Quick ZFC cofinality question
in [Math]
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From: Jim Heckman on 29 Aug 2007 20:40 Can ZF alone prove that every totally ordered set has a cofinal well-ordered subset, or does that require AC? (Typically done by applying Zorn's Lemma to the set of well-ordered subsets, ordered by continuation.) -- Jim Heckman
From: Herman Rubin on 29 Aug 2007 21:38 In article <13dc4f7iafujc6c(a)corp.supernews.com>, Jim Heckman <weu_rznvy-hfrarg(a)lnubb.pbz.invalid> wrote: >Can ZF alone prove that every totally ordered set has a cofinal >well-ordered subset, or does that require AC? (Typically done by >applying Zorn's Lemma to the set of well-ordered subsets, ordered >by continuation.) It is at least false in the Mostowski model in which the universe is linearly ordered, but all well-ordered subsets of the set of individuals aren finite. In the metamathematical way of looking at the model, the individuals look like the rationals. The rational ordering does hold in the model, but for example the subset which is the integers does not exist in the model. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin(a)stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
From: Jim Heckman on 2 Sep 2007 03:27 On 29-Aug-2007, hrubin(a)odds.stat.purdue.edu (Herman Rubin) wrote in message <fb573c$4qve(a)odds.stat.purdue.edu>: > In article <13dc4f7iafujc6c(a)corp.supernews.com>, > Jim Heckman <weu_rznvy-hfrarg(a)lnubb.pbz.invalid> wrote: > > >Can ZF alone prove that every totally ordered set has a cofinal > >well-ordered subset, or does that require AC? (Typically done by > >applying Zorn's Lemma to the set of well-ordered subsets, ordered > >by continuation.) > > It is at least false in the Mostowski model in which the > universe is linearly ordered, but all well-ordered subsets > of the set of individuals aren finite. In the metamathematical > way of looking at the model, the individuals look like the > rationals. > > The rational ordering does hold in the model, but for example > the subset which is the integers does not exist in the model. Thanks. I was hoping for something more elementary, though, since I know next to nothing about model theory. My naive-set-theory intuition is having trouble seeing how any model of ZF can fail to have infinite (within the model) well-ordered sets -- starting with omega. -- Jim Heckman
From: Butch Malahide on 2 Sep 2007 04:35 On Aug 29, 7:40 pm, "Jim Heckman" <rot13(reply-to)@none.invalid> wrote: > Can ZF alone prove that every totally ordered set has a cofinal > well-ordered subset, or does that require AC? (Typically done by > applying Zorn's Lemma to the set of well-ordered subsets, ordered > by continuation.) I'm no expert, but I'd bet that without AC you'd have a hard time proving even that every set of real numbers (usual ordering) has a cofinal well-ordered subset.
From: Butch Malahide on 2 Sep 2007 04:50
On Sep 2, 2:27 am, "Jim Heckman" <rot13(reply-to)@none.invalid> wrote: > On 29-Aug-2007, hru...(a)odds.stat.purdue.edu (Herman Rubin) > wrote in message <fb573c$4...(a)odds.stat.purdue.edu>: > > > > > > > In article <13dc4f7iafuj...(a)corp.supernews.com>, > > Jim Heckman <weu_rznvy-hfr...(a)lnubb.pbz.invalid> wrote: > > > >Can ZF alone prove that every totally ordered set has a cofinal > > >well-ordered subset, or does that require AC? (Typically done by > > >applying Zorn's Lemma to the set of well-ordered subsets, ordered > > >by continuation.) > > > It is at least false in the Mostowski model in which the > > universe is linearly ordered, but all well-ordered subsets > > of the set of individuals aren finite. In the metamathematical > > way of looking at the model, the individuals look like the > > rationals. > > > The rational ordering does hold in the model, but for example > > the subset which is the integers does not exist in the model. > > Thanks. I was hoping for something more elementary, though, since I > know next to nothing about model theory. My naive-set-theory > intuition is having trouble seeing how any model of ZF can fail to > have infinite (within the model) well-ordered sets -- starting with > omega. I believe the models Professor Rubin is referring to are not models of ZF. They are models of a set theory which, unlike ZF, allows the existence of so-called individuals (aka atoms or urelements), objects which are not sets. Omega is an infinite well-ordered set, but it's not a set of individuals, it's a set of sets. |