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ZF's axiom of regularity implies that no infinite descending sequence of sets $x_1 \ni x_2 \ni x_3 \ni \cdots$ exists. Precisely this theorem asserts the non-existence of a map from $\mathbb{N}$ to sets satisfying the stated property. This may be seen to be distinct from the conceptually simpler statement that no sets satisfying the given statement exist; i.e. the statement, expressed in infinitary logic, that $\neg\exists x_1 \exists x_2 \exists x_3 \cdots (x_1 \ni x_2 \wedge x_2 \ni x_3 \wedge \cdots)$.

What I am wondering is, can a model of ZF fail to satisfy the second statement? Stated intuitively, can a model of set theory contain no infinite descending sequences of sets as objects in the model, and yet in actual fact contain an infinite descending sequence of sets?

I'm thinking, in this connection, of the fact that some models of ZF are countable, yet still satisfy the statement that there is no bijection between the rationals and the reals. From outside the model, in the "real world," we would say that there is a bijection between the model's rationals and the model's reals; yet no such bijection is an object of the model, and so from within the model Cantor's theorem seems to hold. Can an analogous situation obtain with respect to the absence of infinitely descending sequences?

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Yes. If there are models of $\sf ZF$ then there are non-well founded models of $\sf ZF$, which are also known as non-standard models.

To see this, given a model $M$ of $\sf ZF$ if it is non-standard then we are done. If it is a standard model, then we can take an ultrapower of this model by a free ultrafilter over $\Bbb N$. It is not hard to show that such model has to be non-standard. The reason is that we can consider the following functions in the ultrapower: $$f_n(k)=\begin{cases}0 & k<n\\k-n & n\leq k\end{cases}$$

What is interesting is that if there are models of $\sf ZF$ to begin with, then it is consistent that there are only non-standard models of $\sf ZF$ in the universe, but no standard ones. Therefore the consistency strength of the theory $\sf ZF$ with "There exists a standard model of set theory" is strictly stronger than that of $\sf ZF$ with "There exists a model of set theory".

We can in fact say more, e.g. the existence of a countable standard model does not imply the existence of an uncountable standard model, as shown here.

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Related. –  Andres Caicedo Aug 27 '13 at 22:57
    
@Asaf, does the fact that there are only non-standard models of $\mathsf{ZF}$ in the universe follows from the fact that $\mathsf{ZF}$ cannot prove the existence of a transitive set-model of $\mathsf{ZF}$?; as if the model was well-founded, then its transitive collapse would be an isomorphism. –  Camilo Arosemena Aug 28 '13 at 12:33
    
@Camilo: In my mind transitive and well founded are the same thing for this exact reason. Yes. –  Asaf Karagila Aug 28 '13 at 13:06

You can get a proper class model of $ZFC$ that is not well-founded. Recall a class model is $(B,E)$ where $B$ and $E$ are definable class where $E$ is interpreted as a binary relation. Being well-founded means $E$ is a well-founded relation in $V$.

Let $U$ be an ultrafilter over some set. Let $(Ult(V,U), E_U)$ be the ultrapower with respect to $U$. A easy exercise (Proposition 5.3 in $\textit{The Higher Infinite}$):

$U$ is $\omega_1$-complete if and only if $E_U$ is well-founded.

Note that having a $\omega_1$-complete ultrafilter is a very strong condition which implies the existence of measureable cardinals.

So if $U$ is any nonprincipal not $\omega_1$-complete ultrafilter, then $(Ult(V,U), E_U)$ is not well-founded. By Los's Theorem, ultrapowers of $V$ are models of $ZFC$. In particular, $Ult(V,U)$ models the foundation axiom. So $Ult(V,U)$ thinks that it is well-founded; however, $V$ knows that the definable relation $E_U$ is not well-founded.

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