Using the Definition of Dedekind-Infinite to Replace the Axiom of Infinity

The axiom of infinity in ZFC (see this wikipedia link) allows us to construct the natural numbers. But I think we can drop that axiom and replace it with this one:

Axiom of Infinity: There exists a set $$M$$ and a non-surjective injective mapping $$\sigma: M \to M$$.

We can use this to construct the natural numbers.

Select a function $$\sigma$$ that injects a set $$M$$ into itself but does not hit all the elements of $$M$$. Assume $$0 \in M$$ is not in the range of $$\sigma$$.

Consider the family of all subsets of $$R$$ of $$M$$ with the following properties:

$$\tag 1 0 \in R$$
$$\tag 2 \text{The image of } R \text{ under } \sigma \text{ is contained in } R$$

So if we restrict $$\sigma$$ to $$R$$ and change the target to $$R$$, we have another function with the properties we are examining.

One can easily check that the intersection of two 'restriction sets' $$R$$ and $$S$$ in this family also satisfies both (1) and (2). In fact, we can intersect all of the subsets satisfying (1) and (2) and the result is a minimal such set

So assume that our function $$\sigma: M \to M$$ has no viable restrictions to any proper subsets.

It is easy to see that the only element not in the range of $$\sigma$$ is $$0$$.

Now let $$S \subset M$$ have the following properties:

$$\tag 3 0 \in S$$
$$\tag 4 \text{If } n \in S \text{ Then } \sigma(n) \in S$$

Then, since $$M$$ is minimal, $$S$$ must be equal to $$S$$.

But we have shown that $$\sigma$$ is a 'successor' function (hello Peano!) and (3) and (4) is the induction scheme.

We have shown that $$M$$ is the set of natural numbers, $$\mathbb N$$.

Question 1: Does this work indeed make sense in the ZFC setting?

Question 2: If yes, has this already been examined?

Question 3: If 1=yes/2=no, does it change anything in ZFC?

• – Asaf Karagila Sep 28 '18 at 7:07
• – Asaf Karagila Sep 28 '18 at 7:21
• @AsafKaragila If you put those links along with several sentences that would most likely have been my accepted answer. I'm now shocked that my question was not marked duplicate and down-voted... – CopyPasteIt Sep 28 '18 at 8:25
• To be fair, I haven't read your question in full. I had just woken up, and now I'm still busy. It might get closed later on during the day. Or, if you have a preference, I believe you can close your own question as a duplicate. – Asaf Karagila Sep 28 '18 at 8:26
• @AsafKaragila I'll leave that in the hands of the 'ordinal priests'. – CopyPasteIt Sep 28 '18 at 8:45

The answer is that it makes sense and doesn't change anything. Precisely, over ZF-Inf (= ZF without the axiom of infinity), your statement (call it "$$(*)$$") is equivalent to the usual axiom of infinity.

One direction is immediate: obviously the axiom of infinity implies $$(*)$$.

Here's a sketch of the other direction. Fix $$m\in M\setminus ran(\sigma)$$. For an ordinal $$\alpha$$ say that a map $$e:\alpha\rightarrow M$$ is a good embedding if:

• $$e(0)=m$$ (or $$\alpha=0$$).

• If $$\beta+1<\alpha$$, then $$e(\beta+1)=\sigma(e(\beta))$$.

Now call an ordinal $$\alpha$$ finitary (avoiding the word "finite" for obvious reasons) if it has a maximal element and no limit elements: that is, if $$(i)$$ $$\alpha=\beta\cup\{\beta\}$$ for some ordinal $$\beta$$ (unless $$\alpha$$ is empty), and $$(ii)$$ for every nonempty $$\gamma\in\alpha$$ we have that $$\gamma=\theta\cup\{\theta\}$$ for some ordinal $$\theta$$. The key point is this: we can show that for every $$k\in M$$ there is at most one finitary ordinal $$\alpha$$ such that $$k$$ is the image of the maximal element of $$\alpha$$ under some good embedding, and conversely every finitary ordinal admits exactly one good embedding into $$M$$. This lets us apply Replacement to get the set of all finitary ordinals; but we can show that this is exactly $$\omega$$.

So why use the axiom of infinity as it's written?

Well, note the use of Replacement above. There are very good reasons to want to be able to drop Replacement from time to time, so an axiom which requires Replacement to use in the obvious ways is immediately suspicious. Basically, by being so explicit the axiom of infinity can save us unnecessary axiomatic overhead, while apparently-reasonable alternatives may not do so.

What you're doing works, after a fashion.

But it will be cumbersome to work with, because even though your axiom claims that sets satisfying your criteria for $$M$$ and $$\sigma$$ exists, it doesn't in itself give you any way to point unambiguously to a particular $$M$$ and $$\sigma$$ as part of a logical formula. So the statement of every theorem that mentions integers would have to be wrapped boilerplate:

Theorem. Suppose $$M$$ is a set and $$\sigma$$ is an injection $$M\to M$$ and $$0\in M\setminus \sigma(M)$$. Then (bla bla bla bla), where $$S$$ is the smallest subset of $$M$$ that contains $$0$$ and is closed under $$\sigma$$.

The sensible way to use your proposed axiom would be to use it once and for all to prove that some particular distinguished representation of $$\mathbb N$$ exists -- for example the usual $$\omega$$! (You can produce $$\omega$$ from your axiom by applying Replacement to your $$S$$).

• It appears that the only sensible way of using the axiom is in 'naive set theory'. where the approach/proof is certainly of interest. For example, we can simply 'see' the natural numbers as urelements, oblivious to the inner workings and complexities of ZF. – CopyPasteIt Sep 28 '18 at 9:52