0
$\begingroup$

Source: Patrick Suppes: Axiomatic Set Theory

My question is why do we always bother with the routine of showing that the abstract property/formula implies the membership in the bigger set we're separating over? Doesn't the axiom schema alone suffice, on its own, as justification for the existence of the new set we're trying to build (No additional steps needed)?

$\endgroup$
0
1
$\begingroup$

Suppose you're trying to construct some set $\{x:P(x)\}$. The axiom of separation says that for any set $A$, you can construct the set $\{x\in A:P(x)\}$. But this is not necessarily equal to the set you actually want: it only contains the elements of $A$ that satisfy $P$, not everything that satisfies $P$. To be sure this set really does contain everything that satisfies $P$, you have to additionally prove that everything that satisfies $P$ is in $A$.

$\endgroup$
1
  • 1
    $\begingroup$ To add to this answer: consider what happens if you apply separation to the formula "$x\not\in x$." For any specific set $A$ we can get $\{x\in A: x\not\in x\}$, but no single $A$ contains every set satisfying "$x\not\in x$" (which is good since otherwise we'd run into Russell's paradox). $\endgroup$
    – Noah Schweber
    Apr 19 at 3:21

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.