I seem to have some basic confusion about how to understand finite ordinals when one allows the existence of some non-standard integers.
In ZF the axiom of infinity says that there exists an inductive set. We then define $\omega$ as the intersection of all inductive sets, i.e., the smallest inductive set (Jech, p. 13). Jech p. 20 also says $\omega$ is the smallest nonzero limit ordinal. In particular, it is an ordinal, namely it is transitive and well ordered wrt the $\in$ relation. Members of $\omega$ are called finite ordinals.
On the other hand, by the compactness theorem of first order logic, there should exist a non-standard model of ZF where there is an ordinal $c\in\omega$ that is larger than all the "standard" natural numbers $0, 1, 2, \dots$ ($c$ is "finite" by definition.)
Does that mean there is no such thing as the set of "non-standard" finite ordinals? Because if there were, there would be a smallest one, but there isn't a smallest one as each non-standard finite ordinal has a predecessor, which is also non-standard. And that would also violate the well-orderedness of $\omega$. Or equivalently the "standard" finite ordinals do not form a set? What does that mean exactly? That there exist no first order formula that can be used to describe these "sets", otherwise $\omega$ would not be the smallest inductive set?
But the principle of proof by induction still has to work here? (a property P holds for 0 and when it holds for $n$, it also holds for $n+1$; then it has to hold for all finite ordinals, even the non-standard ones). That is very confusing as there seems to be no way to go from the standard natural numbers to the non-standard ones.
Best Answer
You're considering a non-standard model $(M, \in_M)$ of ZF which contains an ordinal $c \in_M \omega_M$ that is larger than all standard natural numbers $0_M, 1_M, \dots$
Remember: here $0_M$ denotes the unique set $k \in M$ satisfying $\forall x \in M. x \not\in_M k$. Similarly, $1_M = 0_M \cup \{0_M\}$ denotes the unique set $k \in M$ so that $\forall x \in M. x \in_M k \leftrightarrow x = 0_M$. Finally, $\omega_M$ denotes the unique set $k \in M$ satisfying the definition of "$k$ is the intersection of all inductive sets" when all the variables in that definition range over elements of $M$, and $\in$ denotes $\in_M$.
All in all, the set $\omega_M$ that $M$ believes to be the set of natural numbers has more elements than the standard set of naturals $\mathbb{N}$.
Of course, there is no first-order formula in the language of ZF that could be used to define the "set of standard naturals". If you know the compactness theorem, you know why. And as you note in your argument, your model $M$ definitely does not contain a set whose $\in_M$-elements are precisely the standard natural numbers.
Yes, induction is a theorem of ZF so it works in any model of ZF. Inside $M$, the principle of induction asserts the following: if $P \in M$, and
all hold, then $P = \omega_M$.
If there was a set of standard naturals $S \in M$, this would allow you to conclude that $\omega_M$ was equal to that set of standard naturals $S$. But since there is no set of standard of standard naturals in $M$, the principle of induction does not allow you to conclude anything surprising.
Instead of using a set $P$, you could state the principle of induction using a ZF-formula $\varphi(x)$, but this doesn't allow you to conclude anything surprising either: as noted above, there is no formula in the language of ZF that can be used to define the set of standard naturals.