- Thread starter AplanisTophet
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I already did one better and pointed out that they have no bearing on the proof so we can scrap them, keep them, articulate as to their meaning, etc. It doesnâ€™t matter. The ellipses are not worth addressing because they have absolutely no bearing on the proof. Change it from a $T$ sequence to an $S$ sequence if you want too and it wonâ€™t change anything mathematically, just your choice of letter. The ellipses are no different here. Why so hard to understand?Another way to go here would be for you to challenge yourself to clearly articulate the meaning of your ellipses notation. But you can't get out of your own way.

That triplet shows up on the second iteration but fits no rule, so $f((2,1,0)) = \{\}$.What if triplet is {2,1,0}

That is a big leap of faith. If it doesn't fit any rule then extrapolate a new rule from existing ones.That triplet shows up on the second iteration but fits no rule, so $f((2,1,0)) = \{\}$.

I already lost after rule 4. Would you be kind enough to post a summarized version?

I donâ€™t know what youâ€™re asking. We certainly are not extrapolating any new rule from $(2,1,0)$. It suffices that triplet never fits any rules, but since the rules can be anything you want for any particular $T$ sequence, we could have a different $T$ sequence where that particular triplet fits every rule too. I want $f((2,1,0))$ to equal $\{\}$ for my $T$ sequences above and the fact that it does is certainly not an inconsistency or anything.That is a big leap of faith. If it doesn't fit any rule then extrapolate a new rule from existing ones.

I already lost after rule 4. Would you be kind enough to post a summarized version?

I think Maschke may be throwing up a smoke screen here.Another way to go here would be for you to challenge yourself to clearly articulate the meaning of your ellipses notation. But you can't get out of your own way.

I believe the ellipses in each of the rules and in the definition of function $f$ were put there in an effort to make it easier to understand. They do follow directly from the opening remarks, so I'm wondering if Maschke read this part:

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 ordinalsâ€™ pattern, so we might then say that $4$ is implied by $1, 2, 3, \dots$ or, similarly, $1, 2, 3, \dots \implies 4$. ... We can also say that $1, 2, 3, \dots \implies \omega$...

The OP actually does what Maschke suggests accidentally and omits the ellipses further down given this typo:

It is true that the use of the ellipses is not necessary. It has been acknowledged that they could be removed in light of what appeared initially to be a helpful suggestion by Maschke. If Maschke believes the OP is unintelligible based on the use of ellipses, the error resides with Maschke as that is simply not true....so we say $3, 2, 1 \implies 0$...

There are some more typos and clarifications that could be made. Adding parenthesis and replacing $B$ with $\bigcup B$ in the following quote would help:

With my suggestions, it would read: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 $T$.

I confirm that $T$ is an enumeration of $\omega^2$ after the three given rules. Any element of $\omega^2$ can be proven to exist within the sequence using a finite chain of steps. For example, $\omega \cdot 2 + 1$ can be proven to exist using the following method: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 removes any redundant elements from $\bigcup B$ before potentially well ordering them so that we can add them to $T$.

$(1,2,3)$ implies $\omega$ appears in the sequence by Rule 3.

$(0,1,\omega)$ implies that $\omega + 1$ appears by Rule 2.

$(0,1,\omega+1)$ implies that $\omega + 2$ appears by Rule 2.

$(\omega, \omega+1, \omega+2)$ implies that $\omega + \omega = \omega \cdot 2$ appears by Rule 3.

$(0,1,\omega \cdot 2)$ implies $\omega \cdot 2 + 1$ appears by Rule 2.

Surely that does not hold for $\alpha$ a limit ordinal, as I believe I stated earlier. I have no objection to this statement for a successor ordinal. OP should so state for clarity and not make us fill in his missing details, which are legion in my opinion.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.

ps -- Oh I think I see your point. If $\alpha = \omega$ and $\beta = 47$ then in fact $\gamma = \omega$, since $47 + \omega = \omega$ and no other choice of $\gamma$ will do. I imagine this always works in the general case.

I must amend my original statement to say that that OP's missing details are legion, but apparently this isn't one of them.

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Here we only consider $f((1,2,3))$ and the definition of function $f$ leaves no ambiguity as to the difference between â€œ1,2,3â€ and â€œ1,2,3, ...â€ for purposes of the OP. I think thatâ€™s the main bullet point youâ€™re missing.No, I don't see that at all. There's a huge difference between "1, 2, 3" and "1, 2, 3, ..."

If there is anything else you think is wrong, Iâ€™m happy to address it. Enumerating $\epsilon_0$ was inspired by you due the insights into proving the Peano axioms consistent you provided in the other thread. We have a theory that enumerates ordinals in the $T$ sequence model, so perhaps there is hope youâ€™ll remain interested?

Well thanks ISP, I couldn't have said it better myself. :wink: Hopefully Maschke comes around eventually.

I assume Maschke has made perfect sense of everything though, as has everyone else, so we can probably start to wrap up this thread.

**The Goal From Here:**

I want to discuss a transfinite sequence $(T^{\alpha})_{\alpha \in Ord}$ compiled by adding rules to the above (or some similar) $T$ sequences.

**When adding rules, I want to accomplish a few things:**

1) I want to make it clear that each $T$ sequence can reference the number of rules used to generate it. Let $T^{\alpha}$ be a $T$ sequence generated using $\alpha$ rules, where $\alpha$ is some ordinal. An uncountable $\alpha$ would require some clarification and for $T^{\alpha}$ to be a transfinite sequence, presumably.

2) Each $T$ sequence may contain only ordinals, but not every $T$ sequence is an enumeration of an ordinal. Equivalently, $\{ x : x \in T \}$ may or may not be an ordinal based on the rules used to generate $T$. I wish to create a transfinite sequence $(T^{\alpha})_{\alpha \in Ord}$ where $\{ x : x \in T^{\alpha} \}$ is an ordinal for each element of the sequence. To do this, it must be ensured that each rule, when added to the previous rules, results in a new $T$ sequence that again enumerates some ordinal.

3) Make it clear that $\bigcup_{\gamma < \alpha} \{ x : x \in T^{\gamma} \} = \{x : x \in T^{\alpha - 1} \}$ whenever $\alpha$ is not a limit ordinal.

4) A new rule $\alpha$ should imply an ordinal that is equal to $\bigcup_{\gamma < \alpha} \{ x : x \in T^{\gamma} \}$ so as to extend the sequence:

$$\text{Rule } \alpha \text{ : } a, b, c, \dots \implies \{ \bigcup_{\gamma < \alpha} \{x : x \in T^{\gamma} \} \}, \text{ where } a,b,c < \bigcup_{\gamma < \alpha} \{ x : x \in T^{\gamma} \}$$

**The Question I Have Now Is:**

Does each point 1 - 4 above sound reasonable?

I assume Maschke has made perfect sense of everything though, as has everyone else, so we can probably start to wrap up this thread.

I want to discuss a transfinite sequence $(T^{\alpha})_{\alpha \in Ord}$ compiled by adding rules to the above (or some similar) $T$ sequences.

1) I want to make it clear that each $T$ sequence can reference the number of rules used to generate it. Let $T^{\alpha}$ be a $T$ sequence generated using $\alpha$ rules, where $\alpha$ is some ordinal. An uncountable $\alpha$ would require some clarification and for $T^{\alpha}$ to be a transfinite sequence, presumably.

2) Each $T$ sequence may contain only ordinals, but not every $T$ sequence is an enumeration of an ordinal. Equivalently, $\{ x : x \in T \}$ may or may not be an ordinal based on the rules used to generate $T$. I wish to create a transfinite sequence $(T^{\alpha})_{\alpha \in Ord}$ where $\{ x : x \in T^{\alpha} \}$ is an ordinal for each element of the sequence. To do this, it must be ensured that each rule, when added to the previous rules, results in a new $T$ sequence that again enumerates some ordinal.

3) Make it clear that $\bigcup_{\gamma < \alpha} \{ x : x \in T^{\gamma} \} = \{x : x \in T^{\alpha - 1} \}$ whenever $\alpha$ is not a limit ordinal.

4) A new rule $\alpha$ should imply an ordinal that is equal to $\bigcup_{\gamma < \alpha} \{ x : x \in T^{\gamma} \}$ so as to extend the sequence:

$$\text{Rule } \alpha \text{ : } a, b, c, \dots \implies \{ \bigcup_{\gamma < \alpha} \{x : x \in T^{\gamma} \} \}, \text{ where } a,b,c < \bigcup_{\gamma < \alpha} \{ x : x \in T^{\gamma} \}$$

Does each point 1 - 4 above sound reasonable?

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Best question ever based on the immediately preceding post:

Without any modifications to the $T$ sequence model so as to allow for a transfinite $T$, we could still start a regular sequence with $\omega_1$ rules: $T^{\omega_1}$. That would be the largest possible countable ordinal, correct?

Without any modifications to the $T$ sequence model so as to allow for a transfinite $T$, we could still start a regular sequence with $\omega_1$ rules: $T^{\omega_1}$. That would be the largest possible countable ordinal, correct?

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