Cardinality once more
in [Logic]
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From: zuhair on 2 Jan 2010 20:41 On Jan 2, 7:10 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > On Jan 2, 4:56 pm, zuhair <zaljo...(a)gmail.com> wrote: > > > Cardinality is a one place function on sets such that for every set > > x , every set y > > we have > > > Card(x)=Card(y) <-> Exist f (f:x-->y, f is bijective) > > > Lets call this: the *basic requirement for Cardinality*. > > Yes, as I said, that is a well-known idea, going back at least to > Tarski. > > > Now Cardinality can be *defined* or can be a *primitive* concept. > > > With defined cardinality here I am referring specifically to a > > definition by a formula in first order logic with identity and epsilon > > membership as the sole primitives i.e. > > in FOL(=,e). > > > This as you said takes the form > > > Card(x)=y <-> Phi (x,y) > > > So lets say that Phi(x y) is a *cardinality defining formula* if card > > (x) defined after this formula will satisfy the basic requirement for > > Cardinality. > > Yes, I said I'm with you there, still a basic set theoretical path. > > > remember Phi(x y) is a formula in FOL(e,=) > > Right. > > I shouldn't have said 'card' can be in Phi. > > So I just corrected my post in this way: > > Is there a formula Phi (in the language of set theory) such that: > > ZF(minus regularity)+"card(x) = y <-> ((AxE!y Phi(x y) & Phi(x y)) or > (~AxE!y Phi(x y) & y=0))" > = > ZF(minus reguarity)+Tarski's axiom ? > > > Now my question was the following: > > > Is their a cardinality defining formula Phi(x y) such that > > > ZF(minus Reg.) + "For all x Exist y ( y=card(x) ) were card(x) <-> Phi > > (x y)" > > > is Equi-interpretable with > > > ZF(minus Reg.) + Tarski's axiom. > > Your formulation is not correct. > > (1) You ignored what I reminded you of: > > card(x) <-> Phi(x y) > > is not well formed. Sorry as you said in the other post, this was a typo I meant Card(x)=y <-> Phi (x y) > > (2) If I'm not mistaken (I'd welcome someone more expert commenting), > equi-interpretability would be of concern if the two theories have > different languages. But your theories do have the same language. > Whether you consider 'card' as primitive in one language but > "defined" (yours is not in a correct definitional form) and thus added > to another language, they end up both being the same language. So, if > I'm not mistaken, to ask a substantive equi-interpretability question, > you need one theory to have 'card' in its language' and the other > theory not to mention 'card' at all (if that theory is adding 'card' > as DEFINED, then any axiom would not NEED to use the symbol 'card'). I need to check that issue. We have one theory were the language contain an extra primitive sybmol that is | | denoting cardinaity i.e. its language has three primitives e,=,| |. The other theory contain a defined which as you said we can dispense with it altogether. How come these two theories have the same Language, while the first contain addition primitive symbol? They seem to have two different languages, that's why I said Equi- interpretable instead of saying Equivalent. Zuhair > > Now, please, take it a step at a time. > > We know what one theory is. ZF(-regularity)+Tarski's axiom, with > 'card' as primitive. > > But you must reformulate your definition of 'card' for the other > theory. Then you can also add any axioms you want with 'card' in it > (though, of course, with 'card' DEFINED, those axioms can also be > written withOUT 'card'). > > Thanks for the welcome back. I hope all is well with you these days. > > (I actually need not to post so I can concentrate on other things > right now, but I couldn't resist this time as I saw a basic mistake > you need to correct before you can advance with your question. But now > that I've posted, I'm probably headed back to the same old time-eating > posting routine!) > > MoeBlee
From: zuhair on 2 Jan 2010 20:44 On Jan 2, 7:20 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > On Jan 2, 6:10 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > > > card(x) <-> Phi(x y) > > > is not well formed. > > Maybe you typoed? Maybe you meant > > card(x) = y <-> Phi(x y). > > So your other theory is?: > > ZF(-regularity)+"AxEy(y = card(x) & Axy(card(x) = y <-> Phi(x y))" > > (Some of those quantifiers might be redundant, but I put them in for > clarity.) > > But, as I mentioned, the part about Ey y = card(x) doesn't add > anything substantive. So you might as well just make it: > > Axy(card(x) = y <-> Phi(x y)) > > Thus your theory: > > ZF(-regularity)+"Axy(card(x) = y <-> Phi(x y))" Yes, but let us be clear here, this theory is in FOL(e,=), I can dispense with the symbol card(x) altogether. > > MoeBlee
From: zuhair on 2 Jan 2010 20:59 On Jan 2, 8:44 pm, zuhair <zaljo...(a)gmail.com> wrote: > On Jan 2, 7:20 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > > > > > > > On Jan 2, 6:10 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > > > > card(x) <-> Phi(x y) > > > > is not well formed. > > > Maybe you typoed? Maybe you meant > > > card(x) = y <-> Phi(x y). > > > So your other theory is?: > > > ZF(-regularity)+"AxEy(y = card(x) & Axy(card(x) = y <-> Phi(x y))" > > > (Some of those quantifiers might be redundant, but I put them in for > > clarity.) > > > But, as I mentioned, the part about Ey y = card(x) doesn't add > > anything substantive. So you might as well just make it: > > > Axy(card(x) = y <-> Phi(x y)) > > > Thus your theory: > > > ZF(-regularity)+"Axy(card(x) = y <-> Phi(x y))" > > Yes, but let us be clear here, this theory is in FOL(e,=), I can > dispense with > the symbol card(x) altogether. Let me repeat the question again without using this symbol in the defined version. Can we have a formula Phi(x y) in FOL(=,e) such that ZF(-regularity) + A x1 x2 y ((Phi(x1 y) & Phi(x2 y)) <-> x1 equinumerous to x2) is Equi-interpretable with ZF(-Regularity) + Tarski axiom. Zuhair > > > > > > > MoeBlee- Hide quoted text - > > - Show quoted text -- Hide quoted text - > > - Show quoted text -
From: zuhair on 2 Jan 2010 21:51 On Jan 2, 8:59 pm, zuhair <zaljo...(a)gmail.com> wrote: > On Jan 2, 8:44 pm, zuhair <zaljo...(a)gmail.com> wrote: > > > > > > > On Jan 2, 7:20 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > > > > On Jan 2, 6:10 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > > > > > card(x) <-> Phi(x y) > > > > > is not well formed. > > > > Maybe you typoed? Maybe you meant > > > > card(x) = y <-> Phi(x y). > > > > So your other theory is?: > > > > ZF(-regularity)+"AxEy(y = card(x) & Axy(card(x) = y <-> Phi(x y))" > > > > (Some of those quantifiers might be redundant, but I put them in for > > > clarity.) > > > > But, as I mentioned, the part about Ey y = card(x) doesn't add > > > anything substantive. So you might as well just make it: > > > > Axy(card(x) = y <-> Phi(x y)) > > > > Thus your theory: > > > > ZF(-regularity)+"Axy(card(x) = y <-> Phi(x y))" > > > Yes, but let us be clear here, this theory is in FOL(e,=), I can > > dispense with > > the symbol card(x) altogether. > > Let me repeat the question again without using this symbol in the > defined version. > > Can we have a formula Phi(x y) in FOL(=,e) such that > > ZF(-regularity) + A x1 x2 y ((Phi(x1 y) & Phi(x2 y)) <-> x1 Even more precisely: ZF(-Regularity) + "AxE!y Phi(x y) & Ax1x2y ((Phi(x1 y) & Phi(x2 y)) <-> x1 equinumerous to x2)" > is Equi-interpretable with > > ZF(-Regularity) + Tarski axiom. > > Zuhair > > > > > > MoeBlee- Hide quoted text - > > > - Show quoted text -- Hide quoted text - > > > - Show quoted text -
From: David Libert on 3 Jan 2010 02:56
David Libert (ah170(a)FreeNet.Carleton.CA) writes: > zuhair (zaljohar(a)gmail.com) writes: >> On Jan 1, 4:37=A0am, zuhair <zaljo...(a)gmail.com> wrote: [Deletion] >>> Next step we define for every x, the set of all sets hereditarily >>> subnumerous to d(x) (i.e. the set of all sets that are subnumerous >>> (injective) to d(x) were every member of their transitive closures is >>> also subnumerous to d(x) ) >>> >>> Lets denote this set as H(d(x)) >>> >>> Then we define cardinality as: >>> >>> Cardinality (x) is the class of d(x) and of all subsets of H(d(x)) >>> that are equinumerous to x. > > How do we show this cardinality is non-empty? If it can become empty > for 2 nonisomorphic x then we get the old problem of nonisomorphic > sets being assigned the same cardinality. > > Inspired by your definition above I make these definitions. > > Define a set is 0-well-orderable <-> it is well-orderable in the > usual sense. > > For alpha an ordinal define inductively from 0 base case above > a set x is alpha-well-orderable <-> > it x is isomorphic to > some set y such that for every member z of y > there exists beta < alpha such that y is beta-well-orderable. > > > If your Cardinality(x) above is nonempty, then x is isomorphic to > some set y a subset of H(d(x)). > > Each element z of such a y is a member of H(d(x)), and so is subnumerous > to ordinal d(x) and hence well-orderable, ie 0-well-orderable in my new > terminology. > > So x is isomoprphic to y with all members 0-well-orderable. > > So x is 1-well-orderable. > > ZF proves AC <-> every set is 0-well-founded. > > Is it possible to have a ZF model with some set x which is not > 1-well-founded? Those last 2 references to well-founded were supposed to be well-orderable. > I don't know. But if it were such x would have empty Cardinality > in the new definition. > > So a proof would have to rule out this possibility. Show from ZF > or whatever extra axioms you want to use every set is 1-well-orderable. Since posting the above, I think I have found proofs. I haven't checked all details, but think they are probably ok. ZF proves for every set x x is rank(x)-well-orderable. Namely the isomorphic set can be x itself, and the isomorphism can be the identity. This proof uses regularity. So over ZF - regularity there remains the possibility of a set x not alpha-well-orderable for any ordinal alpha. In fact below I will claim there are models like this. I give some defintions for ZF and ZF - regularity. So I don't assume for all x there exists alpha x is alpha-well-orderable. For an x which is is alpha-well-orderable for some ordinal alpha, define wo-rank(x) to be the least alpha such that x is alpha-orderable. For x not alpha-well-orderable for any alpha (ie a case that hasn't yet been ruled out for ZF - regularity) define wo-rank(x) = OR, ie all ordinals, just a formal symbol. For a model of ZF or ZF - regularity, define the wo-rank of the model to be the least alpha if it exists (an ordinal in the model) such that for every set x in the model exists ordinal beta in the model beta < alpha and x is beta-well-orderable. If every set x in a ZF or ZF - regularity model has an ordinal wo-rank but they are unbounded in ordinals as x varies, define the wo-rank of the model to be OR. If a ZF - regilarity model has some set x such that for all ordinals alpha in the model x is not alpha-well-orderable, define wo-rank of the model to be OR + 1 . (Just a formal symbol). Then my new constructions. Start from a ZFC base model. We will be constructing symmetric models, using forcing. For any alpha an ordinal we can construct ZF and ZF - regularity models with wo-rank exactly alpha. We can construct ZF and ZF - regularity models with wo-rank OR. We can construct ZF - regularity models with wo-rank OR + 1. -- David Libert ah170(a)FreeNet.Carleton.CA |