September 24th, 2019, 06:40 PM  #1 
Senior Member Joined: Jun 2014 From: USA Posts: 620 Thanks: 52  T Sequences  Explicit Enumeration of $\epsilon_0$ Introduction: Making a Sequence $T$ based on “The Rule of Three” The goal of this thread is to develop the basic model for what I refer to as a "$T$ sequence" and then expand on the basic model so as to create an enumeration of the ordinal $\epsilon_0$. Each $T$ sequence is a listing of ordinals that is iteratively generated based on a list of rules. After developing a basic $T$ sequence model with only three rules that happens to be an enumeration of $\omega^2$, I will then add additional rules so as to (try to) create the explicit enumeration of $\epsilon_0$. The primary means of generating the sequence $T$ is through the use of a function $f$. In general, function $f$ is going to be a function that takes as input a threemember sequence of ordinal numbers (an ordered triplet) and returns the set of all ordinals that are a continuation of any recognizable pattern or limit formed by the triplet. The term ‘recognizable pattern or limit’ is defined for this purpose as one that appears in a sequence of rules (see “Rules” below). Brief Explanation of Function $f$: To get started, consider the ordered triplet $1, 2, 3$ as the initial segment of some pattern of ordinals that could be continued or a limit that could be approached: $1, 2, 3, \dots$. What would come next? The number $4$ is a continuation of the ordinals’ pattern, so we might then say that $4$ is implied by $1, 2, 3, \dots$ or, similarly, $1, 2, 3, \dots \implies 4$. The number $4$ is not the only thing that may be implied by $1, 2, 3, \dots$, however. We can also say that $1, 2, 3, \dots \implies \omega$. If there are any other ordinal numbers that happen to be universally agreeable continuations or limits for the threemember sequence $1, 2, 3, …$, we could consider them too. In this case, I am unaware of any. The triplet $1, 2, 3$ is being used as an example because it implies two ordinals, $4$ and $\omega$ (and so $f((1,2,3)) = \{4, \omega \}$), whereas other triplets may imply one or none. The order of the triplet is important too. If we instead consider $3, 2, 1$ we find that $0$ is a continuation of the ordinals’ pattern (instead of $4$), so we say $3, 2, 1 \implies 0$. To summarize, we are starting to form some basic rules that collectively are going to define the term ‘recognizable pattern or limit’. Rules: Where $a, b, c, \dots$ are ordinals, the following rules are used to define function $f$. $$\text{Rule 1 : }a, a1, a2, \dots \implies \{ a3 \} \text{, where } a \geq 3$$ $$\text{Rule 2 : }0, 1, a, \dots \implies \{a+1\}$$ $$\text{Rule 3 : }a, a+1, a+2, \dots \implies \{ a+3, a + \omega \}$$ Define function $f$: $$f((a,b,c)) = \bigcup \{ x : a, b, c, \dots \implies x \}, \text{ where } a, b, \text{ and } c \text{ are ordinals and } x \text{ is a set defined by the above rules}$$ Define function $g$: For any set of ordinals, $A$, let $g(A)$ be the set of all ordered triplets that can be made from the elements of $A$: $$g(A) = \{ (a,b,c) : a,b,c \in A \}$$ Define $X_{Ord}$: $$X_{Ord} = X \setminus \{ x \in X : x \text{ is not an ordinal} \} \text{ for any set } X$$ Define the sequence $T$: Define a sequence $T = t_1, t_2, t_3, \dots$ via iterations where: Step 1) $t_1 = 1, t_2 = 2,$ and $t_3 = 3$. 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., $A = \{1, 2, 3 \}$ on the first iteration. b) Let $B = \{ f((a,b,c))_{Ord} : (a,b,c) \in g(A) \}$. Using the previous elements of the sequence $A$, this step creates a set $B$ of all the new sets of ordinals implied by letting function $f$ range over $g(A)$. The use of the $X_{Ord}$ function in the definition of $B$ may be unnecessary for this particular $T$ sequence. c) Let $C = \bigcup B \setminus A$ and let $c_1, c_2, c_3, \dots$ be an enumeration of $C$ that is also well ordered if $C \in \mathbb{N}$. This step shaves off any redundant elements of $B$ before potentially well ordering them so that we can add them to $T$. d) If $C \in \mathbb{N}$, then set $t_n = c_1, t_{n+1} = c_2, t_{n+2} = c_3, \dots, t_{n+j1} = c_j$. e) If $C = \mathbb{N}$ (not applicable for this particular $T$ sequence), then let $T’ = t’_1, t’_2, t’_3, \dots$, be a subsequence of the remaining undefined elements of $T$ and set $t’_n = c_1, t’_{n+2} = c_2, t’_{n+4} = c_3, \dots$. Step 3) Proceed to the next undefined index $j$ in $T$, set $n = j$, and repeat step 2. The first few elements of $T$ with the above three rules would be: $$T = 1,2,3,0,4,\omega,5, \omega + 1, 6, \omega + 2,7,\omega+3,\omega \cdot 2,8,\omega+4,\omega \cdot 2 + 1, \dots$$ Where $T$ is generated one iteration at a time, the first iteration takes each ordered triplet that may be comprised from the ordinals $1, 2,$ and $3$ (there are six possible triplets) and tests each triplet to see if its ordering matches one of the rules. We have $f((3,2,1)) = \{0\}$ by Rule 1 and $f((1,2,3)) = \{4, \omega \}$ by Rule 3. The iteration process then takes the union of $\{0\}$ and $\{4, \omega \}$, orders the union to produce a countable sequence that is also well ordered (because it is a finite sequence), and adds the sequence to the initial undefined elements of $T$. We then start the second iteration by taking each ordered triplet that may be comprised from the ordinals $1,2,3,0,4,\omega$ and seeing which new ordinals the rules imply. Here we get the same old $f((3,2,1)) = \{0\}$ and $f((1,2,3)) = \{4, \omega \}$ in addition to $f((0,1,4)) = \{5\}, f((2,3,4)) = \{5\},$ and $f((0,1,\omega)) = \{\omega + 1\}$. Any duplicate ordinals are removed during the iteration process so as to refrain from adding them to $T$ more than once. If we keep going with these three rules, the $T$ sequence will become an enumeration of $\omega^2$. Note that no rule exists that is capable of taking three ordinals less than $\omega^2$ and implying a set containing an ordinal greater than or equal to $\omega^2$. Rule #1 could produce $\omega^2$ given the triplet $(\omega^2 + 3, \omega^2 + 2, \omega^2 + 1)$, but since there are no ordinals greater than $\omega^2$, Rule 1 is not capable of inserting $\omega^2$ into the sequence either. Adding Rules Rule 4 will lead to $T$ becoming an enumeration of $\omega^2 \cdot 2$ (assuming my calculations are correct for all of these): $$\text{Rule 4 : }a \cdot b, a \cdot (b+1), a \cdot (b+2), \dots \implies \{a \cdot (b + \omega) \}$$ Rule 5 will lead to $T$ becoming an enumeration of $\omega^{\omega}$ $$\text{Rule 5 : }a+b \cdot c, a+b \cdot (c+1), a+b \cdot (c+2), \dots \implies \{a+b \cdot (c+\omega)\}$$ Rule 6 will lead to $T$ becoming an enumeration of $\omega^{\omega} \cdot 2$: $$\text{Rule 6 : }a^b, a^{b+1}, a^{b+2}, \dots \implies \{a^{b+\omega}\}$$ Rule 7 will lead to $T$ becoming an enumeration of $\omega^{\omega^2}$: $$\text{Rule 7 : }a+b^c, a+b^{c+1}, a+b^{c+2}, \dots \implies \{a+b^{c+\omega}\}$$ Rule 8 will lead to $T$ becoming an enumeration of $\omega^{\omega^{\omega}} \cdot 2$: $$\text{Rule 8 : }a + b^{c + d \cdot e}, a + b^{c + d \cdot (e+1)}, a + b^{c + d \cdot(e+2)}, \dots \implies \{a + b^{c + d \cdot(e+\omega)}\}$$ Rule 9 will lead to $T$ becoming an enumeration of $\epsilon_0 = \omega^{\epsilon_0}$: $$\text{Rule 9 : }a + b^{c+d^e}, a + b^{c+d^{e+1}}, a + b^{c+d^{e+2}}, \dots \implies \{ a + b^{c+d^{e+w}}\}$$ Last edited by AplanisTophet; September 24th, 2019 at 07:13 PM. 
September 27th, 2019, 10:29 AM  #2 
Senior Member Joined: Jun 2014 From: USA Posts: 620 Thanks: 52 
The calcs do get a little tricky but hopefully the above is an enumeration of $\epsilon_0$. If anyone sees an error or is unable to make sense of the $T$ sequence model I would love to hear about it. If the above looks good, the goal will be to consider $T^{\alpha}$ for any (countable) ordinal $\alpha$. The ordinal $\alpha$ will represent the number of rules employed by the $T$ sequence. An uncountable $\alpha$ would obviously require some clarification as $T$ would have to become a transfinite sequence, etc. 
September 30th, 2019, 03:54 PM  #3  
Senior Member Joined: Aug 2012 Posts: 2,412 Thanks: 755  Quote:
And if you find yourself getting frustrated, can you please criticize the speech and not the speaker? As in: I find your exposition incoherent. As opposed to personal remarks. If so: Quote:
(2) What if $a$ is a limit ordinal in Rule 1? (3) In Rule 2, suppose $a = 2$, giving $0, 1, 2, \dots \implies \{3\}$; but in Rule 3, if $a = 0$ then $0, 1, 2, \dots \implies \{3, \omega\}$. How do you account for this inconsistency? Last edited by Maschke; September 30th, 2019 at 04:12 PM.  
September 30th, 2019, 05:29 PM  #4  
Senior Member Joined: Jun 2014 From: USA Posts: 620 Thanks: 52  Quote:
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2) If $a$ is a limit ordinal, then $a1, a2,$ and $a3$ all equal $a$. 3) On the second iteration $g(A)$ will contain the triplet $(0,1,2)$. We will then get $f((0,1,2)) = \bigcup \{ \{3\}, \{3, \omega\} \} = \{ 3, \omega \}$ where $\{3,\omega\} \in B$ and $3, \omega \in \bigcup B$.  
October 1st, 2019, 04:30 PM  #5  
Senior Member Joined: Aug 2012 Posts: 2,412 Thanks: 755  Quote:
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When I said earlier (and keep saying) that your exposition is incoherent, I don't mean that it becomes so in the details after it's all done. I mean right from the getgo. Your first three rules have no meaning. I pointed out that two of them are inconsistent. You can't subtract from a limit ordinal. And you haven't defined your ellipses notation except to say that it's a "semantic preference." Whose, exactly? Sorry, this is all a nonstarter. It's not that your entire exposition falls apart. It's that it never gets started at all. Rule 1 is nonsense as stated since you don't define your notation. How about drilling down and telling me what the ellipses mean in Rule 1? The standard meaning of the notation 1, 2, 3, ... is the entire sequence of positive integers. If you mean something else, please define it with precision. Last edited by Maschke; October 1st, 2019 at 04:35 PM.  
October 1st, 2019, 05:56 PM  #6  
Senior Member Joined: Jun 2014 From: USA Posts: 620 Thanks: 52  Quote:
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https://en.wikipedia.org/wiki/Ordinal_arithmetic Quote:
No two rules are inconsistent with each other as, unlike axioms, that would be impossible for a $T$ sequence. I told you what the result of $f((0,1,2))$ would be. Did you not understand? Each rule has it's own localized variables where setting $a$ equal to $2$ for purposes of applying rule 2 has no bearing on what $a$ is for rule 3. The ordered triplet 0,1,2,... obviously doesn't fit Rule 1, but it does fit Rule 2 and Rule 3. You might then ask, how many rules can one triplet fit? Uncountably many if you like would be the answer, but we can also have uncountably many rules where any single triplet can only fit a finite number though too. I digress though, as that would be a discussion for when we consider $T^{\alpha}$ for any ordinal $\alpha$. Rule 1 never does assert that we subtract a finite ordinal from a limit ordinal (though we can as explained above and it's perfectly fine) unless I want to consider triplets like $(\omega, \omega, \omega)$, in which case Rule 1 would then have $f((\omega, \omega, \omega)) = f((\omega, \omega1, \omega2)) = \{\omega3\} = \{\omega\}$ because $\omega = \omega\beta$ for any ordinal $\beta < \omega$. Quote:
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If you still don't understand what the ellipses mean after my explanations in the OP, above post, and in this post, just take them all out of the proof and try again. It really won't make a difference, and, since it has no bearing on the underlying mathematics, your assertion that my work is a nonstarter because of the ellipses makes no sense to me.  
October 1st, 2019, 07:02 PM  #7  
Senior Member Joined: Aug 2012 Posts: 2,412 Thanks: 755  Quote:
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I honestly don't think this is going to be productive for me. I'm going to let it go. Sorry I couldn't be of more help.  
October 1st, 2019, 07:10 PM  #8 
Senior Member Joined: Jun 2014 From: USA Posts: 620 Thanks: 52  
October 1st, 2019, 07:47 PM  #9 
Senior Member Joined: Aug 2012 Posts: 2,412 Thanks: 755  
October 1st, 2019, 11:11 PM  #10 
Senior Member Joined: Oct 2009 Posts: 884 Thanks: 340  Aplanis is right here though. The theorem he states is correct. For $\beta\leq \alpha$, there is indeed a unique $\gamma$ such that $\alpha = \beta + \gamma$. This is usually not denoted as subtraction though, but he could define it likes this if he wishes.


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$epsilon0$, enumeration, explicit, sequences 
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