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If $\alpha$ is a countable ordinal, is $\alpha + 1$ countable? Proof or counterexample.That would be the largest possible countable ordinal, correct?

Oh thank goodness. I was going to check you for a pulse.If $\alpha$ is a countable ordinal, is $\alpha + 1$ countable? Proof or counterexample.

I thought maybe you would ask if I had a method for specifying what the $a,b,c$ are for each new rule where all I really gave above was that $a,b,c < \bigcup_{\gamma < \alpha} \{ x : x \in T^{\gamma} \}$:

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:

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

This is what I want:

Then I have a real serious question: What is $T^{\omega_1}$ if we make no modifications to the $T$ sequence model? It will generate a sequence because it has to. What will be the supremum of that sequence? I think it must be $\omega_1$.You might then ask, how many rules can one triplet fit? ...we can also have uncountably many rules where any single triplet can only fit a finite number... I digress though, as that would be a discussion for when we consider $T^{\alpha}$ for any ordinal $\alpha$.

When I leave the thread you beg me to come back; and the moment I do, you're a snide asshole. And you've done this several times in this thread already. Are you a passive-aggressive jerk in real life too? Now those ARE personal remarks.Oh thank goodness. I was going to check you for a pulse.

When you get your notation straight, I'll respond to your exposition. As it is, Rule 1 is ambiguous and meaningless, and Rules 2 and 3 contradict one another as I noted earlier. As long as you choose not to address these points there's nothing else for me to do here.

Your exposition is vague and mostly incoherent. If you don't want to fix it, that's not my problem.

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Rule 1 is my favorite. There is no such thing as a meaningless and ambiguous rule, though you could add rules to particular $T$ sequences that add nothing to the sequence and, if you want to refer to such rules as meaningless and ambiguous, I would agree. As it stands, there is nothing meaningless or ambiguous about Rule 1, or at least nothing you've successfully conveyed as such. I told you I'm all ears if you actually find something wrong and I genuinely hope you find this interesting.When you get your notation straight, I'll respond to your exposition. As it is, Rule 1 is ambiguous and meaningless, and Rules 2 and 3 contradict one another as I noted earlier. As long as you choose not to address these points there's nothing else for me to do here.

As I told you previously, it's impossible for two rules to contradict each other. I went on to show you what $f((0,1,2))$ was because you questioned how $a$ could be $2$ when applying rule 2 while $a$ could be $0$ when applying rule 3. I told you that each rule has its own local variables, which leaves no room for ambiguity. You now continue to assert that rules 2 and 3 contradict each other while ignoring everything I so patiently explained to you without any indication as to whether you even read it.

If you want to say my work is wrong after I've taken every effort to try and address your comments only to have you ignore me, then that makes you a crabby old goat I might speculate, but also one who, if you actually cared enough to clarify your own comments as opposed to just hound on me to clarify mine, might offer up something worth listening to. That's what I'm hoping for anyways.

I haven't said it's wrong. I've said it's incoherent. It's "not even wrong."If you want to say my work is wrong ...

https://en.wikipedia.org/wiki/Not_even_wrong

I have asked you several times to elucidate your ellipses notation and all you said was that it made no difference. That's nonsense as I've told you several times. Your notation doesn't even make sense by your own admission.Rule 1 is my favorite. There is no such thing as a meaningless and ambiguous rule ...

I gave you explicit examples showing that rules 2 and 3 produce contradictory results. You're unable or unwilling to engage with the examples I gave.As I told you previously, it's impossible for two rules to contradict each other.

When I respond to you, you insult me. When I don't respond to you, you reference my handle and beg me to return. You are locked into some kind of passive-aggressive loop that I attribute to your recent life changes. I ask again: Are you like this with the people in your life? Earlier in the thread you insulted me and when I called you on it you said it was a joke. That's the classic pattern of an emotional abuser.that makes you a crabby old goat

In short, it's become

By the way, at least make a half-hearted or even quarter-hearted effort to prove or disprove that if $\alpha$ is a countable ordinal, so is $\alpha + 1$. The fact that you have not even a minimal intuition about this makes it hard to justify spending any time trying to unpack your convoluted posts.

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I haven't said it's wrong. I've said it's incoherent. It's "not even wrong."

https://en.wikipedia.org/wiki/Not_even_wrong

If a simple enumeration of $\omega^2$ is "not even wrong" then that's pretty cool. I don't think that's the case here though, so maybe you could try just a tad harder and explain why you think something so basic to me is so "not even wrong" to you. I am genuinely interested.

I am under the genuine impression that it makes sense to most everyone but you. Here's your version then without the ellipses, as I told you:I have asked you several times to elucidate your ellipses notation and all you said was that it made no difference. That's nonsense as I've told you several times. Your notation doesn't even make sense by your own admission.

Rules:Where $a, b, c, \dots$ are ordinals, the following rules are used to define function $f$.

$$\text{Rule 1 : }a, a-1, a-2\implies \{ a-3 \} \text{, where } a \geq 3$$

$$\text{Rule 2 : }0, 1, a \implies \{a+1\}$$

$$\text{Rule 3 : }a, a+1, a+2 \implies \{ a+3, a + \omega \}$$

Define function $f$:

$$f((a,b,c)) = \bigcup \{ x : a, b, c \implies x \}, \text{ where } a, b, \text{ and } c \text{ are ordinals and } x \text{ is a set defined by the above rules}$$

I took the exact triplet that you said was contradictory with rules 2 and 3 and showed you exactly what happens with that triplet. There was no contradiction, clearly. You still seem to think there is, but I can't guess as to how you are reaching that conclusion and I literally don't know. I'm trying.I gave you explicit examples showing that rules 2 and 3 produce contradictory results. You're unable or unwilling to engage with the examples I gave.

You send me PM's full of profanity and often post nonsense in my threads, which to me is actually kind of entertaining especially when my threads are usually a little bit goofy to begin with, but as I told you, I really am interested in the $T$ sequence model here too so I'm torn.When I respond to you, you insult me. When I don't respond to you, you reference my handle and beg me to return. You are locked into some kind of passive-aggressive loop that I attribute to your recent life changes. I ask again: Are you like this with the people in your life? Earlier in the thread you insulted me and when I called you on it you said it was a joke. That's the classic pattern of an emotional abuser.

In short, it's becomeunpleasantto interact with you. And when I try to ignore the personal crap and talk about the math, you refuse to engage on my specific concerns. There is simply no point to my putting energy into this.

I don't care how unpleasant you get. I'll always treasure our times together. You do know a lot more about math than me in general too so if you cared to make a genuine attempt to help as opposed to just troll me, I would still love it.

Of course there is no greatest countable ordinal. The fact that your intuition didn't pick up on my intentions for that post makes it so much fun to converse with you, but at times unproductive as I'm trying to be serious in this thread too and I'm not the kind of person to take math too seriously too often.By the way, at least make a half-hearted or even quarter-hearted effort to prove or disprove that if $\alpha$ is a countable ordinal, so is $\alpha + 1$. The fact that you have not even a minimal intuition about this makes it hard to justify spending any time trying to unpack your convoluted posts.

I can speak only for myself, but mostly everything in this thread is incomprehensible to me.I am under the genuine impression that it makes sense to most everyone but you. Here's your version then without the ellipses, as I told you:

Thank you. Are you able to state why?I can speak only for myself, but mostly everything in this thread is incomprehensible to me.

Explanation of Rule 1:Rules:Where $a, b, c, \dots$ are ordinals, the following rules are used to define function $f$.

$$\text{Rule 1 : }a, a-1, a-2, \dots \implies \{ a-3 \} \text{, where } a \geq 3$$

$$\text{Rule 2 : }0, 1, a, \dots \implies \{a+1\}$$

$$\text{Rule 3 : }a, a+1, a+2, \dots \implies \{ a+3, a + \omega \}$$

Define function $f$:

$$f((a,b,c)) = \bigcup \{ x : a, b, c, \dots \implies x \}, \text{ where } a, b, \text{ and } c \text{ are ordinals and } x \text{ is a set defined by the above rules}$$

Within the class of all ordered triplets of ordinals, there will be one and only one triplet $(a,b,c)$ for each ordinal $\gamma$ in the class of all ordinals where $a,b,c = a,a-1,a-2$ and $\gamma = a â€“ 3$. If the input for function $f$ is any triplet of the form $a,b,c = a, a-1, a-2$, then Rule 1 states that $a-3$ will be an element of $f(a,b,c)$.

Explanation of Rule 2:

Within the class of all ordered triplets of ordinals, there will be one and only one triplet $(a,b,c)$ for each ordinal $\gamma$ in the class of all ordinals that are not limit ordinals where $a,b,c = 0,1,\gamma - 1$ and $\gamma = c + 1$. If the input for function $f$ is any triplet of the form $a,b,c = 0, 1, \gamma - 1$, then Rule 2 states that $c + 1$ will be an element of $f(a,b,c)$.

Explanation of Rule 3:

Rule 3 states that any triplet $(a,b,c)$ in the class of all ordered triplets of ordinals where $(a,b,c) = (a, a+1, a+2)$ will result in both $a+3$ and $a+\omega$ being elements of $f(a,b,c)$.

Now what in the bleep could you possibly not understand about that you eggheads (sorry... just trying to play the part of a crank).

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