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My textbooks says that there might be elements in $\omega$ that are not standard integers. But I have difficulty imagining how this can be, because if there is an element that is greater than all standard integers, then we can obtain an infinite descending chain by taking successive predecessors of this element.

For the above two facts to be compatible, the only possiblity is that the elements from the infinite descending chain does not form a set. Is this the case?

Jiu
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  • You can see e.g. this post as well as this one – Mauro ALLEGRANZA Jun 23 '21 at 14:20
  • @Jiu: You conclusion is correct. If ZF is inconsistent - it does not have a model - so there is no descending chain, no $\omega$ and no sets. If ZF is consistent then in any model, every non-empty internal subset of $\omega$ must have an $\in$-minimum element. I think this can be proven from the axiom of infinity without using the axiom of foundation - but not sure. –  Jun 23 '21 at 17:00

2 Answers2

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if there is an element that is greater than all standard integers, then we can obtain an infinite descending chain by taking successive predecessors

There are non-standard models of ZF or of ZFC, which are things that satisfy the axioms but in which the members of the model may be things other than sets and the membership relation may be something other than the usual membership relation.

In some non-standard models the members of the model are indeed sets and the membership relation is the membership relation among sets, but not all subsets of the model are members of the model. Those subsets that are members of the model are "internal sets" and those that are not are "external sets."

Your proposed infinite descending chain would be an external set. The proposition that asserts the non-existence of infinite descending chains would be true in this model because there is no internal set that is such an infinite descending chain.

Consider, for example, the proposition that if $a\in\omega$ then there is no one-to-one correspondence between $\{0,1,2,\ldots,a\}$ and $\{0,1,2,\ldots,a,a+1\}.$ If $a$ is a nonstandard member of $\omega,$ then that proposition is true in the nonstandard model because all of the one-to-one correspondences between those sets are external.

In particular, this explains the paradoxical fact that ZFC has countable models: a theorem asserts the uncountability—the nonexistence of an enumeration—of the power set of $\omega.$ Within a countable model, that proposition is true because there is no internal set that is such an enumeration.

Michael Hardy
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  • Would be good to also add that a model whose elements are sets and whose membership relation is the real one can’t have a nonstandard $\omega$, (At least not if the real membership relation is well-founded.) – spaceisdarkgreen Jun 23 '21 at 16:18
  • @spaceisdarkgreen : I'm not sure what you mean. The set that plays the role of $\omega$ in such a nonstandard model is not isomorphic to the usual $\omega$ if you admit external mappings. – Michael Hardy Jun 23 '21 at 16:30
  • Here is what I am saying precisely. Working in a ZF or ZFC background theory, if $(M,\in)$ is a model of ZF (where $\in$ is the restriction of the membership relation for our background theory), then $(\omega^M,\in)$ is isomorphic to $(\omega,\in)$ and so has no infinite decreasing sequences. – spaceisdarkgreen Jun 23 '21 at 16:42
  • @spaceisdarkgreen : But it can have external infinite decreasing sequences. – Michael Hardy Jun 23 '21 at 17:11
  • @spaceisdarkgreen : Consider this question: In what senses can it be isomorphic? Is there some one-to-one correspondence between them WITHIN either of the two models? Clearly not. So you must mean there is some set external to the two models that is an isomorphism between the two $\omega\text{s}.$ But that is false, since one of them has some external infinite decreasing sequences and the other does not. – Michael Hardy Jun 23 '21 at 17:13
  • What two models? There's only one model here, $M$. The isomorphism of $\omega^M$ onto $\omega$ is just the Mostowski collapse of $M.$ The point is that $(M,\in)$ can't have any infinite decreasing $\in$-chains, precisely because the $\in$ relation is (externally) well-founded. – spaceisdarkgreen Jun 23 '21 at 17:18
  • @spaceisdarkgreen : Before getting into anything else about what you're saying, if what you say is true, then what are we talking about here? How can there be non-standard members of $\omega,$ as referred to in the original posting, if there can't be a nonstandard $\omega\text{?}$ Do you mean only that in such a non-standard $\omega$ the membership relation is something other than the actual membership relation? If so, then what about ultrapowers of $\omega\text{?} \qquad$ – Michael Hardy Jun 23 '21 at 17:23
  • This is in reference to $\omega^{(M,E)}$ where $(M,E)$ is a model of ZF. I'm saying if $E$ is (the restriction) of $\in$, then $\omega^M$ is isomorphic to $\omega.$ If $E$ is some other relation, then $\omega^{(M,E)}$ may well be nonstandard. And there are certainly nonstandard models of $Th(\omega,\in)$ or PA or what-have-you, and certainly there are models of $ZF$ with nonstandard naturals (necessarily with $E\ne\in$) provided there are any models of ZF at all. – spaceisdarkgreen Jun 23 '21 at 17:31
  • @spaceisdarkgreen : ok, So that would mean that with ultrapowers of $\omega,$ the membership relation $E$ is not ${\in}. \qquad$ – Michael Hardy Jun 23 '21 at 17:35
  • @spaceisdarkgreen : So specifically, where does the abovementioned $E$ differ $\text{from }{\in}\text{?} \qquad$ – Michael Hardy Jun 23 '21 at 17:36
  • In an ultrapower of $(\omega,\in)$ by some ultrafilter $U$ on some set $A$ we have $fEg$ if and only if ${a\in A: f(a)\in g(a)}\in U.$ (where $f$ and $g$ are (equivalence classes of) functions $A\to \omega.$). So that's what $E$ is there. – spaceisdarkgreen Jun 23 '21 at 17:44
  • @spaceisdarkgreen : I know that "that's what $E$ is there", but it had not occurred to me that that $E$ is not coextensive with $\in.$ And when I said "where" I meant "where", i.e. at which point does it differ? $\qquad$ – Michael Hardy Jun 23 '21 at 18:02
  • @spaceisdarkgreen : BTW, for proper typesetting, notice this difference: $$ \begin{align} & E\ne\in \ & E\ne{\in} \end{align} $$ By civilized norms, the second form is correct. It is coded as E\ne{\in}. As soon as I saw what you typed, it was obvious that that was the remedy to the incorrect result that appeared. The reason is that in the latter form the {\in} is treated as the thing that comes after the =, so that the spacing to the left and right of = is the amount appropriate for an = between two things, whereas otherwise \in is treated as part of the binary relation symbol. – Michael Hardy Jun 23 '21 at 18:07
  • @spaceisdarkgreen : Apparently to many mathematicians who are experienced LaTeX users things like this not conspicuous. That is a fact that I would not have suspected until I saw it happen. – Michael Hardy Jun 23 '21 at 18:08
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    I don't understand the question. The ultrapower is a completely different set from $\omega$ and the relation is completely different from $\in.$ Not sure what it would mean to be coextensive. They could be isomorphic (e.g. if $U$ is principal). (And btw thanks for the typesetting tip... it was bothering me but decided not to bother figuring out how to fix it in the context of a comment.) – spaceisdarkgreen Jun 23 '21 at 18:09
  • @spaceisdarkgreen : ok, I'll treat it as an exercise to explain precisely what I meant and to understand this particular $E. \qquad$ – Michael Hardy Jun 23 '21 at 18:11
  • @spaceisdarkgreen : Initial guess: $E$ is less extensive than $\in. \qquad$ – Michael Hardy Jun 23 '21 at 18:56
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Yes. If $M$ is a non-standard model of ZF(C) with $\omega^M$ not well-founded, it follows that every $A\in M$ such that $M\models A\subseteq \omega$ will still necessarily be well-founded.

It follows that if $x_n\in M$ is a strictly descending sequence in $\omega^M$, then for any $A\in M$, $A\subseteq^M\omega^M$ with $x_n\in^M A$, we must have some $x\in^M A$ with $x<^M x_n$ for all $n$.

tomasz
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