\}$$
$$\text{Rule 21 : }a_{0}, a_{1}, a_{\omega}, \dots \implies \{a_{\omega_1}\}$$
$$\text{Rule 22 : }a_{1}, a_{\omega}, a_{\omega_1}, \dots \implies \{a_{\omega_2}\}$$
$$\text{Rule 23 : }a_{\omega}, a_{\omega_1}, a_{\omega_2}, \dots \implies \{a_{\omega_{\omega}}\}$$

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 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 $c_1, c_2, c_3, \dots, c_j$ be a well ordering of the elements of $C$ by their natural order.
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 $|C| = |\mathbb{N}|$, then set $t_n = c_1, t_{n+2} = c_2, t_{n+4} = c_3, \dots$.

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 $\omega_1$? For this purpose, I am interested in how $T$ grows.

#### AplanisTophet

The above should make it clear what Iâ€™m doing for $T$ overall, but is still a working draft (making a $T$ was never the original aim for this thread). I will be updating a few of the rules that will add even more items to $T$ because Iâ€™m finding it interesting. Itâ€™s going to be a regular pandoraâ€™s box of ordinals. I wonâ€™t be able to tell what ordinals, if any, arenâ€™t in $T$ and then would ask for a little help again.

I can also see models where there are an infinite number of ordinals implied by the rules at each iteration instead of just a finite number, but Iâ€™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.

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 them all.

If you find this, I hope itâ€™s interesting!

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#### AplanisTophet

This redundant post should help a lot if anyone revisits as it is the latest version of a working draft. Each draft improves upon the list of rules, generally, in an attempt to enumerate as many ordinals as possible. There are three questions in bold at the bottom which anyone can feel free to answer at any time!

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

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 meeting 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: $1, 2, 3, \dots$. What would come next? The number $4$ is a continuation of the ordinalsâ€™ 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 $1, 2, 3, â€¦$, 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 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: 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 $Tâ€™ = tâ€™_1, tâ€™_2, tâ€™_3, \dots$, be a subsequence of the remaining undefined elements of $T$ and then set $tâ€™_n = c_1, tâ€™_{n+2} = c_2, tâ€™_{n+4} = c_3, \dots$.

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. Itâ€™s going to be a regular pandoraâ€™s box of ordinals. I soon wonâ€™t be able to tell what ordinals, if any, arenâ€™t in $T$ and then will ask others here if they can find any:

A second question arises: What ordinals do not appear in $T$?

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 Iâ€™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 itâ€™s interesting!

#### AplanisTophet

Anyone see a problem with any of these quick and easy theorems?

Theorem 1: If $X$ is some countable ordinal and $S^X = s_1, s_2, s_3, \dots$ is a (transfinite) sequence having order type $X$ of successive limit ordinals, then $s_x > x$ for all $s_x \in S^X$.

Theorem 2: If $X$ is some countable ordinal and $S^X = s_1, s_2, s_3, \dots$ is a (transfinite) sequence having order type $X$ of successive limit ordinals, then there exists an $s_x \in S^X$ where $s_x \geq X$.

Theorem 3: Let $X$ be some countable ordinal and $S^X = s_1, s_2, s_3, \dots$ be a (transfinite) sequence having order type $X$ of successive limit ordinals. For each $s_x$ (where $x \in X$), let $p(s_x) = s_{x,1}, s_{x,2}, s_{x,3}, \dots$ be a (transfinite) sequence of successive limit ordinals having order type $x$ where $s_{x,1} = s_1$. Then there exists some $s_{x,j}$ (where $x,j \in X$) such that $s_{x,j} \geq X$ and $s_{x,j} \in p(s_x)$ for some $x \in X$.

Theorem 4: Let $X$ be some countable ordinal and $S^X = s_1, s_2, s_3, \dots$ be a (transfinite) sequence having order type $X$ of successive limit ordinals. For each $s_x$ (where $x \in X$), let $p(s_x) = s_{x,1}, s_{x,2}, s_{x,3}, \dots$ be a (transfinite) sequence of successive limit ordinals having order type $x$ where $s_{x,1} = s_1$. Let $s_{xâ€™,j}$ (where $xâ€™,j \in X$) such that $s_{xâ€™,j} \geq X$ and $s_{xâ€™,j} \in p(s_{xâ€™})$ for some $xâ€™ \in X$. Let $R = r_1, r_2, r_3, \dots$ be a (non-transfinite) sequence of ordinals contained in $X$ that is unbounded in $X$. Then there exists some $r_i = q_x$ where $q_x \in X$, $q_x \geq s_{xâ€™}$, and $s_{xâ€™,j} \in p(q_x)$.

If not, then does this make sense?:

Define Function $k$:
For any countable ordinal $X$, let $k(X) = k_1, k_2, k_3, \dots$ be a (potentially finite or transfinite) sequence of successive limit ordinals where $x \in X$ for each $k_x$, $k_1 = \omega$, and each $k_x$, where $x > 1$, is the successor limit ordinal of $k_{x-1}$.
$$\text{E.g., } k(\omega + 2) = \omega, \omega \cdot 2, \omega \cdot 3, \dots, \omega^2, \omega^2 + 1$$

Define T-Sequence Iteration Counter Variable $j$:
The sequence $T$ starts off as $1, 2, 3$ and is then generated one iteration at a time. The first 23 Rules are iteration independent, can appear in any order, and will always result in the same sequence $T$. It will become helpful to count the iterations for future rules, however, so I will add a counter variable $j$ to the function $f$ (e.g., $f((1,2,3), j) = \{0, 4, 5, \omega\}$ for any iteration $j$ under the first 23 rules, so the first iteration sets $t_4 = 0, t_5 = 4, t_6 = 5,$ and $t_7 = \omega$, but future rules may state that $f((1,2,3),j)$ implies additional ordinals depending on the value of $j$).

Function $v$:
For any $T$ sequence iteration $j$, let $v(j)$ equal the largest ordinal less than $\omega_1$ added to $T$ by iteration $j$. E.g., where iteration $j = 1$ adds the ordinals $0, 4, 5,$ and $\omega$, we have $v(1) = \omega$ because $\omega$ is greater than $0, 4,$ and $5$ and is also less than $\omega_1$ under the standard $<$ ordering.

The Sets $E^i$:
Let $E^i$ be a countable set of ordinals for each $i \in \mathbb{N}$, where each $E^i$ is such that:
$$E^i = \{ k_x : k_x \in k(v(j)) \}$$
$$\begin{matrix} & E^1 = e_{1,1}, e_{1,2}, e_{1,3}, \dots \\ & E^2 = e_{2,1}, e_{2,2}, e_{2,3}, \dots \\ & E^3 = e_{3,1}, e_{3,2}, e_{3,3}, \dots & \text{(where } i,j \in \mathbb{N} \text{ for each } e_{i,j} \text{)}\\ & \vdots \end{matrix}$$

Function $w$:
Let $w : \mathbb{N} \rightarrow \mathbb{N} \times \mathbb{N}$ be bijective such that the following order is used to enumerate $\mathbb{N} \times \mathbb{N}$:
$$\begin{matrix} \mathbb{N}_1 \text{ : } & 1_{1,1} & \rightarrow & 2_{1,2} & & 3_{1,3} & \rightarrow & 4 & \dots \\ & & \swarrow & & \nearrow & & \swarrow & & \nearrow \\ \mathbb{N}_2 \text{ : } & 1_{2,1} & & 2_{2,2} & & 3 & & 4 \\ & \downarrow & \nearrow & & \swarrow & & \nearrow\\ \mathbb{N}_3 \text{ : } & 1_{3,1} & & 2 & & 3 \\ & & \swarrow & & \nearrow \\ \mathbb{N}_4 \text{ : } & 1 & & 2 \\ \vdots & \downarrow & \nearrow \end{matrix}$$

Function $u$:
Let $u : \mathbb{N} \times \mathbb{N} \rightarrow (E^i)_{i \in \mathbb{N}}$ be bijective where $u(n_{i,j}) = e_{i,j}$.

Define the Sequence Q:
Let $Q = q_1, q_2, q_3, \dots$ such that each $q_{i} = u(w(i))$.

Add New Rule 24 for Iterations $>$ 1:
$$\text{Rule 24 : } 1, 2, 3, \dots \implies \{ q_1, q_2, q_3, \dots, q_{j-1} \}, \text{ where } j > 1$$

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#### AplanisTophet

Continuing from the Above Post After 24 Rules:

Let $T^{\omega_1}$ be a subsequence of $T$ containing only elements less than $\omega_1$.

We cannot have $sup(T^{\omega_1}) = \omega_1$, because then $T^{\omega_1}$ would be unbounded in $\omega_1$ (and then we have issues from earlier in the thread, continuum hypothesis being solved and all). Therefore, $sup (T^{\omega_1}) \in \omega_1$ and $|sup (T^{\omega_1})| = \omega$.

Where $T^{\omega_1}$ is unbounded in $sup(T^{\omega_1})$ by definition (needs verification), Rule 24 will ensure a sequence of limit ordinals is added to $T^{\omega_1}$ that is unbounded in $k(sup(T^{\omega_1}))$ by theorem 4. Further, the sequence of limit ordinals $k(sup(T^{\omega_1}))$ must contain an element greater than $sup(T^{\omega_1})$ by theorem 2, resulting in a contradiction.

I believe this means that after 24 Rules, $T^{\omega_1}$ cannot be unbounded in $sup(T^{\omega_1})$ if $sup(T^{\omega_1}) \in \omega_1$. That means $sup(T^{\omega_1})$ would be uncountable then (with confinality $\omega_1$), correct? Help Maschke!? If that is the case, then $sup(T^{\omega_1}) = \omega_1$, correct? But that also results in a contradiction (unless the continuum hypothesis is solved, whooopie!!!).

Oh boy, I'm stuck again.

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#### AplanisTophet

Anyone see a problem with any of these quick and easy theorems?

Theorem 2: If $X$ is some countable ordinal and $S^X = s_1, s_2, s_3, \dots$ is a (transfinite) sequence having order type $X$ of successive limit ordinals, then there exists an $s_x \in S^X$ where $s_x \geq X$.
I was thinking about this one a bit. What if $X = \omega^{\omega}$ or, equivalently, does $k(\omega^{\omega} + 1) = \omega, \omega \cdot 2, \omega \cdot 3, \dots, \omega^{\omega}, \omega^{\omega} + \omega$?

#### AplanisTophet

Anyone see a problem with any of these quick and easy theorems?

Theorem 2: If $X$ is some countable ordinal and $S^X = s_1, s_2, s_3, \dots$ is a (transfinite) sequence having order type $X$ of successive limit ordinals, then there exists an $s_x \in S^X$ where $s_x \geq X$.
I was thinking about this one a bit. What if $X = \omega^{\omega}$ or, equivalently, does $k(\omega^{\omega} + 1) = \omega, \omega \cdot 2, \omega \cdot 3, \dots, \omega^{\omega}, \omega^{\omega} + \omega$?
In exploring why my conjectured theorem 2 failed, I came across the Veblen functions and hierarchy:

https://en.wikipedia.org/wiki/Fixed-point_lemma_for_normal_functions

https://en.wikipedia.org/wiki/Veblen_function

I am still very much interested in what $sup(T^{\omega_1})$ equals after the first 23 rules. I'm not sure my 24th rule above added any new elements to the $T$ sequence.

Can anyone help with my original question? Is the following statement true:

If $X$ is a countable ordinal, then there exists an ordered triplet of ordinals $(a,b,c)$ in the class of all such triplets such that $f((a,b,c)) = X$ after 23 Rules.

#### AplanisTophet

Let's try this with a $T$ sequence enumerating only countable ordinals. Any help? This is easier. Thank you!

Question:

For any countable ordinal $X$, is there an ordered triplet of ordinals $(a,b,c)$ in the class of all ordered triplets of ordinals such that $X \in f((a,b,c))$ (where $f((a,b,c))$ is defined below)?

If so, then what is the first countable ordinal that doesn't appear in the sequence $T$ defined below and is that ordinal equal to $\{ t_i : t_i \in T\}$?

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

The primary means of generating a $T$ sequence 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 meeting a sequence of rules (see â€œRulesâ€ below).

Rules: Where $a, b, c, \dots$ are ordinals (comprised from the alphabet $\Sigma = \{0 \dots 9,\omega,<, +, \cdot\}$)
$$\text{Rule 1 : }a, a-1, a-2, \dots \implies \{ a-3 \} \text{, where } a \geq 3$$
$$\text{Rule 2 : }0, 1, a, \dots \implies \{a+1\}$$
$$\text{Rule 3 : }0, 2, a, \dots \implies \{a+2\}$$
$$\text{Rule 4 : }a, a+1, a+2, \dots \implies \{ a+3, a + \omega \}$$
$$\text{Rule 5 : }a \cdot b, a \cdot (b+1), a \cdot (b+2), \dots \implies \{ a \cdot (b+3), a \cdot (b + \omega) \}$$
$$\text{Rule 6 : }a^{b}, a^{b+1}, a^{b+2}, \dots \implies \{a^{b+3}, a^{b + \omega}\}$$
$$\text{Rule 7 : }a^{b \cdot c}, a^{b \cdot (c+1)}, a^{b \cdot (c+2)}, \dots \implies \{a^{b \cdot (c+3)}, a^{b \cdot (c + \omega)}\}$$
$$\text{Rule 8 : }a, s(a), s(s(a)), \dots \implies \{ s(s(s(a))), a^{b^{b^{b^{\vdots}}}}\} \text{, where } s(a) = a^b$$

Define function $f$:

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

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 shaves off any redundant elements of $B$ 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 $tâ€™_n = c_1, tâ€™_{n+2} = c_2, tâ€™_{n+4} = c_3, \dots$.

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

Assuming my calculations are correct (it gets tricky!!!), the first few elements of $T$ would be:
$$T = 1,2,3,0,4,5,\omega,6,7, \omega + 1, \omega +2,8,9, \omega + 3,\omega+4,\omega \cdot 2,10,11,\omega+5,\omega+6, \omega \cdot 2 + 1, \omega \cdot 2 + 2, \dots$$

The first iteration would generate $0$ based on Rule 1 and $(3,2,1)$. It would also generate $4$ and $5$ based on Rules 2 and 3 and $(1,2,3)$. Finally, it would generate $\omega$ based on Rule 4 and $(1,2,3)$. These are then added to $T$ to complete the first iteration.

#### Maschke

Let's try this with a $T$ sequence enumerating only countable ordinals. Any help? This is easier. Thank you!
Just wanted to mention why I'm not engaging.

* It's good that you're interested in the ordinals. However:

* Micrm@ss knows a lot more about ordinals than I do and he thinks your exposition makes no sense.

* From what little I've followed, your exposition makes no sense.

* You claim you've proved or disproved CH. When I challenged you on this point you did not respond. I see no connection between anything you've written and CH.

* Your idea of continuation is not well defined.

* There is no rule of 3 or pattern of 3. You're just hallucinating all of this as far as I can see; and more decisively, as far as Micrm@ss can see.

All the best.

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#### Maschke

ps -- Also part of my feelings about this thread involve a bit of disappointment. The ordinals are near and dear to my heart. Initially you had terrific insight that in the absence of the axiom of choice, the cofinality of $\omega_1$ might be (in some model) $\omega$. Saying it that way is equivalent to all the other stuff we've got going on in your initial argument. It's a good insight and I learned a lot and got new insight into the first uncountable ordinal, one of my favorite mathematical objects.

So it was with a great sense of personal disappointment that I watched you veer into ... well do you remember the movie Apocalypse Now, when the General sends Willard on a mission to assassinate Kurtz. The General says, of Kurtz: "His thinking has become unsound." That's how I felt about this thread. It started out really good then became ... unsound.

I'd much rather chat about the basics of ordinals. I like to contemplate the countable ordinals up to $\epsilon_0$, which is the largest countable ordinal I sort of grok. $\epsilon_0$ is the the upward limit of the successive power towers $\omega^\omega, \omega^{\omega^\omega}$, etc. That is, it's the set of all power towers with a finite number of powers.

We can visualize it as a countably infinite power tower of $\omega$'s. As such, you can see that if you added yet one more power, it would still be the same. That is,

$\omega^{\epsilon_0} = \epsilon_0$.

Define $f : On \to O$n as the class map (ie not a function on a set, but rather a "moral function" on the class of ordinals, denoted $On$) defined by $f(\alpha) =\omega^\alpha$. Then $\epsilon_0$ is the least fixed point of the ordinal exponential function $f$. That's equivalent to the other definition.

[Actually I've always been fuzzy as to how they legitimize functions on proper classes but I take it on faith].

The other factoid I know about $\epsilon_0$ is that Gerhard Gentzen proved in 1936 that Peano arithmetic is consistent if we can assume the validity of transfinite induction up to $\epsilon_0$. So it turns out that ordinals are useful in measuring the consistency strength of various theories.

https://en.wikipedia.org/wiki/Gentzen's_consistency_proof

Now, that's the kind of topic I had in mind when you showed up and said something remarkably insightful about the first uncountable ordinal. And then my hopes were dashed. I hope you can see why I haven't the heart or the interest to attempt to untangle your exposition.

I will say that you lost me way back when you introduced your, what do you call it, the function that finds the next ordinal given three others. You think your idea is obvious but in fact it's incoherent.

Well I hope you're having a good evening.

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