Re: [math-fun] Generalizing strong induction
<< This is the so-called epsilon induction, and it requires not the Axiom of Choice but the Axiom of Restriction. Suppose that for any set A, if P(a) holds for each a in A, then P(A) holds. Then P(A) holds for all sets. The theories of transfinite induction and well-founded relations both generalize induction; see Thomas Jech's book on set theory.
I think we'd also have to suppose P(Ø) to conclude P(A) holds for all sets. Otherwise we could prove any P that's always false holds for all sets. --Dan
Dan Asimov objected:
<< This is the so-called epsilon induction, and it requires not the Axiom of Choice but the Axiom of Restriction. Suppose that for any set A, if P(a) holds for each a in A, then P(A) holds. Then P(A) holds for all sets. The theories of transfinite induction and well-founded relations both generalize induction; see Thomas Jech's book on set theory.
I think we'd also have to suppose P(Ø) to conclude P(A) holds for all sets.
Nope! It's the base-case-free induction. If A is the empty set, then P(a) clearly holds for all none of its members -- the empty conjunction is true. I remember learning that trick from Gerald Sacks as an undergraduate, and walking around in a delighted haze at the cleverness for the rest of the day... --Michael Kleber kleber@brandeis.edu
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Michael Kleber