This is how I interpret his post.You don't get it and you don't want to listen.

I want you to writeone single thing. Like Rule 1. Forget all the metaphysics about natural continuations of sequences or whatever it is you're doing. Give me the raw mathematical formalism,starting with the very first thing there is to know.

Then let me digest that and ask questions.

One idea. A few lines of text or a definition of some formula. Like Rule 1, but now that you've retracted your ellipses notation.

I'll make this concrete for you.

I will not read any post you write that is over 150 words in length.

You do recall I hope that a few posts back I asked you to "clarify or retract" the ellipses. I see you've retracted them. But please remember how often and how vehemently you have thought that it's "obvious what they are," and then later "obvious that there's no difference," then finally abandoned altogether.

So get some self-awareness and some humbleness.

Your lengthy posts arenot tethered to rationalityin my opinion. I don't seem to be able to communicate to you that your exposition is incoherent and meaningless; and your belief that it's practically obvious is a delusion. Not a mathematical confusion of some sort. A genuine break with reality that should generate in you -- I'll be blunt -- concern.

I want you to writeone single thingin 150 words or less. Something I can ask questions about and receive answers.

You essentially recursively build up a function as follows:

Take $\mathcal{A}_0 = \{1,2,3\}$.

Assume that $\mathcal{A}_n$ is defined, then we define according to rule 1,2,3:

1) $x\in \mathcal{B}_n$ if and only if there is some $a\in \mathcal{A}_n$ such that $a, a-1, a-2\in\mathcal{A}_n$ and such that $x=a-3$.

2) $x\in \mathcal{C}_n$ if and only if there is some $a\in \mathcal{A}_n$ such that $x = a+1$

3) $x\in \mathcal{D}_n$ if and only if there is some $a\in \mathcal{A}_n$ such that $a, a+1, a+2\in \mathcal{A}_n$ and such that $x=a+\omega$.

Then we define

$$\mathcal{A}_{n+1} = \mathcal{A}_n\cup \mathcal{B}_n \cup \mathcal{C}_n\cup \mathcal{D}_n$$

Example:

$$\mathcal{A}_0 = \{1,2,3\}$$

We have

$$\mathcal{B}_0 = \{0\},~\mathcal{C}_0 = \{4\},~\mathcal{D}_0 = \{\omega\}$$

Thus $\mathcal{A}_1 = \{0,1,2,3,4,\omega\}$.

Next, once we take for an ordinal $x$, the level of $x$ to be defined as the least $n$ such that $x\in \mathcal{A}_n$. There are finitely many ordinals of level $n$. We then order the ordinals on their levels.

For example:

Of level 0: 1,2,3

Of level 1: 0, 4, $\omega$

Final sequence: 1,2,3, 0, 4, $\omega$, (here come the ordinals of level 2),...

I think this is what he WANTS to do since what he wrote makes little sense. But he writes things very differently and I think he makes several mistakes in his post.

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