October 27th, 2019, 12:32 AM  #1 
Senior Member Joined: Jun 2014 From: USA Posts: 644 Thanks: 55  Having Sum Fun Counting Ordinals
This is a draft. Knowing me, it is probably nonsensical and filled with errors. I'm trying to enumerate some very large countable ordinal assuming of course that math isn't broken and I can't enumerate $\omega_1$ itself. ... And no, jic, I don't think math is broken. Why? Do you? I'll take a moment to step back from this and at some point return to see what corrections, comments, etc., I care to make. Feel free to look at it funny in the meantime. Define $t(\alpha)$ for any ordinal $\alpha$: For $\alpha \geq 2$, 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 $\alpha$. 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) $\zeta < \alpha \implies min\{ \phi_{\zeta}^{1}(b) : \exists k \in b \text{ where } k \geq \phi_{\zeta}^{1}(b)\} < min\{ \phi_{\alpha}^{1}(b) : \exists k \in b \text{ where } k \geq \phi_{\alpha}^{1}(b)\}$. Define function $f$: $$f(x) = \phi_{t^{1}(x)}^{1}(x)$$ 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 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) Let $A = \{ t_i \in T : i < n \}$. E.g., $A = \{0, 1, 2 \}$ on the first iteration. b) 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. c) 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 $T$. d) If $C \neq \aleph_0$, then set $t_n = c_1, t_{n+1} = c_2, t_{n+2} = c_3, \dots$. e) If $C = \aleph_0$, then 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. Set $n = j$, increase the iteration counter by letting $m = m + 1$, and then repeat step 2. Last edited by AplanisTophet; October 27th, 2019 at 12:46 AM. 
October 27th, 2019, 02:03 PM  #2  
Senior Member Joined: Aug 2012 Posts: 2,424 Thanks: 759  Quote:
You idea would generalize better and be more clear if you regard each tuple as a function from a given ordinal to some set. Indeed that's exactly what a highschool tuple is. A point $(x_0, y_0)$ in the Euclidean plane is a particular function $\{x, y \} \to \mathbb R$, namely the one that sends $x$ to $x_0$ and $y$ to $y_0$; and the coordinate plane is the set of all such functions. So now we have no trouble making sense of an "ordinalindexed tuple," namely that it's a function from some ordinal to some set. Now you have not told us what the range of this function is. What are the elements of the tuples? Are they ordinals too? I'll assume they are. So an "ordinalindexed tuple" is a function $f : On \to On$, where $On$ is the class of ordinals. I confess that I still don't quite believe in functions defined on proper classes. Where do they live? But set theorists have some legit way to make sense of it so no matter. Does that seem like what you have in mind? Last edited by Maschke; October 27th, 2019 at 02:39 PM.  
October 28th, 2019, 12:19 PM  #3  
Senior Member Joined: Jun 2014 From: USA Posts: 644 Thanks: 55  Quote:
Don't forget about $\omega$lets though either. I want that function normal. I'm not sure what to say about the rest of your post as a result. Last edited by AplanisTophet; October 28th, 2019 at 12:22 PM.  
October 28th, 2019, 12:43 PM  #4  
Senior Member Joined: Aug 2012 Posts: 2,424 Thanks: 759  Quote:
 
October 28th, 2019, 02:41 PM  #5  
Senior Member Joined: Jun 2014 From: USA Posts: 644 Thanks: 55 
These are some potential questions and clarifications that I've noted as being possibly necessary. There may be more of course: 1) Typo  My definition of $t(\alpha)$ headlines “for any ordinal $\alpha$” and then starts by restricting to $\alpha \geq 2$. This particular work only makes use of $2 \leq \alpha < \omega_1$ as a domain for function $t$, so I should change the headline for my definition. 2) For the definition of $t$, I don’t need to use doublets, triplets, quadruplets, and so on, but it’s easier because $a=b \implies \{a,b\}=\{b,a\} = \{a\} = \{b\}$ whereas $a=b$ does not $\implies (a,b)=(b,a)=(a)=(b)$. 3) I could define $C$ to make sure there are never any duplicate elements in $T$ perhaps... Right now there are never duplicate elements if the cardinality of $C \neq \aleph_0$ on each iteration. I think that if anyone questions anything it will be the following things: 1) When $\alpha \geq \omega$, I think someone may question whether there must be an element $\kappa$ such that $\kappa \in \phi_{\alpha}(\kappa)$. Fodor’s lemma requires that there must be such an element I believe (Fodor’s lemma would have $\kappa$ map to a constant if $\phi$ were ‘fully regressive’ instead of just ‘almost regressive’), so I don’t think that is an issue. 2) The following line is the one that should cause some stir and that I expect people to question for accuracy. As of right now I still stand by it, but why wouldn’t I (I’m a selfprofessed crank)?: “The restriction of $g(A)$ to $A \cap \omega_1$ ensures that $B$ remains countable for this particular $T$ sequence.” Quote:
That said, your only complaint about my work is the same in every thread lately and it's always something to do with my notation. It's not notation this time absent any typo so don't start that. I will assist you if you like in figuring out my work, but that is a courtesy on my behalf because I don't think you can assist me. Feel free to prove me otherwise of course. I would prefer to do that over PM, but I can't stop you from posting in my threads like you typically do. Last edited by AplanisTophet; October 28th, 2019 at 02:46 PM.  
October 28th, 2019, 03:27 PM  #6 
Senior Member Joined: Aug 2012 Posts: 2,424 Thanks: 759  Fuck you.
Last edited by Maschke; October 28th, 2019 at 03:30 PM. 
October 29th, 2019, 11:13 AM  #7 
Senior Member Joined: Jun 2014 From: USA Posts: 644 Thanks: 55 
They might be able to edit that out once this becomes the most famous math forum thread ever. 
October 29th, 2019, 01:12 PM  #8  
Newbie Joined: Oct 2019 From: SV USA Posts: 5 Thanks: 1  Quote:
“The restriction of $g(A)$ to $A \cap \omega_1$ ensures that $B$ remains countable for this particular $T$ sequence.” If $B$ is countable for each iteration then $T$ remains a regular sequence of only order type $\omega$ after $\omega_1$ iterations. Each iteration adds elements to $T$, so there are at least $\omega_1$ ordinals in $T$. That is a contradiction because it implies $T$ enumerates an ordinal with cardinality greater than or equal to $\omega_1$. At the same time, I think $B$ must be countable for each iteration, so I am confused. Hopefully this helps others help. Sorry I don't understand myself. Last edited by ISP; October 29th, 2019 at 01:14 PM.  
October 29th, 2019, 08:50 PM  #9  
Senior Member Joined: Jun 2014 From: USA Posts: 644 Thanks: 55  I hope this is the last draft after having stepped back for a while to see what I could fix and clarify. Please help now and thank you!!!! ISP pinpoints the issues: Quote:
The following is the issue that needs addressing: Quote:
... 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 $a=b$ 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) $\zeta < \alpha \implies min\{ \phi_{\zeta}^{1}(b) : \exists k \in b \text{ where } k \geq \phi_{\zeta}^{1}(b)\} < min\{ \phi_{\alpha}^{1}(b) : \exists k \in b \text{ where } k \geq \phi_{\alpha}^{1}(b)\}$. Define function $f$: $$f(x) = \phi_{t^{1}(x)}^{1}(x)$$ 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 $t_1 = s_1, t_2 = s_2, t_3 = s_3, \dots$. 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 $C \neq \aleph_0$, then set $t_n = c_1, t_{n+1} = c_2, t_{n+2} = c_3, \dots$. f) If $C = \aleph_0$, then 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. Set $n = j$, increase the iteration counter by letting $m = m + 1$, and then repeat step 2. Last edited by AplanisTophet; October 29th, 2019 at 09:49 PM. Reason: fix iteration counter  
October 29th, 2019, 09:59 PM  #10 
Senior Member Joined: Jun 2014 From: USA Posts: 644 Thanks: 55  Fixed. This should do it. Sorry for taking my sweet time working the bugs out. We're down to just that one assertion in the immediately preceding post. Thanks again for any help!


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