I am currently trying to understand a little bit of axiomatic set theory, following Enderton's fun book "Elements of Set Theory" and am a little unclear about the set of natural numbers as defined in Chapter 4.
Firstly, let me note that the axioms in the book preceding the study of the natural numbers are the axiom of extensionality, the empty set axiom, pairing, unions, power sets and the axiom for constructing subsets.
At the beginning of Chapter 4 is given the definition of the successor $a^{+}=a \cup \{a\}$ of a set $a$. Enderton then defines $0=\emptyset$, $1=0^{+}$, $2=1^{+}$ and so on.
A set $a$ is called inductive if $\emptyset \in A$ and $a \in A \implies a^{+} \in A$. The axiom of infinity then asserts the existence of an inductive set, and the set of natural numbers $\omega$ is defined to be the intersection of all inductive sets. The existence of this set follows from the axioms mentioned.
Now clearly each set $0,1,2 \ldots$ belongs to the set $\omega$ since it contains $0=\emptyset$ and is closed under successors.
My question: is the converse true? Is every element of $\omega$ obtained from $0=\phi$ by applying the successor operation to $0$ finitely many times?
I presume that this can be deduced, but as far as I can tell it is not addressed in the book.
Note: If I had a way of constructing the "set" $X=\{0,1,2 ,\ldots n,n^{+},\ldots \}$ then it would be inductive, by construction, and therefore contain $\omega$ but I do not know how to construct the aforementioned set from the axioms given. So an equivalent question is: how can I construct $X$? (Perhaps the later axiom of replacement might help somehow?)
Grateful for any help!
Best Answer
Yes, the converse is true, but only for a somewhat ridiculous reason: "finite" means "in bijection with an element of $\omega$".
However, it is possible to construct (assuming $ZFC$ is consistent) a model of $ZFC$ in which $\omega$ has an element that is intuitively infinite; this model would be ill-founded, because the immediate predecessors of that element would form a decreasing $\in$-chain, but the key is that the set of those immediate predecessors is not in the model. So as far as the model is concerned, the Axiom of Regularity is satisfied.