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November 3rd, 2019, 02:22 AM  #11 
Senior Member Joined: Jun 2014 From: USA Posts: 644 Thanks: 55 
I assume the whole thing is too much for anyone to want to read, so I'll make it easier. Does anyone see a problem with the following definitions for function $t$ and the functions $(\phi_{\alpha})_{2\leq\alpha<\omega_1}$? Does anyone familiar with Fodor's lemma care to discuss whether there must be a minimum element $\kappa \in \omega_1$ where $\kappa \in \phi_{\omega}(\kappa)$? Define $t(\alpha)$ for any ordinal $\alpha \geq 2$: Let $t(\alpha)$ equal a doublet of variables $(a,b)$ if $\alpha = 2$, a triplet of variables $(a,b,c)$ if $\alpha = 3$, a quadrulplet of variables $(a,b,c,d)$ if $\alpha = 4$, and so on, for any ordinal $\alpha$. Note that this notation is used to avoid confusion as $a=b$ does not imply $(a,b) = (b,a) = (a) = (b)$, whereas it does imply $\{a,b\} = \{b,a\} = \{a\} = \{b\}$ in general. Define $(\phi_{\alpha})_{2 \leq \alpha < \omega_1}$: Let each element of $(\phi_{\alpha})_{2 \leq \alpha < \omega_1}$ be almost regressive such that: 1) $\phi_{\alpha} : \omega_1 \setminus \{0\} \rightarrow \{ t(\alpha) : a,b,c,\dots \in t(\alpha) \implies a,b,c,\dots < \omega_1 \}$ is bijective, 2) $a,b,c,\dots \leq \kappa$ for each $a,b,c,\dots \in \phi_{\alpha}(\kappa)$, and 3) $\zeta < \alpha \implies \min\{ \phi_{\zeta}^{1}(b) : \exists k \in b \text{ where } k \geq \phi_{\zeta}^{1}(b)\} < \min\{ \phi_{\alpha}^{1}(b) : \exists k \in b \text{ where } k \geq \phi_{\alpha}^{1}(b)\}$. Last edited by skipjack; November 3rd, 2019 at 03:20 AM. Reason: grammar 
November 4th, 2019, 04:23 AM  #12 
Senior Member Joined: Jun 2014 From: USA Posts: 644 Thanks: 55 
I wish Maschke knew more about ordinals... Hey ISP, could you do what Maschke usually does? 
November 4th, 2019, 04:41 AM  #13 
Senior Member Joined: Oct 2009 Posts: 905 Thanks: 354  
November 4th, 2019, 04:49 AM  #14  
Newbie Joined: Oct 2019 From: SV USA Posts: 5 Thanks: 1  Quote:
Your notation stinks bud. This whole function $t$ thing is just ****ing goofy to start. People think $t(2) = (a,b)$ where $a$ and $b$ are distinct, but what you mean is for $a$ and $b$ to take a range that you don't introduce until your definition of the functions $\phi$. You mean for $a$ and $b$ to be bounded by $\omega_1$ in the case of doublets. In your definition of each $\phi_{\alpha}$, you use a set $\{ t(\alpha) : a,b,c,\dots \in t(\alpha) \implies a,b,c,\dots < \omega_1 \}$. In English, what you mean in the case of $\phi_2$ is that it is a bijection from $\omega_1 \setminus \{0\}$ onto the set of all doublets such that each doublet contains only elements of $\omega_1$. Your functions $f$, $k$, and $h$ all rely on the definition of function $t$, but given the above clarification, they become sensible if you clarify what you mean by $t^{1}$. What you mean to assert, generally, is that if $x$ is a doublet then $t^{1}(x) = 2$. If $x$ is a triplet, then $t^{1}(x) = 3$, and so on. I understand why you will want to keep your notation. You are looking for a way to make it so others understand it without having to change it. You shouldn't want to keep notation that others can't understand. (Full disclosure: I am AplanisTophet, and yes I will eventually switch to this username full time. For now, ISP can stand for internet sock puppet, but given my full disclosure please tolerate any silly conversations I may have with myself for this thread only. I am of course lonely here in this forum, listening to some good tunes.)  
November 4th, 2019, 04:51 AM  #15  
Senior Member Joined: Jun 2014 From: USA Posts: 644 Thanks: 55  Quote:
It's not an insult. There's a lot I don't know about a lot of things. I'm no mechanic for example. Why care if someone knows about ordinals?  
November 4th, 2019, 06:38 AM  #16 
Senior Member Joined: Jun 2014 From: USA Posts: 644 Thanks: 55 
Assuming my sock puppet doesn't come through... https://math.stackexchange.com/quest...ffodorslemma Above, Andrés E. Caicedo helps clarify the notation and points out that an assumption as to CH may be necessary for determining whether there must be some minimum $\kappa \in \omega_1$ where $\kappa \in \phi_{\omega}(\kappa)$. I assume we'll get an answer there. Notation Guide: To understand the use of function $t$ within the definition of each function $\phi_{\alpha}$, consider first $\phi_2$. The set of all doublets, where each element of each doublet is an element of $\omega_1$, is the English translation for: $$\{ t(2) : a,b,c,\dots \in t(2) \implies a,b,c,\dots < \omega_1 \} = \{ (a,b) : a,b \in \omega_1 \}$$ The function $\phi_2$ therefore bijects $\omega_1 \setminus \{0\}$ onto $\{(0,0),(0,1),(1,0),(1,1),(0,2),(a \in \omega_1, b \in \omega_1),...\}$. The function $t^{1}(x)$ will return $2$ for any doublet $x$, $3$ for any triplet $x$, $4$ for any quadruplet $x$, and so on depending on the order type of $x$. 
November 6th, 2019, 04:57 AM  #17 
Senior Member Joined: Jun 2014 From: USA Posts: 644 Thanks: 55 
This version is meant to avoid assuming the continuum hypothesis and clean up the notation. As usual, if I'm posting it here for the first time, it's a work in progress so feel free to say how awful it is. I hope I at least fixed the notation. Where $a,b$ are ordinals: 1) $a=b$ does not imply $(a,b) = (a) = (b)$. 2) $a \neq b$ does not imply $(a,b) = (b,a)$. Define $t(\alpha)$ for any ordinal $\alpha \geq 2$: Let $t(\alpha)$ equal a doublet of ordinals $(a,b)$ if $\alpha = 2$, a triplet of ordinals $(a,b,c)$ if $\alpha = 3$, a quadruplet of ordinals $(a,b,c,d)$ if $\alpha = 4$, and so on, for any ordinal $\alpha$. Similarly, let $t^{1}(x)$ equal $2$ for any doublet of ordinals $x$, $3$ for any triplet of ordinals $x$, $4$ for any quadruplet of ordinals $x$, and so on as determined by the order type of $x$. Use of set builder notation: Consider the set of all doublets of ordinals such that each element of each doublet is a member of $\omega_1$. It will become helpful to use setbuilder notation in the following manner to define such a set: $$\{t(2) : a,b \in t(2) \implies a,b \in \omega_1 \} = \{ (a,b) : a,b \in \omega_1 \} = \{(0,0),(0,1),(1,0),(a \in \omega_1,b \in \omega_1),\dots\}$$ Define the sets $P$ and $P'$: $$P = \{ t(\omega) : a,b,c,\dots \in t(\omega) \implies a,b,c,\dots \, \text{is a computable sequence}, a \neq b \neq c \neq \dots, \, \text{and } a,b,c,\dots \in \omega \} \equiv P' = \{ p \in \mathcal{P}(\omega) : p = \aleph_0 \, \text{and } p \, \text{is computable} \}$$ Define the functions $(v_{\alpha})_{\alpha \in \omega_1}$ over index $P'$: $$v_{\alpha}(p) = \{ \omega \cdot \alpha + p' : p' \in p \}, \text{ for any } p \in P', \alpha \in \omega_1$$ Define the set $Q$: $$Q = \{ v_{\alpha}(p) : p \in P' \, \text{and } \alpha \in \omega_1 \}$$ Define the functions $(r_{\alpha})_{\omega < \alpha < \omega_1}$: Let $r_{\alpha} : \alpha \rightarrow \omega$ be bijective. Define 'Likewise Computable': Let $t(\alpha) = ((\omega \cdot \beta + a)_0, (\omega \cdot \beta + b)_1, (\omega \cdot \beta + c)_2, \dots)$ be likewise computable for any ordinal $\alpha$ if and only if $\omega < \alpha < \omega_1$, $\beta \in \omega_1$, and an $\omega$type ordering by index $( (\gamma)_{r_{\alpha}^{1}(0)}, (\beta)_{r_{\alpha}^{1}(1)}, (\mu)_{r_{\alpha}^{1}(2)},\dots )$ of the $\alpha$type ordering $( a_{r_{\alpha}(0)}, b_{r_{\alpha}(1)}, c_{r_{\alpha}(2)},\dots )$ is an element of $P$. Define $(\phi_{\alpha})_{2 \leq \alpha < \omega_1}$: Let each element of $(\phi_{\alpha})_{2 \leq \alpha < \omega_1}$ be almost regressive such that: 1) $\alpha < \omega \implies \phi_{\alpha} : \omega_1 \setminus \{0\} \rightarrow \{ t(\alpha) : a,b,c,\dots \in t(\alpha) \implies a,b,c,\dots < \omega_1 \}$ is bijective, 2) $\alpha \geq \omega \implies \phi_{\alpha} : \omega_1 \setminus \{0\} \rightarrow \{ t(\alpha) : a,b,c,\dots \in t(\alpha) \implies a,b,c,\dots < \omega_1 \, \text{and } (t(\alpha) \in P \, \text{or } t(\alpha) \, \text{is likewise computable})\}$ is bijective, 3) $a,b,c,\dots \leq \kappa$ for each $a,b,c,\dots \in \phi_{\alpha}(\kappa)$, and 4) $\zeta < \alpha \implies \min\{ \phi_{\zeta}^{1}(b) : \exists k \in b \text{ where } k \geq \phi_{\zeta}^{1}(b)\} < \min\{ \phi_{\alpha}^{1}(b) : \exists k \in b \text{ where } k \geq \phi_{\alpha}^{1}(b)\}$. Define function $f$: $$f(x) = \phi_{t^{1}(x)}^{1}(x), \, \text{where, given } x \, \text{has order type } \alpha, 2 \leq \alpha < \omega, x \in P, \, \text{or } x \, \text{is likewise computable}$$ Define $k(\alpha)$ for any ordinal $\alpha$: $$k(\alpha) = \{ x : f(x) = \alpha \}$$ Define $h(\alpha)$ for any ordinal $\alpha$: $$h(\alpha) = min\{ t^{1}(x) : x \in k(\alpha) \text{ and } \forall y \in x(y < \alpha) \}$$ Define function $g$: For any set of ordinals, $A$, let $g(A)$ be the set of all ordered doublets, triplets, quadruplets, and so on, that can be comprised from the elements of $\omega_1 \cap A$: $$g(A) = \{ t(\alpha) : t(\alpha) \setminus (A \cap \omega_1) = \emptyset \text{ and } 2 \leq \alpha < \omega_1 \}$$ Define the sequence $T$: Define a (potentially transfinite) sequence $T = t_1, t_2, t_3, \dots$ over $\omega_1$ iterations where: Step 1) Let $t_1 = 0, t_2 = 1, t_3 = 2$, and iteration counter $m = 1$. Step 2) Each $t_n$, where $n \geq 4$, is defined by the previous elements of the sequence. Starting with $n = 4$: a) If $m$ is a countable limit ordinal and $T$ is of order type $\omega$, free up space in $T$ by first letting $(s_n)_{n \in \mathbb{N}}$ be defined for odd indexes and undefined for even indexes: $s_{n \cdot 2  1} = t_{n}$. Then, set $t_1 = s_1, t_2 = s_2, t_3 = s_3, \dots$. Finally, if $m = \omega$, set $t_j$ undefined for any index $j > i$ where $t_j = t_i$ and $i,j < \omega$. b) Let $A = \{ t_i \in T : i < n \}$. E.g., $A = \{0, 1, 2 \}$ on the first iteration. c) Let $B = \{ f(x) : x \in (g(A) \setminus \{ x \in g(A) : t^{1}(x) > h(f(x))\} ) \}$. Using the previous elements of the sequence $A$, this step creates a set $B$ of all the new ordinals implied by letting function $f$ range over $g(A)$. The restriction of $g(A)$ to $A \cap \omega_1$ ensures that $B$ remains countable for this particular $T$ sequence. d) Let $C = B \setminus A$ and let $c_1, c_2, c_3, \dots$ be an enumeration of $C$ that is also well ordered if $C \neq \aleph_0$. This step removes any redundant elements from $B$ before potentially well ordering them so that we can add them to $T$. e) If $C < \aleph_0$ or if a transfinite $T$ is desired, set $t_n = c_1, t_{n+1} = c_2, t_{n+2} = c_3, \dots$ and then proceed to step 3. If a transfinite $T$ is not desired, proceed to substep f. f) Let $T’ = t’_1, t’_2, t’_3, \dots$ be a subsequence of the remaining undefined elements of $T$ and set $t’_1 = c_1, t’_3 = c_2, t’_5 = c_3, \dots$. Step 3) Let $j$ be the first ordinal such that $t_j$ is undefined. If $j>n$, set $n = j$. Increase the iteration counter by letting $m = m + 1$, and then repeat step 2. 
November 6th, 2019, 06:32 AM  #18 
Senior Member Joined: Jun 2014 From: USA Posts: 644 Thanks: 55 
This version of the question is meant to avoid assuming the continuum hypothesis (as is done in the link below) and clarify the notation. https://math.stackexchange.com/quest...ffodorslemma Question: Based on the definitions below for function $t$ and the functions $(\phi_{\alpha})_{2\leq\alpha<\omega_1}$, must there be a minimum element $\kappa \in \omega_1$ where $\kappa \in \phi_{\omega}(\kappa)$? Note that there is some $\kappa \in \omega_1$ where $\kappa \in \phi_{\alpha}(\kappa)$ for each $\alpha < \omega$ due Fodor's lemma (Fodor would have each element of $\{ \kappa : \kappa \in \phi_{\alpha}(\kappa) \}$ map to a constant for each $\alpha$ in $\omega$ should $\phi_{\alpha}$ be fully regressive instead of just almost regressive). Where $a,b$ are ordinals: 1) $a=b$ does not imply $(a,b) = (a) = (b)$. 2) $a \neq b$ does not imply $(a,b) = (b,a)$. Define $t(\alpha)$ and $t^{1}(\alpha)$ for any ordinal $\alpha \geq 2$: Let $t(\alpha)$ equal a doublet of ordinals $(a,b)$ if $\alpha = 2$, a triplet of ordinals $(a,b,c)$ if $\alpha = 3$, a quadruplet of ordinals $(a,b,c,d)$ if $\alpha = 4$, and so on, for any ordinal $\alpha$. Similarly, let $t^{1}(x)$ equal $2$ for any doublet of ordinals $x$, $3$ for any triplet of ordinals $x$, $4$ for any quadruplet of ordinals $x$, and so on, as determined by the order type of $x$. Use of set builder notation: Consider the set of all doublets of ordinals such that each element of each doublet is a member of $\omega_1$. It will become helpful to use setbuilder notation in the following manner to define such a set: $$\{t(2) : a,b \in t(2) \implies a,b \in \omega_1 \} = \{ (a,b) : a,b \in \omega_1 \} = \{(0,0),(0,1),(1,0),(a \in \omega_1,b \in \omega_1),\dots\}$$ Define the sets $P$ and $P'$: $$P = \{ t(\omega) : a,b,c,\dots \in t(\omega) \implies a,b,c,\dots \, \text{is a computable sequence}, a \neq b \neq c \neq \dots, \, \text{and } a,b,c,\dots \in \omega \} \equiv P' = \{ p \in \mathcal{P}(\omega) : p = \aleph_0 \, \text{and } p \, \text{is computable} \}$$ Define the functions $(v_{\alpha})_{\alpha \in \omega_1}$ over index $P'$: $$v_{\alpha}(p) = \{ \omega \cdot \alpha + p' : p' \in p \}, \text{ for any } p \in P', \alpha \in \omega_1$$ Define the set $Q$: $$Q = \{ v_{\alpha}(p) : p \in P' \, \text{and } \alpha \in \omega_1 \}$$ Define the functions $(r_{\alpha})_{\omega < \alpha < \omega_1}$: Let $r_{\alpha} : \alpha \rightarrow \omega$ be bijective. Define 'Likewise Computable': Let $t(\alpha) = ((\omega \cdot \beta + a)_0, (\omega \cdot \beta + b)_1, (\omega \cdot \beta + c)_2, \dots)$ be likewise computable for any ordinal $\alpha$ if and only if $\omega < \alpha < \omega_1$, $\beta \in \omega_1$, and an $\omega$type ordering by index $( (\gamma)_{r_{\alpha}^{1}(0)}, (\zeta)_{r_{\alpha}^{1}(1)}, (\mu)_{r_{\alpha}^{1}(2)},\dots )$ of the $\alpha$type ordering $( a_{r_{\alpha}(0)}, b_{r_{\alpha}(1)}, c_{r_{\alpha}(2)},\dots )$ is an element of $P$. Define the functions $(\phi_{\alpha})_{2 \leq \alpha < \omega_1}$: Let each element of $(\phi_{\alpha})_{2 \leq \alpha < \omega_1}$ be almost regressive such that: 1) $\alpha < \omega \implies \phi_{\alpha} : \omega_1 \setminus \{0\} \rightarrow \{ t(\alpha) : a,b,c,\dots \in t(\alpha) \implies a,b,c,\dots \in \omega_1 \}$ is bijective, 2) $\alpha \geq \omega \implies \phi_{\alpha} : \omega_1 \setminus \{0\} \rightarrow \{ t(\alpha) : a,b,c,\dots \in t(\alpha) \implies a,b,c,\dots \in \omega_1 \, \text{and } (t(\alpha) \in P \, \text{or } t(\alpha) \, \text{is likewise computable})\}$ is bijective, 3) $a,b,c,\dots \leq \kappa$ for each $a,b,c,\dots \in \phi_{\alpha}(\kappa)$, and 4) $\zeta < \alpha \implies \min\{ \phi_{\zeta}^{1}(b) : \exists k \in b \text{ where } k \geq \phi_{\zeta}^{1}(b)\} < \min\{ \phi_{\alpha}^{1}(b) : \exists k \in b \text{ where } k \geq \phi_{\alpha}^{1}(b)\}$. Define function $f$: $$f(x) = \begin{cases} \phi_{t^{1}(x)}^{1}(x) & \text{if, given } x \, \text{has order type } \alpha, 2 \leq \alpha < \omega, x \in P, \, \text{or } x \, \text{is likewise computable} \\ \text{empty string} & \text{otherwise} \\ \end{cases}$$ Define $k(\alpha)$ for any ordinal $\alpha \in \omega_1$: $$k(\alpha) = \{ x : f(x) = \alpha \}$$ Define function $h$: $$h(\alpha) = \begin{cases} min\{ t^{1}(x) : x \in k(\alpha) \text{ and } \forall y \in x(y < \alpha) \} & \text{if } \alpha \in \omega_1 \\ 1 & \text{otherwise} \\ \end{cases}$$ Define function $g$: For any set of ordinals, $A$, let $g(A)$ be the set of all ordered doublets, triplets, quadruplets, and so on, that can be comprised from the elements of $\omega_1 \cap A$: $$g(A) = \{ t(\alpha) : t(\alpha) \setminus (A \cap \omega_1) = \emptyset \text{ and } 2 \leq \alpha < \omega_1 \}$$ Define the sequence $T$: Define a (potentially transfinite) sequence $T = t_1, t_2, t_3, \dots$ over $\omega_1$ iterations where: Step 1) Let $t_1 = 0, t_2 = 1, t_3 = 2$, and iteration counter $m = 1$. Step 2) Each $t_n$, where $n \geq 4$, is defined by the previous elements of the sequence. Starting with $n = 4$: a) If $m$ is a countable limit ordinal and $T$ is of order type $\omega$, free up space in $T$ by first letting $(s_n)_{n \in \mathbb{N}}$ be defined for odd indexes and undefined for even indexes: $s_{n \cdot 2  1} = t_{n}$. Then, set $t_1 = s_1, t_2 = s_2, t_3 = s_3, \dots$. Finally, if $m = \omega$, set $t_j$ undefined for any index $j > i$ where $t_j = t_i$ and $i,j < \omega$. b) Let $A = \{ t_i \in T : i < n \}$. E.g., $A = \{0, 1, 2 \}$ on the first iteration. c) Let $B = \{ f(x) : x \in (g(A) \setminus \{ x \in g(A) : t^{1}(x) > h(f(x))\} ) \}$. Using the previous elements of the sequence $A$, this step creates a set $B$ of all the new ordinals implied by letting function $f$ range over $g(A)$. The restriction of $g(A)$ to $A \cap \omega_1$ ensures that $B$ remains countable for this particular $T$ sequence. d) Let $C = B \setminus A$ and let $c_1, c_2, c_3, \dots$ be an enumeration of $C$ that is also well ordered if $C \neq \aleph_0$. This step removes any redundant elements from $B$ before potentially well ordering them so that we can add them to $T$. e) If $C < \aleph_0$ or if a transfinite $T$ is desired, set $t_n = c_1, t_{n+1} = c_2, t_{n+2} = c_3, \dots$ and then proceed to step 3. If a transfinite $T$ is not desired, proceed to substep f. f) Let $T’ = t’_1, t’_2, t’_3, \dots$ be a subsequence of the remaining undefined elements of $T$ and set $t’_1 = c_1, t’_3 = c_2, t’_5 = c_3, \dots$. Step 3) Let $j$ be the first ordinal such that $t_j$ is undefined. If $j>n$, set $n = j$. Increase the iteration counter by letting $m = m + 1$, and then repeat step 2. Last edited by AplanisTophet; November 6th, 2019 at 06:35 AM. Reason: Add Question 
November 6th, 2019, 10:11 AM  #19  
Senior Member Joined: Jun 2014 From: USA Posts: 644 Thanks: 55 
The following definitions need a slight tweaking for $\phi_{\omega}$ to be bijective (they need to be defined over $\omega \leq \alpha < \omega_1$ instead of merely over $\omega < \alpha < \omega_1$): Quote:
Define the functions $(r_{\alpha})_{\omega \leq \alpha < \omega_1}$: Let $r_{\alpha} : \alpha \rightarrow \omega$ be bijective. Define 'Likewise Computable': Let $t(\alpha) = ((\omega \cdot \beta + a)_0, (\omega \cdot \beta + b)_1, (\omega \cdot \beta + c)_2, \dots)$ be likewise computable for any ordinal $\alpha$ if and only if $\omega \leq \alpha < \omega_1$, $\beta \in \omega_1$, and an $\omega$type ordering by index $( (\gamma)_{r_{\alpha}^{1}(0)}, (\zeta)_{r_{\alpha}^{1}(1)}, (\mu)_{r_{\alpha}^{1}(2)},\dots )$ of the $\alpha$type ordering $( a_{r_{\alpha}(0)}, b_{r_{\alpha}(1)}, c_{r_{\alpha}(2)},\dots )$ is an element of $P$.  
November 8th, 2019, 07:47 PM  #20  
Senior Member Joined: Jun 2014 From: USA Posts: 644 Thanks: 55 
The above needs to be changed because $\phi_{\omega}$ and higher cannot be bijective as written. Glad to see you guys read it! Quote:
Define the set $P$: $$P = \{ t(\omega) : a,b,c,\dots \in t(\omega) \implies a,b,c,\dots \, \text{is a computable sequence} \, \text{and } a,b,c, \dots \in \{0,1\} \} \equiv \, \text{the set of computable binary sequences}$$ Define the functions $(r_{\alpha})_{\omega \leq \alpha < \omega_1}$: Let $r_{\alpha} : \alpha \rightarrow \omega$ be bijective. Define 'Likewise Computable': Let $t(\alpha) = ((\beta + a)_0, (\beta + b)_1, (\beta + c)_2, \dots)$ be likewise computable for any ordinal $\alpha$ if and only if $\omega \leq \alpha < \omega_1$, $\beta \in \omega_1$, and an $\omega$type ordering by index $( (\gamma)_{r_{\alpha}^{1}(0)}, (\zeta)_{r_{\alpha}^{1}(1)}, (\mu)_{r_{\alpha}^{1}(2)},\dots )$ of the $\alpha$type ordering $( a_{r_{\alpha}(0)}, b_{r_{\alpha}(1)}, c_{r_{\alpha}(2)},\dots )$ is an element of $P$.  

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