Non-standard elements of $\omega$ in ZF 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?
 A: 
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.
A: 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 $n$.
