September 8th, 2019, 12:48 AM  #51  
Senior Member Joined: Oct 2009 Posts: 865 Thanks: 328  Quote:
$$f:A\rightarrow \bigcup A$$ such that $f(x)\in x$. Is this what you mean? If so, what is the $A$ here? If not, what do you mean with "the choice function across ordinal indexes $i$ for each countable $s_i$"? Define it please.  
September 8th, 2019, 03:40 PM  #52 
Senior Member Joined: Jun 2014 From: USA Posts: 569 Thanks: 43 
I think you found it. I was thinking that if all $s_i$ are countable (which adheres to the definition of $\omega_1$ as you point out) there would have to be some uncountable $i$ before the countable union of countable sets theorem could be negated. This is my first take. I’ll think about it a little more on and off too. That was tricky and I definitely have learned a lot (e.g., confinality kicking in again, and I hadn’t heard of the term before this thread). Thank you.

September 9th, 2019, 02:18 AM  #53  
Senior Member Joined: Jun 2014 From: USA Posts: 569 Thanks: 43  Quote:
I am curious as to trying at least one sequence here. It’s going to be based on what I’ll call the rule of three. That is, every time the previous elements of the sequence show a pattern of three ordinals that we could typically follow with "$\dots$" and have it be universally understood as implying the standard continuation of the ordinals’ pattern (e.g., 1, 2, 3 appearing anywhere in the sequence implies that $\omega$ would appear; $\omega^\omega, \omega^{\omega^\omega}, \omega^{\omega^{\omega^\omega}}, \dots \implies $ the existence of $\omega^{\omega^{\omega^{\omega^{\vdots}}}}$, etc.) we’ll jump a limit ordinal: $$\text{Call this sequence } T \text{ for “testing.”}$$ $$T = 1, 2, 3, \omega, \omega+1, \omega+2, \omega \cdot 2, \omega \cdot 2 + 1, \omega \cdot 2 + 2, \omega \cdot 3, \omega^2, …$$ I trust you see how I arrive at $\omega^2$, $\omega^{\omega^{\omega^{\omega^{\vdots}}}}$, etc? As long as the pattern is obvious and could be rigorously defined should there be sufficient motivation, I'm happy here. What is the first $i \in \omega_1$, call it $i'$, such that $i'$ is provably greater than every element of the above sequence? If there is no such $i' \in \omega_1$ given the statement that “every $i \in \omega_1$ is countable” then we have issues. I suspect there must be such an $i'$ though as indicated, so how to solve for it or prove it exists given the statement that such an $i'$ must be itself countable? Anyone know? Last edited by AplanisTophet; September 9th, 2019 at 02:20 AM.  
September 11th, 2019, 05:19 AM  #54 
Senior Member Joined: Jun 2014 From: USA Posts: 569 Thanks: 43 
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 König's Theorem. I was thinking I could apply it here. Can I?: https://en.wikipedia.org/wiki/K%C3%B...em_(set_theory) I really do appreciate the help and I find this all interesting. I hope you do too. 
September 11th, 2019, 05:31 AM  #55  
Senior Member Joined: Oct 2009 Posts: 865 Thanks: 328  Quote:
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.  
September 11th, 2019, 09:37 AM  #56  
Senior Member Joined: Jun 2014 From: USA Posts: 569 Thanks: 43  Quote:
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 $t_n = t_{n1} + 1$, 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 threemember 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 $4$. It’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 “rules” that will help us define $T$. For this purpose, let $a, b, c, \dots$ be ordinals so that, e.g.: $$1, 2, 3, \dots \implies \omega$$ $$a+1, a+2, a+3, \dots \implies a + \omega \text{ or, similarly, } a, a+1, a+2 \implies a + \omega \text{ too.}$$ $$a, a^2, a^3, \dots \implies a^{\omega}$$ $$a^a, a^{a^a}, a^{a^{a^a}}, \dots \implies a^{a^{a^{a^{\vdots}}}}$$ $$a_a, a_{a_a}, a_{a_{a_a}}, \dots \implies a_{a_{a_{a_{\vdots}}}}$$ . . . 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 threemember sequence to imply more than a finite number of new ordinals be added to the sequence at any given stage. I’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?: Quote:
 
September 11th, 2019, 10:56 AM  #57 
Senior Member Joined: Jun 2014 From: USA Posts: 569 Thanks: 43 
PS  I think we could argue that $\omega +1$ appears with my more refined definition of $T$, but to be certain we could instead choose to only add the greatest ordinal implied by the previous ones (so only add $\omega$ after $1,2,$ and $3$ instead of $0$ and $4$ too). Either way works. I want it to grow, obviously, and not get bogged down in the little ordinals. If an element of $T$ were to increase past $\omega_1$, I would want the function to have the ability to go backwards too, which is why I pointed out $3,2,1 \implies 0$. We can set it up either way. That said, to answer my questions, we just need to allow for the addition of the ordinals that make $T$ grow and, if you tell me it's greater than $\omega_1$, I would then make it so we can accommodate the $0$ and the $4$, etc. 
September 12th, 2019, 07:55 AM  #58 
Senior Member Joined: Jun 2014 From: USA Posts: 569 Thanks: 43 
E.g., we would have $0,1,\omega \implies \omega + 1$, so my refined definition of $T$ would still get there... I approach this as though the set of all $(a,b,c, d)$, where all are ordinals and $d$ is a universally agreeable continuation of the pattern made by $a,b,c$, is what it is as opposed to being something for me to define. Do you not agree? 
September 12th, 2019, 08:40 AM  #59  
Senior Member Joined: Oct 2009 Posts: 865 Thanks: 328  Quote:
 
September 12th, 2019, 03:31 PM  #60  
Senior Member Joined: Jun 2014 From: USA Posts: 569 Thanks: 43  Quote:
I think the sequence $T$ would be a good one for the lounge perhaps, where people could take turns posting the next element. If you change your mind and find you can actually understand what I'm talking about, let me know! I am sure you have better things to do in general (I do too), but maybe you'll be bored some day and revisit. Thanks again.  

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