My Math Forum T Sequences - Explicit Enumeration of $\epsilon_0$

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October 17th, 2019, 08:45 AM   #61
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Quote:
 Originally Posted by Micrm@ss All sets in ZFC being countable isn't a wrong statement. It follows from the Lowenheim-Skolem-Tarski theorems and is known as Skolem's paradox. In any case, if you think you have a good proof of your assertions, go publish them in a journal.
That is really cool, thank you. However, Skolem's Paradox does not introduce an outright contradiction into set theory whereas $T^{\omega_1}$ being an enumeration of $\omega_1$ is an outright contradiction, so we aren't quite there yet. That is sooooooo close to being an explanation of what is happening here though as I understand it, only I go on to force the actual contradiction that philosophical discussion of the paradox focuses on:

"Philosophical discussion of this paradox has tended to focus on three main questions. First, there's a purely mathematical question: why doesn't Skolem's Paradox introduce an outright contradiction into set theory?"

PS - I'm fairly sure I'll just publish in public math forums, but if an actual mathematician ever wanted to work with my material and publish on my behalf, I would be open to it.... not that I ever expect that to happen.

 October 19th, 2019, 06:09 AM #62 Banned Camp   Joined: Jun 2014 From: USA Posts: 650 Thanks: 55 We are not quite able to make an explicit set of rules for a $T^{\omega_1}$ meeting my above criteria is what I have to assume. I would like to use this straightforward attempt for the purpose of discussing why there may not be an explicit way to define $\omega_1$ rules meeting my above criteria. I suggest that having a familiarity with normal functions and fixed points may be helpful: Generally, we are trying to find a function that takes any ordinal greater than $0$ and returns a specific triplet of ordinals that are all less than it in a fashion where no specific triplet in the class of all triplets is mapped to more than a finite number of times. This seems like it should be easy because there are $a^3$ triplets that can be formed from the elements of any ordinal $a$. Here is a fairly straightforward attempt. Let $Ord$ denote the class of all ordinals and let each $Ord_{\alpha}$ be a subclass of ordinals such that $(Ord_{\alpha})_{\alpha \in Ord}$ forms a partition of $Ord$. The transfinite sequence $(Ord_{\alpha})_{\alpha \in Ord}$ is defined as follows: $Ord_0$ is the class of ordinals that are not limit ordinals. $Ord_{\alpha} = (K, <)$, where $<$ is the standard ordering and $K = \{ x_j : (x_j, <)_{j \in Ord} \text{ well orders } Ord \setminus (\bigcup_{\kappa < \alpha} Ord_{\kappa}) \text{ and } j \text{ is not a limit ordinal} \}$. Note that we need $i,j < \alpha$ for each $(\alpha)_{i,j}$ to be successful. $$\begin{matrix} Ord_0 \text{ : } & 0_{0,0} & 1_{0,1} & 2_{0,2} & \dots & (\omega + 1)_{0,\omega} & (\omega + 2)_{0,\omega+1} & (\omega + 3)_{0,\omega+2} & \dots & (\omega \cdot 2 + 1)_{0,\omega \cdot 2} & (\omega \cdot 2 + 2)_{0,\omega \cdot 2 + 1} & (\omega \cdot 2 + 3)_{0,\omega \cdot 2 + 2} & \dots\\ Ord_1 \text{ : } & (\omega)_{1,0} & (\omega \cdot 2)_{1,1} & (\omega \cdot 3)_{1,2} & \dots & (\omega^2 + \omega)_{1,\omega} & (\omega^2 + \omega \cdot 2)_{1,\omega+1} & (\omega^2 + \omega \cdot 3)_{1,\omega+2} & \dots & (\omega^2 \cdot 2 + \omega)_{1,\omega \cdot 2} & (\omega^2 \cdot 2 + \omega \cdot 2)_{1,\omega \cdot 2 + 1} & (\omega^2 \cdot 2 + \omega \cdot 3)_{1,\omega \cdot 2 + 2} & \dots \\ Ord_2 \text{ : } & (\omega^2)_{2,0} & (\omega^2 \cdot 2)_{2,1} & (\omega^2 \cdot 3)_{2,2} & \dots & (\omega^3 + \omega^2)_{2,\omega} & (\omega^3 + \omega^2 \cdot 2)_{2,\omega + 1} & (\omega^3 + \omega^2 \cdot 3)_{2,\omega + 2} & \dots & (\omega^3 \cdot 2 + \omega^2)_{2,\omega \cdot 2} & (\omega^3 \cdot 2 + \omega^2 \cdot 2)_{2,\omega \cdot 2 + 1} & (\omega^3 \cdot 2 + \omega^2 \cdot 3)_{2,\omega \cdot 2 + 2} & \dots\\ \vdots \\ Ord_{\omega} \text{ : } & (\omega^{\omega})_{\omega,0} & (\omega^{\omega} \cdot 2)_{\omega,1} & (\omega^{\omega} \cdot 3)_{\omega,2} & \dots & (\omega^{\omega} \cdot \omega + \omega^{\omega})_{\omega,\omega} & (\omega^{\omega} \cdot \omega + \omega^{\omega} \cdot 2)_{\omega,\omega + 1} & (\omega^{\omega} \cdot \omega + \omega^{\omega} \cdot 3)_{\omega,\omega + 2} & \dots & (\omega^{\omega} \cdot \omega \cdot 2 + \omega^{\omega})_{\omega,\omega \cdot 2} & (\omega^{\omega} \cdot \omega \cdot 2 + \omega^{\omega} \cdot 2)_{\omega,\omega \cdot 2 + 1} & (\omega^{\omega} \cdot \omega \cdot 2 + \omega^{\omega} \cdot 3)_{\omega,\omega \cdot 2 + 2} & \dots\\ \vdots \\ \exists Ord_{\alpha} \text{ : } & (\alpha)_{\alpha,0} \in Ord_{\alpha} & <------ \text{Oops!!!!}\\ \vdots \end{matrix}$$ The following rules are for a $T$ sequence that enumerates some ordinal I believe. Does anyone know which one? A Veblen ordinal would be my initial guess perhaps...?: $$\text{Rule 1 : } \mu = 3, \beta = 2, \gamma = 1 \implies \{0\}$$ $$\text{Rule } \alpha \text{ : } \mu = 0, \beta = 1, \gamma = \alpha - 1 \implies \{\alpha\}, \text{ where } 2 \leq \alpha < \omega$$ $$\text{Rule } \alpha \text{ : } \mu = 0, \beta = u, \gamma = v \implies \{\alpha_{u,v}\}, \text{ where } \alpha \geq \omega$$ I think any discussion should probably address two questions: Is there, or is there not, an explicit way to create a set of rules meeting my above criteria? If not, is there a way to create a set of rules meeting my above criteria using the axiom of choice or some other non-explicit means? Thanks from idontknow
 October 20th, 2019, 05:35 AM #63 Banned Camp   Joined: Jun 2014 From: USA Posts: 650 Thanks: 55 Question: Where $\mu, \beta, \gamma, \alpha,$ and $\kappa$ are ordinals and $\alpha, \kappa > 0$, does there exist a function $\phi$ such that $\phi(\alpha) = (\mu, \beta, \gamma)$ where: 1) $\mu, \beta, \gamma < \alpha$, and 2) $\alpha \neq \kappa \implies \phi(\alpha) \neq \phi(\kappa)$ I'm trying to let this play out here because I love this place, but I'm sure someone at StackExchange would be happy to tackle this one if you guys are sleeping on the job... Last edited by AplanisTophet; October 20th, 2019 at 05:48 AM.
 October 21st, 2019, 05:49 AM #64 Banned Camp   Joined: Jun 2014 From: USA Posts: 650 Thanks: 55 Looks like Fodor's lemma may be helpful. I suspect we'll get some clarification soon on SE. Right now the question is on hold as I asked for a sec to look up Fodor's lemma myself after Andrés Caicedo directed me to it. Noah Schweber has seen this question too and he had helped me with a previous $T$ sequence formulation (that was quite awful, oops), but since Noah seems quite bright, that is also promising: https://math.stackexchange.com/quest...dinals#3401372
October 21st, 2019, 05:27 PM   #65
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Quote:
 Originally Posted by AplanisTophet The following rules are for a $T$ sequence that enumerates some ordinal I believe. Does anyone know which one?
Did anyone pick up on the fact that it enumerates $\epsilon_0$, and explicitly I might add? I have got to be the funniest crank ever...

Note that $Ord_{\epsilon_0} \implies (\omega^{\epsilon_0} = \epsilon_0)_{\epsilon_0,0} \in Ord_{\epsilon_0}$. That means $\epsilon_0$ is the first ordinal that won't appear in my $T$ sequence with the above rules because the triplet that allows for its inclusion in the sequence, $(0, \epsilon_0,0)$, is not comprised of ordinals less than $\epsilon_0$.

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