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The ordinal numbers are themselves well-ordered. This would lead to the Burali-Forte paradox, except we work around this by saying the ordinals are a "class" and not a "set."

I'm wondering if an alternative approach exists with anti-foundation axioms.

Within these set theories, is it possible for the set of all ordinals to actually exist, as a true set?

The reason I ask is that with the ordinary Burali-Forti paradox, the set of all ordinals cannot exist, or else it would have to contain itself and hence define a new ordinal. But if ill-founded sets are allowed, there is no problem with the set of ordinals containing itself.

I'm sort of envisioning a structure where, as you define the ordinals, everything is nicely well ordered, well founded, etc. Then, only at the very top, when you want to look at the set of all ordinals, does this set turn out to contain itself and be ill-founded, and hence not define a new ordinal. But I'm not sure if there's just another paradox that emerges then.

Mike Battaglia
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    The Burali-Forti paradox is older than axiomatic set theory, and certainly before the axiom of foundation. So... no, the paradox is not dependent on the axiom of foundation. – Asaf Karagila Oct 19 '17 at 22:17
  • The paradox is older than the notion of a proper class, and was resolved by introducing a proper class. The question is if we can resolve it by introducing ill-founded sets instead. I'm saying to consider the set $\Omega$ which contains every ordinal and also contains itself. In other words, $\Omega = {1,2,...,\omega,\omega+1,...,\omega_1,...\omega_2,...\omega_{\omega_{\omega_{...}}},...\text{every ordinal}...,\Omega}$. Can this ill-founded set exist in any set theory with ill-founded sets? – Mike Battaglia Oct 19 '17 at 22:44
  • No, because the class of ordinals is naturally well ordered, and that would contradict that. – Asaf Karagila Oct 20 '17 at 05:51
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    @MikeBattaglia: You are really going to have to look at the detailed proof that there is no set of all well-orderings (or ordinals). See this relatively self-contained post here for motivation of well-orderings and ordinals and proof sketches for the fact (without using Foundation) that there is no set of all von Neumann ordinals. This also implies that there is no set of all well-orderings, otherwise you can easily extract from it the set of ordinals. – user21820 Dec 01 '17 at 16:12
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    And you don't 'resolve paradoxes' by introducing anything. In ZFC you cannot even say the set of ordinals is well-ordered. Rather, you can prove the sentence ( For every ordinals $x,y,z$ we have $x=y$ or $x \in y$ or $y \in x$, and if $x \in y \in z$ then $x \in z$, and if $x \in y$ then $x \ne y$, and every set of ordinals $S$ has a member that is in every other member of $S$. ) where "$x$ is an ordinal" too can be expressed as some sentence. Worse still, proper classes don't 'resolve' it, since there is still no class of all well-ordered classes. – user21820 Dec 01 '17 at 16:28

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Even if we allow $x\in x$ for general sets, it is impossible to have $x\in x$ for ordinals because $\in$ is a well-order for ordinals. In other words, Burali-Forti does not need a Foundation Axiom because for ordinals we have a foundation theorem.

Hagen von Eitzen
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  • Unless I misunderstand, this would only imply that $\Omega$ is not itself an ordinal, right? But we know that this must be true anyway, whether we use ill-founded sets or not. – Mike Battaglia Oct 19 '17 at 21:35
  • @Mike If $\Omega $ is a set, it is an ordinal, isn't it? – Andrés E. Caicedo Oct 19 '17 at 22:01
  • I'm saying to consider the set $\Omega$ which contains every ordinal and also contains itself. In other words, $\Omega = {1,2,...,\omega,\omega+1,...,\omega_1,...\omega_2,...\omega_{\omega_{\omega_{...}}},...\text{every ordinal}...,\Omega}$. $\Omega$ can't be an ordinal by the ordinary definition because if it were, then we'd have $\Omega = \Omega+1$. Ordinary set theory prohibits such an $\Omega$ existing because it isn't well-founded, but we're dealing with anti-foundation axioms. – Mike Battaglia Oct 19 '17 at 22:42
  • @Mike Such a thing cannot exist by comprehension. – Andrés E. Caicedo Oct 20 '17 at 00:55
  • @Hagen von Eitzen could you give some reference or skech of proof? I am unable to show, that if $x={x}$ then relation $\in_x={(y,z)\in x\times x:y\in z}$ does not well order $x$. After all $\in_x ={(x,x)}=x\times x$ and this set has only one nonempty subset (with the least element), so it is well ordered (usually formal definition of well ordering requies reflexivity). The only solution I see is to define well order to be anti-reflexive. – Przemek Dec 02 '17 at 19:47