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This question follows up on a discussion elsewhere at MSE where it was suggested that some set theories do not allow arbitrary separation, while admitting the existence of a universal set.

I have been working for some time with a set theory that allows for arbitrary separation (creating subsets of another set). It is a trivial matter to prove the existence of, for example, the identity function $f(x)=x$ on a given set $S$. Just take the Cartesian product $S\times S$, and select the subset $f=\{(x,y)\in S\times S: x=y\}$. Then, of course, you would have to prove that $f$ is a function. How would this be accomplished without initially invoking arbitrary separation as here?

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Dan Christensen
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  • What exactly do you mean by arbitrary separation? The principle that all sets must exclude some element(s)? – Newb Dec 09 '13 at 05:00
  • I mean axioms (e.g. the Axiom Schema of Separation in ZFC) that, given the existence of one set, enable you to infer the existence of an arbitrary subset of that set, the only restriction on the selection criteria being that it not refer to the name of the subset itself. – Dan Christensen Dec 09 '13 at 05:14

1 Answers1

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The details would depend on the type of separation that is allowed.

  • A common restriction is $\Delta_0$-separation. In this case there is no problem for the identity function (since being a (Kuratowski) ordered pair with equal first and second coordinates is a $\Delta_0$-property).
  • For $\mathsf{NF}$ the strategy is similar: showing that there is a stratified formula which defines the function.

In such theories one cannot expect that every $\mathsf{ZFC}$-provable function is still provably a function.

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