- #1

AplanisTophet

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- 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!