I don't understand your T sequence. I mean, I know what you want to do, but I don't understand what comes after the ... In order to answer your question, I'd have to see a general rule for T: describe T without the ... Otherwise I find it too ambiguous.

Also, when you write sequence, you should know that the usual connotation is that it has countably many elements. I don't know if this is in fact your intention. If it is good. If it is not, you should write transfinite sequence.

I suppose we could try to lay down some a, b, câ€™s when it comes to our $1, 2, 3\text{â€˜s}$. That is, we should probably start by letting the alphabet for the ordinals be the usual $\Sigma = \{0, \dots, 9, \omega, <, +, \cdot \}$. :giggle:

We can start the sequence $T$ by simply saying $t_1, t_2,$ and $t_3$ equate to $1, 2,$ and $3$, respectively. For each $t_n$ thereafter, we are going to be referring to the previous elements of $T$ in order to define $t_n$. The first option will always be $t_n = t_{n-1} + 1$, but this option is overridden when the previous elements of the sequence provide us with some direction.

When $n = 4$, the previous elements equal $1, 2,$ and $3$. All possible three-member orderings of the previous elements (the number of which is finite for any $t_i$) are:

$$1, 2, 3, \dots$$

$$1, 3, 2, \dots$$

$$2, 1, 3, \dots$$

$$2, 3, 1, \dots$$

$$3, 1, 2, \dots$$

$$3, 2, 1, \dots$$

The universal continuation of $1, 2, 3, \dots$ seems to be $\omega$ in this context, but it could also be $4$. Itâ€™s of no matter because we can accept both (so long as they are ordinals). We also have $3, 2, 1, \dots$ continuing to $0$. We can start to come up with some obvious and basic â€œrulesâ€ that will help us define $T$. For this purpose, let $a, b, c, \dots$ be ordinals so that, e.g.:

$$1, 2, 3, \dots \implies \omega$$

$$a+1, a+2, a+3, \dots \implies a + \omega \text{ or, similarly, } a, a+1, a+2 \implies a + \omega \text{ too.}$$

$$a, a^2, a^3, \dots \implies a^{\omega}$$

$$a^a, a^{a^a}, a^{a^{a^a}}, \dots \implies a^{a^{a^{a^{\vdots}}}}$$

$$a_a, a_{a_a}, a_{a_{a_a}}, \dots \implies a_{a_{a_{a_{\vdots}}}}$$

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The further our sequence continues the more rules will exist perhaps, but they will all be defined from previous elements of the sequence, so the fact that our list of rules will have to keep growing is of no concern.

In the case of $t_4$, we find that $0, 4,$ and $\omega$ are all implied per above, so we then go by their natural order and set $t_4 = 0$, $t_5 = 4$, and $t_6 = \omega$. We then go to $t_7$ and repeat the process.

$$T = 1, 2, 3, 0, 4, \omega, \dots$$

I believe it would be impossible for the same three-member sequence to imply more than a finite number of new ordinals be added to the sequence at any given stage. Iâ€™ll note that it would be possible to make this work so long as only a countable number of new ordinals are added if not just a finite number.

Does that help to define $T$ better? Can we go back to my questions here?:

Anyone notice that we could have $3, 2, 1, \implies 0$ too? Knock yourselves out... Anyone figure out the above? What is $sup \bigcup T$? Is it greater than $\omega_1$ or maybe it equals $\omega_1^{CK}$? I'm really confused.

Also, Maschke, I'm having trouble understanding KÃ¶nig's Theorem. I was thinking I could apply it here. Can I?:

https://en.wikipedia.org/wiki/König's_theorem_(set_theory)

I really do appreciate the help and I find this all interesting. I hope you do too.