October 17th, 2019, 08:45 AM  #61  
Banned Camp Joined: Jun 2014 From: USA Posts: 650 Thanks: 55  Quote:
https://plato.stanford.edu/entries/paradoxskolem/ "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 nonexplicit means? 
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 AndrĂ©s 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  
Banned Camp Joined: Jun 2014 From: USA Posts: 650 Thanks: 55  Quote:
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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$epsilon0$, enumeration, explicit, sequences 
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