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For the first infinite von Neumann ordinal $\omega$, I think I understand (per threads here and here) the distinction between "standard" and "non-standard" $\omega$. But I'm wondering how the notion of "standardness" (or non-standardness) can be extended to larger ordinals.

  • Is there a notion of "standardness" that can meaningfully be defined for larger ordinals (like $\omega_{1}$ for example)?
  • If a set universe has non-standard $\omega$ does that necessarily imply non-standardness of larger ordinals?
  • Is it possible for $\omega$ to be standard but some larger ordinal to be non-standard?

My semi-educated guess would be that "standardness" of all von Neumann ordinals flows from the standardness of $\omega$ (if $\omega$ is standard, then it's essentially true by definition that the other ordinals are also standard). But I'm not at all certain of this...

NikS
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    The notion you're looking for is well-foundedness. Because of the rank function, any non-wellfoundedness (i.e. any infinite decreasing membership chain) leads to an infinite decreasing sequence of ordinals. It is possible for an $\omega$-model to not be well-founded... the funny business just starts higher up rather than right after the standard naturals. – spaceisdarkgreen May 13 '23 at 07:02
  • @spaceisdarkgreen: In an $\omega$-model of ZF, I would think that the Axiom of Foundation should rule out such non-wellfoundedness, should it not? – NikS May 13 '23 at 08:31
  • No. Why should foundation rule it out for $\omega$-models when it doesn't for non-$\omega$-models? – spaceisdarkgreen May 13 '23 at 16:37
  • For reference – spaceisdarkgreen May 13 '23 at 21:46
  • @spaceisdarkgreen : Are you saying that in a non-well-founded $\omega$-model of ZF, there can exist an infinite membership chain? That would seem inconsistent with the Axiom of Foundation. For example in Jech’s Set Theory (1st ed., free at archive.org books), sub-chapter 9 starts by stating the Axiom of Regularity (a.k.a. Axiom of Foundation), then says “as a consequence, there is no infinite sequence $x_0 \ni x_1 \ni x_2 \ni \ldots$” – NikS May 14 '23 at 07:08
  • Or am I getting confused somehow between the $\textbf{E}$ relation of the model and the “true” $\in$ of the universe? – NikS May 14 '23 at 07:16
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    @NikS In any non-well-founded model $M$ of ZFC, there exists (from our external point of view), an infinite descending membership chain. That chain is given by a function from our external $\omega$ to $M$. An infinite descending membership chain internal to $M$ would be an element of $M$ which is a function with domain $\omega^M$. Now there is no reason why the existence of the former should imply the existence of the latter, even if our $\omega$ is isomorphic to $\omega^M$. – Alex Kruckman May 14 '23 at 14:16
  • @NikS What Alex said. But the point I'm trying to emphasize is that non-$\omega$-models are ill-founded, so this is nothing new. If the model's natural numbers are not (externally) well-ordered, then there is (externally) an infinite descending membership chain. The new thing here is that $Ord^M$ can fail to be (externally) well-ordered, even if its initial segment $\omega^M$ is. – spaceisdarkgreen May 14 '23 at 16:04
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    @NikS Oh, I think maybe I see the point of confusion. By infinite descending membership chain we mean a sequence $x_n$ of elements of $M$ such that $x_{n+1}Ex_n$ for all $n$. Now, if $M$ is an $\omega$-model, then we could also view this as a map from $\omega^M\to M$ satisfying this property. But there's no guarantee that this mapping corresponds to a set in $M$... a model will generally be missing some of these functions (remember, countable models exist and in that case they clearly have to be missing some). So $M$ can still satisfy foundation. – spaceisdarkgreen May 14 '23 at 17:50
  • @spaceisdarkgreen : Ah, OK. So, to summarize: If we define *”actually” finite* according to the normal notion of counting with the (standard) natural numbers, then 1) In a well-founded model of ZF, every descending membership chain is “actually” finite 2) A non-well-founded model $M$ of ZF (whether $\omega$-model or not) contains descending membership chains which are “actually” infinite but the model doesn’t “think” of such a chain as an “infinite sequence” because $M$ lacks any function which implements a bijective mapping between $\omega^M$ and the sets in the chain…(cont’d) – NikS May 15 '23 at 03:56
  • …3) An ordinal $\alpha$ is “standard” iff: for every descending membership chain comprised of members of $\alpha$, the chain is “actually” finite. Have I got these points 1-3 about right? – NikS May 15 '23 at 03:57
  • @NikS Mostly. It's not really that the model doesn't think of the chain as an infinite sequence so much as it doesn't know about the chain at all. – spaceisdarkgreen May 15 '23 at 04:46
  • @spaceisdarkgreen - To clarify: you mean that the model “knows about” the chain iff it contains a function that implements a bijective mapping between $\omega^M$ and the sets in the chain? (i.e. the chain exists in the model even if the model doesn’t “know” about it in the aforementioned sense of “knowing”) – NikS May 15 '23 at 05:08
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    @NikS A chain is a sequence... it is a function between $\omega$ and $M.$ Yes, by "knows about" I mean this function is "implemented" inside $M.$ In very precise terms, if $f:\omega\to M$ is the chain in question, then is there a $f^M\in M$ whose $E$-elements are exactly ${(n^M, f(n))^M:n\in \omega},$ where $(,)^M$ means $M$'s ordered pairing function and $n^M$ is $M$'s interpretation of the natural number $n$. – spaceisdarkgreen May 15 '23 at 05:37

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