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I'm doing some work in Jech's Set Theory book and I am a little confused about the definition of parameter in the Axiom Schema of Separation. Within a formula, what relationship would a parameter have with the free variables, and bound variables? I'm just fuzzy about what exactly a parameter is and how it is used in the formulas. I appreciate any help. Thanks.

Ethan Bolker
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Newman
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1 Answers1

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Suppose that you want to talk about the set of real numbers which are roots of some function(s).

But set theory does not have an intrinsic sense of what are the real numbers, what are functions and what are roots of a function. But we do know that we can define a set and structure that gives us the real numbers, and we can define functions like $\sin$ or $\cos$.

So when you want to talk about $\{x\in\Bbb R\mid\sin(3x^2-5)=0\}$, you have actually used a lot of parameters: the parameters which tell you what is $\Bbb R$, the function $x\mapsto\sin(3x^2-5)$ and the number $0$ in $\Bbb R$.

This is why we have parameters in the separation schema. So when we think about our universe as "tangible" (read: we work inside a specific model) we can use the objects of the universe to define things which we may or may not have access to from a syntactical point of view.

The same holds for the Replacement schema as well. Only in the case of the Replacement schema we have an interesting situation that we don't really need the parameters, but the proof of that is quite complicated and nontrivial. So it's much easier to just allow parameters and simplify everything.

Asaf Karagila
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  • Your last paragraph sounds like that parameters are really needed for Separation, but that's not the case, right? We can derive every parameterful instance of Separation from parameter-less Separation together with Pairing, Union, and Power Set axioms. – hmakholm left over Monica Apr 15 '17 at 23:17
  • I'm not sure. I think this is false. There was a question about this, I think on MathOverflow, not too long ago. I will look for it later. – Asaf Karagila Apr 16 '17 at 06:38
  • Suppose WLOG there is a single parameter $p$. Create ${{a}\mid a\in A}$ by selection on $\mathcal P(A)$. Add an extra element ${\varnothing,{p}}$, take the power set and select the subsets of the form ${{x},{\varnothing,{y}}}$ that satisfy $\phi(x,y)$. Take the union of the result, select the singletons and take the union once again. This gives you ${a\in A\mid \phi(a,p)}$. – hmakholm left over Monica Apr 16 '17 at 10:22
  • Henning, see https://mathoverflow.net/questions/254953/is-extreme-parameter-free-zfc-equivalent-to-zfc/ also http://math.bu.edu/people/aki/20.pdf – Asaf Karagila Apr 23 '17 at 04:34
  • Hmm, I may be missing something, but those sources don't seem to be saying anything about parameter-free Z. Are we speaking past each other somehow? – hmakholm left over Monica Apr 23 '17 at 11:47
  • @Henning: If you include Replacement, then you can omit Separation entirely with or without parameters. So what's the point? – Asaf Karagila Apr 23 '17 at 11:58
  • Let's consider the following property of an axiom schema X: "In the presence of the non-schema axioms of Z(F)(C), parameter-free X implies full X". I understood the last paragraph of your answer to mean that X=Replacement has this property in contrast to X=Separation. – hmakholm left over Monica Apr 23 '17 at 12:05
  • @Henning: Well, in the usual cases Z is given with Separation. If you know the proof that Induction (over PA) can be taken as parameter free, then the argument is similar. And Replacement is indeed what you need when you want to talk about $\Sigma_n$-truth predicates and Reflection principles. So I would be surprised to learn the same holds for Separation. And if my memory is right, in Azriel's paper he discusses Separation and that it is not enough to be taken as parameter free. – Asaf Karagila Apr 23 '17 at 12:27
  • I sketched an argument in my comment above -- do you think something is wrong with that? – hmakholm left over Monica Apr 23 '17 at 12:31
  • I think that it implicitly uses the parameter when you define ${\varnothing,{p}}$ and use that to select the subsets. I'm not sure. This sort of arguments tend to have a lot of very very subtle points which are very very hard to miss. – Asaf Karagila Apr 23 '17 at 13:40