The Axiom Schema of Separation is: $\forall a \forall \Psi (x) \exists x \forall y (y \in x \leftrightarrow y \in a \space \land \Psi (y))$. If we make $a:=\{\emptyset\}$, $\Psi (x) :\leftrightarrow \lnot y \in x$, $x:=x$ and $y :=\emptyset$, we get the following: $y \in x \leftrightarrow y \in \{\emptyset\} \space \land \lnot y \in x$. This in turn implies: $y \in x \leftrightarrow \lnot y \in x$, which is contradictory and should negate the Axiom due to Modus Tollendo Tollens. Where am I wrong?


1 Answer 1


The genuine axiom scheme of separation requires that the formula used to specify the elements of the "separated" set does not refer to that set.

A correct statement of the scheme is

$$\forall z_1\cdots \forall z_k\forall a\exists x\forall y(y\in x\leftrightarrow y\in a\land \Psi(y,z_1,\ldots,z_k))$$

where $\Psi(y,z_1,\ldots,z_k)$ is a formula whose free variables are drawn from $y,z_1,\ldots,z_k$.

  • 1
    $\begingroup$ ... and the condition is necessari exactly to avoid the above "misuse"; see Patrick Suppes, Axiomatic set theory (1960), page 21 for the same example. $\endgroup$ Sep 22, 2015 at 11:35
  • $\begingroup$ ...hey, nice reference! $\endgroup$
    – mmw
    Sep 22, 2015 at 12:03
  • $\begingroup$ Then, why is the Axiom Schema of Abstraction ($(\forall \psi (x) \exists y \forall a)(a \in y \iff \psi (a))$) replaced by a the Axiom Schema of Separation, when it could only be extended to saying that the property $\psi (x)$ cannot refer to the set being constructed? $\endgroup$ Oct 9, 2015 at 7:34
  • $\begingroup$ Sorry for being so ignorant in this topic and thanks for your answer. $\endgroup$ Oct 9, 2015 at 7:35
  • $\begingroup$ I didn't say that naive comprehension could be rescued by adding that proviso alone. E.g. naive comprehension under the proviso still leaves Russell's paradox. By the way, note that putting a quantification over properties as you do makes the statement no longer a first-order schema but a second-order formula. $\endgroup$
    – mmw
    Oct 9, 2015 at 10:44

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.