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Nov 3, 2023 at 20:36 comment added Mark Saving "Together with AC and foundation, replacement implies weak collection". Note that you don't actually need the axiom of choice here; the proof you outlined works with only foundation and excluded middle.
Jan 22, 2010 at 5:02 vote accept Mike Shulman
Jan 22, 2010 at 2:14 answer added Joel David Hamkins timeline score: 35
Jan 22, 2010 at 2:08 comment added François G. Dorais That's a neat question! I know the answer is generally no in weak set theories, where such things actually do matter. The answer is no without pairing. Namely the class of ordinals satisfies all the axioms except pairing, replacement, and separation, but it does satisfy foundation, collection, union and powerset (which is just the ordinal successor!). I'll have to think about combining pairing and powerset. By the way, it probably won't cause any confusion but "strong collection" already has a standard meaning, which is different from yours.
Jan 22, 2010 at 0:53 comment added Gerhard Paseman Since I am currently unable to comment, I will post as an answer. Suppose in weak collection, we were able to extend phi(x,y) to phi(x,y) and (y in A). I don't know if this needs a parameterized form of collection or what, but it seems one step closer to separation. Perhaps looking at forms of collection involving parameters would help. Gerhard "Ask Me About System Design" Paseman, 2010.01.21
Jan 22, 2010 at 0:50 history edited Mike Shulman CC BY-SA 2.5
Delta_0 separation is okay to assume too
Jan 21, 2010 at 23:13 history asked Mike Shulman CC BY-SA 2.5