TITLE: Can a model of set theory think it is well-founded and in fact not be?
QUESTION [10 upvotes]: 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?

REPLY [4 votes]: 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.