I am having some issues using the Paeno axioms to prove that closure under addition exists within the natural numbers. I think that a large part of my issue stems from my confusion over the notation used by my professor.

We are given that N is defined as:

N := { x $\in$ Z | $\exists$ m $\geq$ 1 s$^m$(0) = x}

I guess I am most confused by what "m" is. Does "m" is just a way of 'counting' the "s's" in the set? Am I overthinking this? Any help would be appreciated!

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    $\begingroup$ $s^1(0)=s(0)=1$, and $s^{m+1}(0)=s(s^m(0))$. In other words, $s^m(0)=m$. Of course, it seems like a little circular of a definition. $\endgroup$ – Hayden Apr 15 '17 at 5:32
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    $\begingroup$ I'm guessing there's some important context you're leaving out here, since that "definition" of $\mathbb{N}$ indeed looks very circular, and is not at all the same as "defining the natural numbers using the Peano axioms". What are the Peano axioms supposed to have to do with this definition of $\mathbb{N}$? $\endgroup$ – Eric Wofsey Apr 15 '17 at 5:36
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    $\begingroup$ How was Z defined? This... could work. You have to make a .... thing... that is 0, and it has successors and so you get a set of ... things... of which m can be the notation of one so that $s^m $ is meaningful, and then you can give these things the name natural numbers.... but I'm not comfortable with that. $\endgroup$ – fleablood Apr 15 '17 at 6:32
  • $\begingroup$ I'd be a tad happier if the definition was $\mathbf N:=\bigcap \{\,A\subseteq \mathbf Z\mid 0\in A, \forall x\in A\colon s(x)\in A\,\}$ $\endgroup$ – Hagen von Eitzen Apr 17 '17 at 9:39
  • $\begingroup$ @HagenvonEitzen: Yeap that's the other 'standard' way of getting the closure under some operations, so it would work after enough work, but I'd argue that the easiest way to get to the integers is still from the naturals, and this backward 'definition' can't be meant to ensure that naturals are a subset of the integers, for two reasons: (1) We can first construct $Z$ to be isomorphic to the desired integers, then replace the appropriate elements with the 'true' naturals; (2) We can't possibly do the same in extending from integers to rationals to reals to complex numbers.. $\endgroup$ – user21820 Apr 17 '17 at 9:50

There is a problem with your professor's definition. We more or less already need the notion of the collection of natural numbers satisfying the Peano Axioms, even before it makes sense to talk about iterating a function, in this case $s$. In fact, the standard way to define iterates is:

Let $s^0$ be the identity function on $\mathbb{Z}$.

Let $s^{n+1} = s^n \circ s$ for every natural number $n$.

How can we possibly do this without the natural numbers? We essentially cannot, because the natural numbers are precisely what we need to use to count the iterations!

The reason I can say that your professor is making a mistake is that the quantification of $m$ is ill-defined; it quantifies over all "$m \ge 1$", but what is that even supposed to mean? It is in fact meaningless unless $m$ is restricted to some kind of number, not to say that "$s^m$" is meaningless unless $m$ is a natural number (or integer if $s$ is invertible; presumably it is not in this case). So it already is necessary to know $\mathbb{N}$ before any of the notation makes sense.

However, if you do things carefully in set theory, you could first construct an inductive set $ω$, and then somehow define the integers $\mathbb{Z}$ without using natural numbers, and then define the natural numbers via the recursion theorem. This is not going to be anywhere as 'simple' as the ill-defined definition quoted. For details of how to do this rigorously in ZFC, see this post. If a different formal system is being used, then he will have to specify it.


As you already have Z as the ordered ring of integers your definition just says N is the set of positive integers. You are asked to prove that the positive integers are closed under addition. In your definition $s$ is the successor function and the exponent $m$ is repeated application, so $m=x$. What you would like to say is given $x,y \in $N you can find $m,n$ such that $s^m(0)=x, s^n(0)=y,$ then $ x+y=s^{m+n}(0)$, so $x+y \in $N. As others have said, this is a strange order of definitions, and whether this is easy to prove will depend on what those definitions are.

  • $\begingroup$ Yes indeed, but observe that the "what you would like to say" part of your answer relies on $m,n$ being positive natural numbers and that $m+n$ is also a positive natural number. Doesn't that sound precisely like what you call "closed under addition"? So it's more likely that the proposed definition is circular than that there is some actual proper mathematics intended by it, at least from the looks of it. Right? =) $\endgroup$ – user21820 Apr 18 '17 at 5:19

Usually the integers are constructed from (i.e. defined in terms of) the natural numbers, not the other way around. Usually the natural numbers are defined using some version of Peano's Axioms. Then the integers are defined in terms in the natural numbers, typically as follows::

$\forall a: [a\in Z \iff a\subset N^2 \land a\neq \emptyset$

$\land \forall b,c:[(b,c)\in a \implies \forall d,e\in N:[(d,e)\in a \iff b+e=c+d]]]$

Then addition on $Z$ is usually defined as a subset of $Z^3$. (I can't remember the exact form.)

Perhaps your professor meant his statement to be a theorem about $N$ and $Z$ rather than a definition of $N$. Even in that case, however, I think the theorem would be poorly stated. It is no wonder you are confused.

  • $\begingroup$ This should be a comment, since it doesn't actually address the question. $\endgroup$ – Noah Schweber Apr 18 '17 at 0:47
  • $\begingroup$ @NoahSchweber See my edits. $\endgroup$ – Dan Christensen Apr 18 '17 at 2:21
  • $\begingroup$ @NoahSchweber I'm guessing the OP's "definition" was actually a theorem. That would make more sense to me. $\endgroup$ – Dan Christensen Apr 18 '17 at 2:51

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