# Another Opus Here

#### AplanisTophet

This is corrected (silly me, sorry):

Assuming my calculations are correct (it gets tricky!!!), the first few elements of $T$ would be:
$$T = 1,2,3,0,4,\omega,5,6, \omega + 1, \omega +2,7,8, \omega + 3,\omega+4,\omega \cdot 2,9,10,\omega+5,\omega+6, \omega \cdot 2 + 1, \omega \cdot 2 + 2, \dots$$

The first iteration would generate $0$ based on Rule 1 and $(3,2,1)$. It would also generate $4$ and $\omega$ based on Rule 4 and $(1,2,3)$. These are then added to $T$ to complete the first iteration.

#### AplanisTophet

Just wanted to mention why I'm not engaging.

* It's good that you're interested in the ordinals. However:

* [email protected] knows a lot more about ordinals than I do and he thinks your exposition makes no sense.

* From what little I've followed, your exposition makes no sense.

* You claim you've proved or disproved CH. When I challenged you on this point you did not respond. I see no connection between anything you've written and CH.

* Your idea of continuation is not well defined.

* There is no rule of 3 or pattern of 3. You're just hallucinating all of this as far as I can see; and more decisively, as far as [email protected] can see.

All the best.
I haven't proved or disproved CH. We knew there was something wrong with my work because I had a logic chain saying I did, so naturally one of the logical steps was flawed. [email protected] led me to see it because he (or she) took the time to wade through what I had done.

I then continued to try and make a sequence of ordinals that enumerated as many ordinals as possible solely for fun. I've continued to learn a lot (normal functions, fixed points, Veblen Hierarchy, Church-Kleene ordinal and Kleene O notation, etc.)

There is a rule of 3. More specifically, there are 8 rules that collectively comprise 'the rule of 3' in the most recent simplified version of my $T$ sequence that enumerates only ordinals less than $\omega_1$. However, any particular $T$ sequence can have any number of rules. I'm just playing with some basic models here.

1 person

#### AplanisTophet

This gives a clear explanation of my attempt to enumerate as many elements of $\omega_1$ as possible:

https://math.stackexchange.com/questions/3363741/question-on-an-explicit-enumeration-of-ordinals

Based on my answer over at stackexchange, we might note that there will always be another rule we could add based on the fixed point of a normal function for a given $T$ sequence that will extend the sequence. To find the rule, we look for the normal function's first fixed point:

$$\text{Let } \phi : \omega_1 \rightarrow \omega_1 \text{ be a normal function where: } \phi(\{t_i : t_i \in T\}) = \{t_i : t_i \in T\}$$

#### Maschke

This gives a clear explanation of my attempt to enumerate as many elements of $\omega_1$ as possible:

https://math.stackexchange.com/questions/3363741/question-on-an-explicit-enumeration-of-ordinals

Based on my answer over at stackexchange, we might note that there will always be another rule we could add based on the fixed point of a normal function for a given $T$ sequence that will extend the sequence. To find the rule, we look for the normal function's first fixed point:

$$\text{Let } \phi : \omega_1 \rightarrow \omega_1 \text{ be a normal function where: } \phi(\{t_i : t_i \in T\}) = \{t_i : t_i \in T\}$$
You got zero agreement on the sanity of your exposition; and specific pushback from Noah Schweber, a professional mathematician. Posting and then answering your own incoherent question does not constitute support for your position.

I grant you that Noah at least understood your question and offered some constructive criticisms. But he did not agree with the overall coherence as of the time of my posting this.

Not saying there isn't a pony in there somewhere. But why not wait for someone to actually agree with you before promoting your SE thread? Did you think nobody would click and see that you generated zero agreement?

Also: If your goal is to "enumerate the ordinals" as you say, what's wrong with the USUAL way of taking successors and limits?

Also also: Noah referred to $\Sigma$, which does not appear in your exposition. So clearly there is ongoing substantive editing going on. Why do you ask us to refer to a work in progress?

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#### AplanisTophet

Whatâ€™s wrong with you? The answer is the answer, itâ€™s posted, Noah gets it.

PS - Noah deleted his answer, which was just a general one on normal functions and fixed points, but didn't pick out the exact supremum of $T$ after 8 Rules.

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#### AplanisTophet

You got zero agreement on the sanity of your exposition
I could easily get a formal education in Mathematics instead of taking the crash and burn approach where I play around with things I know nothing about, reach incorrect conclusions, and then try to learn by figuring out why. Yes, my approach may be a terrible one if I wasn't just doing this for fun, as stated. I called you out because it's obvious that I don't care. I found an ordinal that couldn't possibly be in my $T$ (and every successive $T$ if adding rules that keep adding the least upper bound of $T$ in the class $Ord$).

I see that the elements of $T$ maybe wouldn't comprise an ordinal after the 8 rules I've given or may even need some patching, but since my goal was to try and traverse as many elements of $\omega_1$ as possible as opposed to also trying to enumerate them along the way, it didn't really matter to me.

I may try to come up with a sequence of rules that enumerates every ordinal and, if allowed to be a transfinite sequence of rules of any ordinal order type, would comprise a normal function from $Ord \rightarrow Ord$. This could be useful as a classification mechanism, who knows. Again, if I do it, I would be doing it for fun because it would be interesting to see what limits I could push $T$ sequences to.

If you think the above is a worthwhile endeavor and you seriously want to make sense of the $T$ sequence model so as to find a family of $T$ sequences that classify the ordinals in a useful way as opposed to just troll me Maschke, I'm all ears.

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#### AplanisTophet

What do you mean with "the choice function across ordinal indexes $i$ for each countable $s_i$"? Define it please.
I was confused as to the proper application of the proof that a countable union of countable sets is countable given some choice. If you recall, I was thinking that if both the ordinal indexes $i$ were countable and the sets $s_i$ themselves were countable, that we had a countable union of countable sets across the indexes. I was mistaken assuming the cofinality of $\omega_1$ is $\omega_1$.

However, you did ask me for an explicit sequence. If you'll instead allow me the axiom of choice, then I choose a sequence $T$ as defined here (assume the non-transfinite option):

https://math.stackexchange.com/questions/3425712/question-involving-strength-of-fodors-lemma-no-ch-assumption

You are saying that my sequence cannot be unbounded in $\omega_1$, but I believe that my sequence $T$ is an enumeration of $\omega_1$. One of us must be mistaken and I'm happy to assume it's me considering the likelihood of me getting a Fields Medal isn't too great. :unsure:

I think that what we are saying is that it takes $\omega_1$ iterations (a piecewise function with $\omega_1$ pieces), which doesn't fall under the typical definition of a bijection when we make the standard assumption that "no bijection can exist between $\omega_1$ and $\omega$"?