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August 15

Zermelo set theory

Someone on mathoverflow[1] claims "you need replacement to show the set of Von Neumann ordinals greater than ω+n for all n<ω is a proper class (otherwise it is consistent that it is empty)". Is that right? ZFC without replacement is ZC: is it still true in ZC that all sets can be well-ordered? What would be the order type of the reals if there were no ordinals >= ω+ω? Aren't there plenty of everyday objects like finite trees, that have high countable order types? I had heard that V_ω+ω was a model of ZC and you'd get that by iterating (ω+w times) the powerset operation on ω. But surely you end up with plenty of uncountable ordinals by doing that. So either the MO poster or me is confused. There's a high chance that it's me though, as I have never actually studied this stuff. 2602:24A:DE47:BA60:8FCB:EA4E:7FBD:4814 (talk) 09:26, 15 August 2021 (UTC)[reply]

Moved to the Mathematics section of the Reference desk. --Lambiam 10:35, 15 August 2021 (UTC)[reply]

Whoops, not sure how I put that here by accident. 2602:24A:DE47:BA60:8FCB:EA4E:7FBD:4814 (talk) 17:25, 15 August 2021 (UTC)[reply]