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I have to present about the Peano axioms and the ZFC for my introductory seminar. It's one of the first topics presented, so I can't refer to more advance topics like cardinality, the only things introduced are the ZFC-axioms, the peano axioms and how functions/tupels are defined (and russles antiome, but i don't see it being useful here ;) ).

I would like to give an (impractical) practical example of how Von Neumann-Ordinals and functions work by defining an addition function (a set of ((x,y),z) representing x+y=z or +(x,y)=z).

This is my first idea in predicate logic:$$\exists A:((\emptyset,\emptyset),\emptyset)\in A \wedge (\forall ((x,y),z) \in A:(S(x),y),S(z)) \in A \wedge (x,S(y)),S(z)) \in A)$$

My problem with this definition is that i don't know how i can demand that it's only containing the things i want it to. The Peano axioms eliminate this by having $$S(m)=S(n) \to m=n$$ (i think), but i don't see how i can provide a similar construct here. Ideally i would like to have a set-comprehension, but i am not able to find/create one that does not use the not formally introduced concepts like cardinality.

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  • $\begingroup$ if nobody has any idea that's okay, i can rework the part of my presentation. I was also a bit struggling to asking this on math.stackexchange, but this is not a simple assignment, i have to present about the topic and am free to do this in the way i feel like its the easiest to follow for the other students. $\endgroup$ – Leander Nov 1 '16 at 18:01
  • $\begingroup$ There is a typo. You probably want to say that $\exists A:((\emptyset,\emptyset),\emptyset)\in A \wedge (\forall ((x,y),z) \in A:(S(x),y),(S(z)) \in A \wedge (x,S(y)),(S(z)) \in A)$. Also keep in mind that $S(x) = x \cup \{x\}$. It's then an easy exercise to prove (in ZFC) that for any sets $x,y$ $S(x)=S(y) \implies x = y$. However, your set $A$ may nonetheless contain thinks you don't want in there - namely, it's domain and range may contain other sets than ordinals. $\endgroup$ – Stefan Mesken Nov 1 '16 at 18:13
  • $\begingroup$ "Peano axioms and the ZFC" ... I would do these as separate topics. Trying to put them in a common framework is not for beginners. $\endgroup$ – GEdgar Nov 1 '16 at 18:13
  • $\begingroup$ On first reading I assumed that $A$ should only represent addition on $\omega$. Now I think it's supposed to define ordinal definition in general. Here you run into trouble with $\alpha + \lambda$ for limit ordinals $\lambda$. In order to do that, you need some sort of recursion theorem and its proof will be a bit complicated. It may therefore be a good idea, not to define ordinal definition in detail (since it will take way too much time to do properly) or to restrict yourself to $\omega$. $\endgroup$ – Stefan Mesken Nov 1 '16 at 18:17
  • $\begingroup$ It was meant to define addition ω. @Stefan yeah, there was a typo, thank you! $\endgroup$ – Leander Nov 1 '16 at 18:19
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I don't know about ZFC, VN oridinals, etc., but, if you need to construct the addition function, you could construct the set $A$ as a subset of $N^3$ such that:

$\forall a,b,c:[(a,b,c)\in A \iff (a,b,c)\in N^3 $

$\land \forall d\subset N^3:[\forall e\in N:(e,0,e)\in d$

$\land \forall e,f,g:[(e,f,g)\in d \implies (e,S(f),S(g))\in d]$

$ \implies (a,b,c)\in d]] $

You would then have to prove $A$ is a function such that:

$\forall a,b\in N: A(a,b)\in N$

$\forall a \in N: A(a,0)=a$

$\forall a,b\in N:A(a,S(b))=S(A(a,b))$

It should be easy to translate these statements into the required language of ZFC.

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