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From Set-theoretic definition of natural numbers

n+1=n $\cup$ {n}

i.e.

  • 0 = {}
  • 1 = {{}}
  • 2 = {{},{{}}}
  • etc

It seems to me that a simpler, equally valid definition would be

n+1={n}

i.e.

  • 0 = {}
  • 1 = {0} = {{}}
  • 2 = {1} = {{{}}}
  • etc

Why is the ZF definiton preferred over mine?

(The only "advantage" I can think of is that in ZF, $|n|=n$, but this strikes me as being circular reasoning.)

spraff
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    Try to define "$n<m$" in both cases. Which one is easier? Try to define "$X$ has size $n$". Which one is easier? – Jonathan Schilhan Aug 23 '23 at 09:39
  • A similar comment to Jonathan's: consider that the cardinality of $n$ is $n$ in this construction. In yours, all the naturals are singletons and are all isomorphic. If you want natural numbers to be a special case of ordinals, which you do, then having everything a singleton with no concept of order is useless – FShrike Aug 23 '23 at 12:26
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    Have you tried proving the Peano Axioms using your definition? How is addition of natural numbers defined? How do we get 1+1 = 2 with your definition? As @FShrike said, the Ordinals are a very important Class in set theory, and so it is convenient to have the natural numbers be a special case. – Michael Carey Aug 23 '23 at 16:18
  • With the usual definition, every integer is a transitive set, and the $\lt$ relation simply is membership: $m \lt n \iff m\in n$. With your definition, $m \in n \iff n = m+1$, and you still have to define $m\lt n$ — for example, as $m\in \operatorname{TC}(n) =$ transitive closure of $x$ — and show it’s a total order. – BrianO Aug 24 '23 at 06:50
  • Related: https://math.stackexchange.com/questions/85672/the-history-of-set-theoretic-definitions-of-mathbb-n – Asaf Karagila Aug 25 '23 at 05:21

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