Relation between Axiom of Foundation and $\in$-induction

54 Views Asked by At

At the end of an intro to set theory course, we were introduced the Axiom of Foundation and the Principle of $\in$-induction as one of its consequences. I found it easy to prove that, assuming the Axiom of Choice, $\in$-induction implies the Axiom of Foundation, and I am wondering whether this is also true without Choice. I asked my professor and he spent a good bit of time trying to find out the answer to no avail.

1

There are 1 best solutions below

5
On

Choice has nothing to do with this equivalence, but the law of excluded middle has everything to do with it.

In $\sf ZF$, assuming $\in$-induction, consider the statement "every transitive $y$ such that $x\in y$ is well founded". If all the elements of $x$ satisfy this statement, then any transitive $y$ such that $x\in y$ must be well founded as well. So, the statement holds for all $x$, and easily this implies the Axiom of Foundation.

In the other direction, we can simply translate this to an induction on the rank and appeal to Replacement as necessary.