3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Henning Makholm Apr 15 '17 at 23:17
  • $\begingroup$ 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. $\endgroup$ – Asaf Karagila Apr 16 '17 at 6:38
  • $\begingroup$ 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 $\{a\in A\mid \phi(a,p)\}$. $\endgroup$ – Henning Makholm Apr 16 '17 at 10:22
  • $\begingroup$ Henning, see mathoverflow.net/questions/254953/… also math.bu.edu/people/aki/20.pdf $\endgroup$ – Asaf Karagila Apr 23 '17 at 4:34
  • $\begingroup$ 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? $\endgroup$ – Henning Makholm Apr 23 '17 at 11:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.