This builds on the previous article. In that article we showed that we don’t really know what cardinal numbers are, but we do know how they behave. Similarly we showed that Peano’s axioms do not define natural numbers but instead define any kind of progression. In this article we’ll define what a natural number is.
We begin with some description of the classical Cantor theory of ordinal numbers. This theory depends upon first defining the notion of order type and then designating ordinal numbers as certain special order types, namely, the order types of well-ordered sets.
Loosely speaking, an order type is the set of all simply ordered sets which can be put into 1-1 correspondence with a given simply ordered set such that the order is “preserved”.
First let’s define what a simple order structure is:
\(\check{A} = \langle A, R \rangle\) is a simple order structure if and only if \(R\) is a strict simple ordering of \(A\).
for example a simple order structure could be \(\check{N} = \langle N, < \rangle\) where \(N\) are the natural numbers. In general we use \(\omega\) instead of \(\check{N}\).
Let’s define now what it means to be similar:
\(\check{A} = \langle A, R \rangle\) is similar under \(f\) to \(\check{B} = \langle B, S \rangle\) if and only if:
Then we have that:
\(\check{A} = \langle A, R \rangle\) is similar to \(\check{B} = \langle B, S \rangle\) if and only if there is an \(f\) such that \(\check{A}\) is similar under \(f\) to \(\check{B}\)
Ok so now we can define an order type in intuitive set theory:
\(\tilde{A} = \{ \check{B}: \check{B}\text{ is similar to }\check{A}\}\)
The difficulty with this is exactly the same we encountered with the corresponding definition of cardinal numbers: we cannot prove that the equivalence class \(\tilde{A}\), which is the order type of \(\check{A}\), is not empty. We would have to introduce an extra axiom similar to the one we showed in the previous article.
In view of the difficulties with definition by abstraction of cardinal numbers or order types, it is fortunate indeed that for the special case of ordinal numbers, which are classically order types of well-ordered sets, a device may be adopted which requires no special axioms to support it. The idea is to choose just one representative of each order type which is an ordinal and call this representative the ordinal. That is, in the case of well-orderings we are actually able to construct a definite example of each possible well-ordering, which we cannot do in the case of arbitrary orderings. The definite representative we choose are built up from the empty set:
\(1 = \{0\}\)
\(2 = \{0, \{0\}\} = \{0, 1\}\)
\(3 = \{0, \{0\}, \{0, \{0\}\}\} = \{0, 1, 2\}\)
and so on. Each ordinal has all smaller ordinals as members, and the membership relation \(\in\) provides the appropriate well-ordering.
More formally, let’s define first:
\( \varepsilon A = \{ \langle x, y \rangle : x \in A\ \&\ y \in A\ \&\ x \in y \} \)
Now let’s define what it means to be complete:
\(A\) is complete if and only if every member of \(A\) is a subset of \(A\).
Finally then
\(A\) is an ordinal if and only if \(A\) is complete and \(\varepsilon A\) connects \(A\).
An interesting fact in ZF about ordinals:
There is no set \(A\) such that for every \(x, x \in A\) if and only if \(x\) is an ordinal.
The proof consists in showing that if there were such a set, say \(A\), then it would be an ordinal and thus a member of itself, violating the fact that \(\varepsilon A\) well-orders \(A\). That is, it is absurd to have \(A \in A\).
This theorem shows that in Zermelo set theory the Burali-Forti paradox of the greatest ordinal is blocked by the fact that the set of all ordinals does not exist.
Finally we can define the natural numbers by saying:
\(A\) is a natural number if and only if \(A\) is an ordinal and \(\widetilde{\varepsilon A}\) well-orders \(A\).
where \(\widetilde{\varepsilon A}\) is \(\{\langle x, y \rangle: y\ \varepsilon A\ x\}\) and now we have a proper definition of natural numbers (in ZF).
Content taken from Axiomatic Set Theory by Patrick Suppes