11.1 The Axioms

We normally think of the natural numbers \(\mathbb N\) as starting from \(1\). In this work, we shall, purely for convenience, start with \(0\). From this point, you should ‘pretend’ that you know nothing about the natural numbers!

We can think of the natural numbers from two, apparently different points of view. On the one hand, we use then to count, or measure the size of collections of objects. On the other hand, we use them to rank or order collections of objects into a linear list. In the former case, we refer to the natural numbers as (finite) cardinals, whilst in the latter case, as (finite) ordinals. We shall start off by considering the ordinal properties of the natural numbers and develop mathematically, the intuitive idea that for every natural number, there is a next one.

A simplified version of Peano’s axioms involve the assumption that \(\mathbb N\) is a set and that there is a function \(s:\mathbb N\to\mathbb N\), called the successor function, such that

\(s\) is an injection, that is to say, if \(s(x)=s(y)\) then \(x=y\);
\(s\) is not a surjection - more explicitly, there exists an element \(0\in \mathbb N\) such that \(0\not\in\text{ im }(s)\);
If there is a subset \(A\subseteq\mathbb N\) with the property that \[ 0\in A\text{ and }s(x)\in A\text{ whenever }x\in A \] then \(A=\mathbb N\).

Peano’s original axioms included conditions that relate to equality, “\(=\)”, but for simplicities sake we shall omit these. Axiom 1 is saying that two distinct natural numbers have distinct successors. Axiom 2 says that there is a first natural number. Axiom 3 is normally called the principle of mathematical induction.

Since \(0\in\mathbb N\) then on application of the function \(s\) we see that \(s(0)\in\mathbb N\). Moreover, since \(0\not\in\text{ im }(s)\) by axiom 2 then \(0\not= s(0)\). In a similar way \(s(s(0))\in\mathbb N\) and \(s(s(0))\not=0\). In addition, if \(s(s(0))=s(0)\) then by axiom 1 we could deduce the contradiction that \(s(0)=0\). Hence \(0, s(0)\) and \(s(s(0))\) are all distinct.

Continuing in this fashion we can, using only axioms 1 and 2, deduce the existence of the countably infinite set of natural numbers \[ \{0, s(0), s(s(0)), s(s(s(0))), \ldots\}. \]

One may be excused for thinking that only axioms 1 and 2 are needed to specify the natural numbers, but notice that there are other familiar sets that also satisfy these two axioms, for example the non-negative rationals or the non-negative reals. In addition, to formally determine that the list above doesn’t include repetition (as opposed to the hand-waving argument above), would involve an argument using induction, which is what axiom 3 provides for us. To show that Peano’s axioms characterise the natural numbers as we (think we) know them, we need axiom 3. Let \[ A=\{0, s(0), s(s(0)), s(s(s(0))), \ldots\} \] and notice that \(A\) satisfies the conditions given in axiom 3. Consequently \(A=\mathbb N\).

To show, formally that there is no repetition from elements in \(A\), let \(B\) be the set of elements from the list \[ 0, s(0), s(s(0)), s(s(s(0))), \ldots \] that do not coincide with any other element from the list. Notice that \(0\in B\) because of axiom 2. Suppose that \(x\in B\) and suppose, by way of contradiction, that \(s(x)\not\in B\). Then since, by axiom 2, it follows that \(s(x)\ne0\) then \(s(x) = s(y)\) for some \(y\) in the list. Consequently, by axiom 1, \(x = y\) and so \(x\not\in B\), a contradiction. Hence by axiom 3, \(B=\mathbb N= A\), and the elements in the list are all distinct.

Of course we usually abbreviate the notation and write \(s(0)=1\), \(s(s(0))=2\) etc. It is then clear that \(2=s(1), 3=s(2)\) etc.