Will we still have Burali-Forti paradox if we use Zermelo's definition of ordinals instead of the Von Neuman's definition? Because I think when we define the ordinals as {}, {{}}, {{{}}}, ... (using Zermelo's definition) then the set of ordinals which is { {}, {{}}, {{{}}}, ... } can not contain itself by definition hence prevents the paradox. If I am correct then what is it about Von Neuman's definition that raises Burali-Forti paradox?

  • $\begingroup$ Those were once used as a definition for naturals, but also for ordinals? $\endgroup$ – Hagen von Eitzen Sep 20 '17 at 20:58
  • $\begingroup$ @ Hagen Can't we use Zermelo's definition as the definition of ordinals? $\endgroup$ – K. Smith Sep 20 '17 at 21:02
  • $\begingroup$ You can use that as teh definition of hypre-gizmos if you like, but if youwant to use it as definition of ordinal, the question is: What desirable properties of ordinals would that definition convey? Apparently, Zermelo used a definition very similar to von Neumann's in 1915 (google finds), the idea being that each ordinal is the set of all smaller ordinals $\endgroup$ – Hagen von Eitzen Sep 20 '17 at 21:10

The point of the Burali-Forti paradox is that there is no largest well-ordered set.

Simply because if there is such a set, then by collecting all the initial segments of such a well-ordering (including the set itself), we obtain a well-ordering which is strictly longer, by one point to be precise, and thus we get a contradiction to the maximality of the order.

The actual definition of an ordinal is irrelevant. We just need to be able to run the basic comparison argument on well-orders (i.e. given two well-ordered sets, one of them is isomorphic to an initial segment of the other), and that given a linearly ordered set, we can form the set of its initial segments.

Of course, using the full power of $\sf ZF$, we don't care about arbitrary well-orders, just about the ordinals, so the paradox is seemingly about the class of ordinals. But it really isn't, it's a paradox about having a largest well-ordered set. Just like Cantor's paradox is about having a largest cardinal.

  • $\begingroup$ Got it. Thank you very much. $\endgroup$ – K. Smith Sep 20 '17 at 21:22

Zermelo's definition only works for finite ordinals - what should $\omega$ be? "$\{\{\{\{...\}\}\}\}$" isn't a set (think about the axiom of foundation). Burali-Forti works because we can prove that the set of ordinals (assuming it exists) is an ordinal; this fails for Zermelo ordinals.

  • $\begingroup$ Well, Zermelo's theory did not have Foundation... $\endgroup$ – Asaf Karagila Sep 20 '17 at 21:04
  • $\begingroup$ @AsafKaragila There's still no good notion of infinite Zermelo ordinal, so far as I know, even without Foundation. $\endgroup$ – Noah Schweber Sep 20 '17 at 21:19
  • $\begingroup$ Of course. I was just nitpicking. :) $\endgroup$ – Asaf Karagila Sep 20 '17 at 21:20
  • $\begingroup$ @Noah Thanks for the answer. It made it more clear for me complementing Asaf's answer. $\endgroup$ – K. Smith Sep 20 '17 at 21:26
  • $\begingroup$ One remark that perhaps should be given, is that even if you take into considering this encoding, Zermelo's axiom of infinity required there is a set whose elements are the natural numbers. This is not $\{\{\ldots\}\}$, but rather $\{\varnothing,\{\varnothing\},\{\{\varnothing\}\},\ldots\}$. One could argue that this can represent $\omega$. Indeed, there is no easy way to axiomatize the $<$ membership of this coding using $\in$, and the only reasonable way goes through transitive closures which give us the von Neumann ordinals already. $\endgroup$ – Asaf Karagila Sep 21 '17 at 6:27

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