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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!

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    $\begingroup$ How do you define "finitely many times"? (This might be broken by the existence of $\omega$-inconsistent models…) $\endgroup$
    – Akiva Weinberger
    Commented Feb 24, 2017 at 5:19
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    $\begingroup$ Relevant: non-standard model of arithmetic. $\endgroup$
    – Clive Newstead
    Commented Feb 24, 2017 at 5:22

2 Answers 2

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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.

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  • $\begingroup$ 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. $\endgroup$
    – foundationalNovice
    Commented Feb 26, 2017 at 9:55
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    $\begingroup$ 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. $\endgroup$
    – foundationalNovice
    Commented Feb 26, 2017 at 10:03
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    $\begingroup$ @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. $\endgroup$
    – Reese Johnston
    Commented Feb 26, 2017 at 16:02
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From reading this answer, I think that we can define a natural number as a finite ordinal number then we can construct the real numbers from them then redefine a natural number as a real number that every inductive set contains. Which things we construct and call real numbers is not important. It can be shown that there is a complete ordered field which is unique up to isomorphism and we can define the real numbers with addition, multiplication, and inequality to be any complete ordered field. In some constructions, every real natural number is the finite ordinal number it corresponds to. In other constructions, that is not the case. In the case where every real natural number is the finite ordinal number it corresponds to, we still cannot prove that $\omega$ is a real number. We can only prove that for every natural number, the set of all natural numbers less than it is a real number because the set is in fact that natural number itself. However, we can still construct the intersection of all inductive sets and prove that that set exists. We just can't prove that that set is a real number. We can't disprove it either if all you assume is that the real numbers are a complete ordered field and that every real natural number is the finite ordinal number it corresponds to. For example, we could pick a wierd construction that includes $\omega$ as one of the real numbers. Then you would have to be careful to realize that for each natural number $x$, the statement $\omega > x$ could mean using the inequality relation defined on the ordinal numbers or using the inequality relation defined on the real numbers.

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