 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 $\epsilon_0$. Tags $epsilon0$, enumeration, explicit, sequences Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post mahjk17 Applied Math 4 June 27th, 2012 08:46 PM arron1990 Calculus 12 January 31st, 2012 05:17 AM techmix Math Software 0 September 24th, 2011 12:12 PM khyratmath123 Math Software 1 January 27th, 2010 05:28 PM khyratmath123 Calculus 4 January 10th, 2010 06:16 AM

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