Can a model of set theory think it is well-founded and in fact not be?

Solution 1:

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.

Solution 2:

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.