Wikipedia:Reference desk/Archives/Mathematics/2021 August 15
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August 15
[edit]Zermelo set theory
[edit]- Moved here from the Computing section of the Reference desk. --Lambiam 10:36, 15 August 2021 (UTC)
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 Vω+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)
- Without Replacement, you can't guarantee that the order-type of every wellordered set is a von Neumann ordinal, meaning the representation where α < β just in case α ε β. It would still be a "naive" ordinal, the sort of ordinals Cantor introduced in the Grundlagen or Beiträge or whichever it was. Yes, it does follow from ZC that every set can be wellordered, and the proof is conceptually the same, but it has some extra technical difficulties without the von Neumann ordinals around. --Trovatore (talk) 19:06, 15 August 2021 (UTC)
- Ah nice, that is very interesting. Thanks. 2602:24A:DE47:BA60:8FCB:EA4E:7FBD:4814 (talk) 21:06, 15 August 2021 (UTC)
- Without Replacement, you can't guarantee that the order-type of every wellordered set is a von Neumann ordinal, meaning the representation where α < β just in case α ε β. It would still be a "naive" ordinal, the sort of ordinals Cantor introduced in the Grundlagen or Beiträge or whichever it was. Yes, it does follow from ZC that every set can be wellordered, and the proof is conceptually the same, but it has some extra technical difficulties without the von Neumann ordinals around. --Trovatore (talk) 19:06, 15 August 2021 (UTC)