Confusion regarding the definition of $\omega$ (the set of natural numbers) in a model of ZFC

Question

Suppose that ZFC is consistent. In a model of ZFC, an inductive set is a set $A$ satisfying $\emptyset\in A$ and $n\cup\{n\}\in A$ for every $n\in A$. Suppose that $X$ is the inductive set given by the axiom of infinity. The definition of $\omega$, the set of natural numbers, is in my mind $$\omega = \bigcap_{X'\subset X, X'\text{ is an inductive set}} X'.$$ Now, I've heard that $\omega$ may not be the standard model of Peano arithmetic. But how is this possible? If $\omega$ is a nonstandard model of PA, then it has a proper subset that is the standard model of PA (a property shared by any nonstandard model of PA), and that subset is itself an inductive set!

What parts of my understandings are wrong here? Any help appreciated.

Answer 1

Essential point is that $\mathcal{P}(X)$ doesn’t necessarily include “every subset of $X$ you can imagine”. Rather, its members are just the subsets of $X$ that actually exist in the model. So, in a model with non-standard $\omega$, there is no set whose members are only the initial segment of $\omega$ that’s order-isomorphic to “standard $\mathbb{N}$”

Answer 2

Sure, there's a 'standard cut' of $\omega$, but that's not an element of the model. Models aren't closed under (external) subsets, and $\omega$ is merely the smallest inductive set in the model.

One could ask the same question about the nonstandard model just in the context of arithmetic and get a similar answer. After all, the induction schema essentially says that 'the universe is the smallest inductive set', so when our universe is actually a nonstandard model, we might point to its standard cut, a smaller inductive set and ask why we can't prove by induction that that is the whole universe.

Of course the solution is that the standard cut just can't be first-order definable (i.e. the overspill principle), and thus can't be plugged into the induction schema.

In both cases, we know externally that the standard cut is there, but internally the cut is invisible and the nonstandard model is indistinguishable from the standard one.