My Math Forum Having Sum Fun Counting Ordinals

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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
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Quote:
 Originally Posted by AplanisTophet I wish Maschke knew more about ordinals... Hey ISP, could you do what Maschke usually does?
Man, you're so rude. Maschke really knows a lot about ordinals. It's just types like you who burn out people.

November 4th, 2019, 04:49 AM   #14
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Quote:
 Originally Posted by AplanisTophet I wish Maschke knew more about ordinals... Hey ISP, could you do what Maschke usually does?
LOL, sure.

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
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Quote:
 Originally Posted by Micrm@ss Man, you're so rude. Maschke really knows a lot about ordinals. It's just types like you who burn out people.
Maschke said himself (or herself) he didn't really know anything about ordinals. As he put it, "note to self, learn" or some such when it came to basic ordinal arithmetic.

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...f-fodors-lemma 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 set-builder 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 sub-step 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...f-fodors-lemma 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 set-builder 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 sub-step 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
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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:
 Originally Posted by AplanisTophet 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 $(r_{\alpha})_{\omega \leq \alpha < \omega_1}$:

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
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The above needs to be changed because $\phi_{\omega}$ and higher cannot be bijective as written. Glad to see you guys read it!

Quote:
 Originally Posted by AplanisTophet 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 $P$.
This should work. Vsotvep is reviewing it right now on SE to help me get any mechanical bugs out.

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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