I'm new to set theory and I'm studying how natural numbers are defined.
I know that the natural numbers are defined as
$0$ (zero) = $\emptyset$
$n^+$ = $n \cup$ {$n$}
And one of the most important property of natural numbers is the following:
If $m$ and $n$ are natural numbers, then $m \lt n$ means $m \in n$
It seems to me that the important point in defining natural numbers is the recursive way we use it and not the starting point. Therefore, i'm wondering if it is technically valid to define the natural numbers without starting from the empty-set or not (suppose that there is one crazy man who wants to do it), say:
$0$ (zero) = {apple}
$n^+$ = $n \cup$ {$n$}
Therefore, 1 = $0 \cup $ {$0$} = {apple, $0$}; 2 = {apple, $0$, $1$}; 3 = {apple, $0$, $1$, $2$}. And the above property of natural numbers is always satisfied.
Could you please help me with this question and could you please share some insights about the technical advantage of choosing $\emptyset$ as the starting point (beside of the fact that it is very intuitive) ?
Thank you very much for your help!
The whole point is that the only set whose existence you admit axiomatically is the empty set, so the simplest choice to construct the set of natural numbers. But of course you can start by building the set $A=\{\emptyset,\{\emptyset\}\}$ and use it as your elementary block in the construction. All you need to construct natural numbers is their properties, not their elements, and $\{A, \{A,\{A\}\}, \{A,\{A,\{A\}\}\}\},\dots\}$ has them. You define zero to be $A$ and the union to correspond to the "successor" relation and Peano's axioms are satisfied.