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I am little bite confused. I know that in a ZFC model can be defined the set of natural numbers. I know, for instance, how natural numbers are constructed in a von Neumann Universe. My question is how could I define a natural number by a sentence in the first-order lenguage of theory of sets? How could I construct natural number in arbitrary ZFC model?

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    The natural numbers are the finite ordinals. Or at least, that's one way for using sets to model them. – Asaf Karagila Apr 22 '17 at 12:41
  • So, how do you define 'being a finite ordinal' in a first-order language? –  Apr 22 '17 at 12:51
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    See in my answer here: https://math.stackexchange.com/questions/1308042/what-do-people-mean-by-finite/1308179#1308179 – Asaf Karagila Apr 22 '17 at 12:55
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    You can also just talk about being an element of "the smallest inductive set". – Asaf Karagila Apr 22 '17 at 12:57
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    You can also do something that I find amusing, and define "$x$ is a natural number" as "$x$ is a transitive set of transitive sets, and if $A$ is a set such that $x\in A$ and $A$ is closed under the operation $y\mapsto \bigcup y$, then $\varnothing\in A$." This lets you define a natural number (and arithmetic operations) even when you remove the Axiom of Infinity. – Malice Vidrine Apr 22 '17 at 14:18

2 Answers2

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To answer a question you asked in the comment :

$n\subset m$ is an abbrevition for $\forall x, x\in n\implies x\in m$

"$x\neq \emptyset$" is an abbreviation for $\exists z, z\in x$

"$n$ is transitive" is an abbreviation for $\forall x, x\in n\implies x\subset m$

"$\in$ is a well-ordering on the transitive set $n$" is an abbreviation for "$n$ is transitive and $(\forall x, x\neq \emptyset \implies (x\subset n\implies \exists z, z\in x\land \forall y, y\in x\implies z\in y\lor z=y))\land (\forall z, z\in n\implies \neg z\in z) \land (\forall x\forall y\forall z, (x\in n\land y\in n\land z\in n)\implies ((x\in y\land y\in z)\implies x\in z))$

"$n$ is an ordinal" is an abbreviation for "$\in$ is a well-ordering on the transitive set $n$"

$x=y\cup\{y\}$ is an abbreviation for $y\subset x\land y\in x\land \forall z, z\in x\implies z=y\lor z\in y$

"$n$ is a finite ordinal" is an abbreviation for $n$ is an ordinal and $(n\neq \emptyset \implies \exists z, n=z\cup\{z\})\land (\forall y, y\in n\implies (y\neq \emptyset \implies \exists z, y=z\cup\{z\}))$

So obviously one could write down an explicit formula but: it wouldn't be very useful; it would harly be readable; it would be extremely tedious to do

Maxime Ramzi
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  • Also, the whole definition can be made slightly better by defining ordinals as hereditarily transitive sets instead of transitive sets well-ordered by "$\in$". – Noah Schweber Apr 22 '17 at 17:44
  • Indeed, that was a typo (about $x$ and $n$), and yes, I was thinking about it when writing it but forgot to mention $\emptyset$. I'm editing my answer, thanks for pointing it out – Maxime Ramzi Apr 22 '17 at 17:56
  • As for improving the definition : I know it can, but here I was just giving the most basic one I knew. And for those who are interested, under the axiom of foundation, the condition $\forall u, v\in x, u\in v\lor u=v\lor v\in u$ also works to define ordinals – Maxime Ramzi Apr 22 '17 at 17:59
  • Damn, I don't know why I keep making these. I guess I'll just shut up or I'll make more mistakes, thanks again for correcting that – Maxime Ramzi Apr 22 '17 at 19:03
  • Heh, happens to all of us! – Noah Schweber Apr 22 '17 at 19:04
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Yet another approach: $x$ is a finite ordinal iff $x$ is an element of every inductive set, that is, iff $$\forall z[(\emptyset \in z\wedge\forall w(w\in z\implies w\cup\{w\}\in z))\implies x\in z].$$ That's a bit abbreviated, but hopefully it's clear how to turn it into a genuine formula in the language of set theory.

Noah Schweber
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