>>13517154Forcing makes sense from a completely platonist perspective.
>>13517154That's not what it does retard. All it does is to reinterpret the language of set theory within a specific structure.
>>13517149It's not really recursion, since you're not defying a function. Also comes much earlier than forcing. See rank of well-founded relations.
>>13517089Correct, it's strictly speaking not a rigorous definition. But you can make it rigorous in the following way. Let P(x) denote the property of a set being an ordinal number. Then we want to find a P(x) such that "P(x) <=> (x transitive) and (forall y in x, P(y))".. First suppose we have such a P(x), then we prove that it's unique. Suppose Q(x) also satisfies the same property. Suppose there is a set w with Q(w) but not P(w). Since not P(w), there is an element w' of w with the same property Q(w') but not P(w'). Then the set of all such elements w' is nonempty and hence by the axiom of foundation has a least element w.r.t. set membership. Call it w_l. Then not P(w_l). But for all elements x in w_l, since w is transitive, x belongs to w so by minimality of w_l x satisfies either Q(x) or not P(x), but we know P(x) is true hence Q(x), hence for all elements of w_l, Q(x), hence Q(w_l), a contradiction. Hence there is at most one property P(x) satisfying this condition. It's enough to construct one such property. One definition is "P(x) = (x is transitive) and (x is well-ordered by set membership)". Then you can prove "P(x) <=> (x transitive) and (forall y in x, P(y))". Another more obvious definition is "P(x) = (x is transitive) and (there exists a subset y of x s.t. for all z in x, ((z in y) <=> (for all w in z, w in y))" then this is a well-defined property satisfying "P(x) <=> (x transitive) and (forall y in x, P(y))" which is less ad-hoc.
Maybe other anons have other ideas on how to make the definition rigorous.
