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I hear there might be a countable ill-founded model of ZFC, i.e. one in which there is an infinite decreasing sequence $\alpha_0 \ni \alpha_1 \ni \alpha_2 \cdots$. This is possible when there is no function in the model which corresponds to said sequence.

Is it possible that all such $\alpha_i$ are ordinals of the model?

Damian
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2 Answers2

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In fact, every ill-founded model $M$ of $\mathsf{ZFC}$ contains an infinite descending sequence of ordinals. If $s_0\ni^M s_1\ni^M s_2\ni^M ...$ is a descending sequence in $M$, consider the sequence of ranks of the $s_i$s: letting $\alpha_i=rank(s_i)$, we have $\alpha_0\ni^M \alpha_1\ni^M \alpha_2\ni^M ...$

Put another way, we can tell if a model of $\mathsf{ZFC}$ is ill-founded just by looking at its ordinals.

Noah Schweber
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  • You have used $\ni^M$, please clarify if your answer needs the meta-theory to be set-like. The model should see every $\alpha_i$ as one of its ordinals, let us disregard what the meta-theory thinks of them. Is the answer still the same? – Damian Apr 24 '21 at 16:58
  • @Damian I don't understand your question. "$\in^M$" just refers to $M$'s interpretation of the symbol "$\in$" - there's no metatheoretic issue here. If you prefer, I could have said "for each $n$ we have $M\models \alpha_{n+1}\in \alpha_n$." (And I don't know what "set-like metatheory" even means.) – Noah Schweber Apr 24 '21 at 17:02
  • @Damian Maybe Asaf's answer is relevant to your metatheory concern - the point being that the rank calculation is taking place inside $M$ itself. – Noah Schweber Apr 24 '21 at 17:04
  • By "set-like metatheory" I mean a reasoning about ZFC inside ZFC, so there is an inner as well as outer $\in$. In such scenario it is ambiguous whether an ordinal is an ordinal of the inner or outer ZFC. Maybe it is even both. I wanted to be clear that "being an ordinal" in my original question is not a property of the meta-theory but the model. – Damian Apr 24 '21 at 17:19
  • @Damian Yes, the $\alpha_i$s are $M$-ordinals, not "true" ordinals. This is justified, again, by the internality of the rank construction: for every $x\in M$ there is some $M$-ordinal $\beta$ such that $M$ thinks that $\beta$ is the rank of $x$. – Noah Schweber Apr 24 '21 at 17:48
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The notion of the von Neumann rank is internal. That is, $\sf ZF$ proves that there is a definable function $r$ such that $r(x)=\bigcup\{r(y)\mid y\in x\}$ for all $x$. We can show that the range of this function is exactly the class of ordinals.

Now, if we have an ill-founded model $M$, with a decreasing sequence of elements $(x_n)_{n<\omega}$ (the sequence is not in the model, of course), we can apply this $r$ function pointwise to the sequence and obtain $(r(x_n))_{n<\omega}$ which is a sequence of ordinals of $M$, and by induction it is easy to see that it is indeed a decreasing sequence of ordinals.

Asaf Karagila
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