3
$\begingroup$

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!

  • 2
    How do you define "finitely many times"? (This might be broken by the existence of $\omega$-inconsistent models…) – 2017-02-24
  • 2
    Relevant: [non-standard model of arithmetic](https://en.wikipedia.org/wiki/Non-standard_model_of_arithmetic). – 2017-02-24

1 Answers 1

2

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.

  • 0
    I think I understand your second point: to the first order theory ZFC I can add a schema of axioms asserting that each inductive set admits an element greater than $n = 0^{++\ldots +}$, one axiom for such $n$. In any model of this theory $\omega$ will admit an element larger than $n$ for each such $n$. By the compactness theorem, if a model of ZFC exists a model of this theory exists. So the claim that $\omega$ admits a non-standard element is consistent with ZFC. – 2017-02-26
  • 0
    I cannot say that I really understand your first point though. Admittedly I would have preferred not to use the word "finite" in my original formulation of the question, but I don't know what to replace it by. Expressing the fact that $\omega$ has only the standard elements appears to be outside the capacity of finitary first order logic. I don't really see how I could reformulate my question in a useful way. – 2017-02-26
  • 0
    @foundationalNovice Actually, it sounds like you understand my first point perfectly! The notion of "finite in the standard sense" cannot be formulated in first-order logic, but ZFC (being a first-order theory) can only prove things that can be formulated in first-order logic. So ZFC can only prove what it *thinks* we mean by "every element of $\omega$ is generated by finitely many successor steps"; but if the model is ill-founded, it can be wrong about what "finite" means. So I guess the answer is that you *can't* reformulate your question in a useful way. – 2017-02-26