Is Infinite descending sequence reverse of construction of Zermelo ordinals?

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I wonder how do they deduce "Infinite descending sequence of sets" from "Axiom of Regularity". So, how one imagine and say hey people! there is this "infinite descending chain" and we should avoid it?

So, when i see Zermelo ordinals, i realized that they are infinitely ascending chains.

$$ 0 = \{ \}, $$ $$ 1 = \{0\} = \{\{ \}\}, $$ $$ 2 = \{1\} = \{\{\{ \}\}\},$$ $$ n = \{n−1\} = \{\{\{...\}\}\}, $$

Thus, i think when they(Zermelo et al.) try it in reverse order so bigger number to smaller number they got infinite descending chains like -1,-2,-3,...

Question 1: Would this right or "Infinite descending sequence of set" are some usual thing without requiring this realization by Zermelo et al.?

Question 2: Does Infinitely descending sequence relates to negative numbers(-1,-2,-3,...) this way or not how?

Question 3: So, then they proposed the well ordering theorem this way, right? There should be a minimal element, an maximum is not accepted, right?

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Implementing ill-founded structures in $\mathsf{ZFC}$ - such as $(\mathbb{Z}:<)$ - is straightforward as soon as we realize that there's no reason to look for too simple implementations. For example, we can define $\mathbb{Z}$ in terms of $\mathbb{N}$ and basic set-theoretic operations: we treat an integer as an equivalence class of pairs of natural numbers with respect to the relation $$(a,b)\sim (c,d)\quad\iff\quad a+d=c+b$$ (intuitively, $(x,y)$ means $x-y$). Similarly, rationals are defined in terms of pairs of integers, reals are defined in terms of sets of rationals, and so forth. The point is to get away from the idea that order has to be generated by $\in$: while sometimes we can make that work, we don't want to commit to it too much.

Incidentally, nobody really uses the Zermelo ordinals anymore except in very rare situations; the von Neumann ordinals are the things to focus on first. For example, they generalize to the transfinite more naturally, and their ordering is simpler to define (it's just $\in$).

The basic takeaway is that we don't need ill-founded sets for anything: a rich enough structure of well-founded sets can implement ill-founded structures.


As to why we actively focus on well-founded sets, this comes down to a philosophical point. To avoid Russell's paradox, we want some intuition about the universe of sets which constrains what sort of comprehension (= set formation) we can have. There are various ways to do this (consider e.g. $\mathsf{NFU}$ or $\mathsf{GPK^+_\infty}$), but the $\mathsf{ZFC}$ approach is to start with the idea of a cumulative hierarchy. And we trivially only get well-founded sets that way.

  • This is not to say that by ruling out ill-founded sets we escape Russell's paradox! Adding axioms never makes a theory more consistent. Rather, the idea of the cumulative hierarchy - which trivially rules out ill-founded sets as a byproduct - motivates a particular set of axioms, the $\mathsf{ZFC}$-axioms or some variant thereof ($\mathsf{ZF}$, $\mathsf{Z}$, $\mathsf{ZC}$, ...), in which Russell's paradox fails to go through.

That said, there's no rule saying that we can't consider versions of $\mathsf{ZFC}$ without regularity/foundation. The relevant term here is antifoundation axiom; the best known one is Aczel's, but I personally find Boffa's equally interesting (although it leads to some rather odd results - see e.g. here).