Cardinality once more
in [Logic]
|
From: zuhair on 3 Jan 2010 04:06 On Jan 3, 2:56 am, ah...(a)FreeNet.Carleton.CA (David Libert) wrote: > David Libert (ah...(a)FreeNet.Carleton.CA) writes: > > zuhair (zaljo...(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. There is no such possibility, that is very clear. Cardinality according to the new definition is always not empty, because it will always contain d(x) by definition, and for every set x we have d(x), that's clear. However there are lots of reasons why this definition doesn't work, that's why I already shied away from it. But as to the question weather cardinalities can be empty according to this definition, the answer is straightforwards actually. IT CANNOT. That's clear. Zuhair > > > 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 ah...(a)FreeNet.Carleton.CA
From: zuhair on 3 Jan 2010 10:22 On Jan 3, 2:56 am, ah...(a)FreeNet.Carleton.CA (David Libert) wrote: > David Libert (ah...(a)FreeNet.Carleton.CA) writes: > > zuhair (zaljo...(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 ah...(a)FreeNet.Carleton.CA To add to my previous reply that there cannot be an empty cardinality. The reason why I said that is because of the very basic methodology adopted here. I said we ought to define a function d(x) on all sets in ZF such that all equinumerous sets have the same d(x) and for any two sets x and y if x is strictly subnumerous to y then d(x) is strictly subnumerous to d(y), and the same thing applies with strict supernumerous. Now my definition of d(x) as an ordinal was erroneous, so I must drop this property, so what I am saying here if we can define such a function, then this mean that for every set x there exist d(x) Now the cardinality of x was defined as the set that has d(x) as a member and all hereditarily strictly subnumerous to d(x) sets that are equinumerous to x (I already defined what I meant by hereditarily) So there is no way whereby these cardinals can be empty because they will always contain d(x). Anyhow, as I said this definition doesn't work as a cardinal outside Choice. So it better be dropped. Zuhair
From: MoeBlee on 3 Jan 2010 17:10 On Jan 2, 5:41 pm, zuhair <zaljo...(a)gmail.com> wrote: > How come these two theories have the same Language, while the first > contain > addition primitive symbol? One theory is in a language that has 'card' as primitive. (If you properly form by only definitions), then the other theory is a theory extended by definitions. The language of the theory extended by definitions includes 'card' but its source language (the language of the the theory that was extended by definitions) does not include 'card'. > They seem to have two different languages, that's why I said Equi- > interpretable instead of saying Equivalent. That's fine IF your second theory is TRULY merely an extension by definition. By the way, 'equivalent' is actually 'equal' in this sense, since theories in the same language with all the same theorems are just the same theory (if we take 'theory' in the sense of 'a set of sentences closed under entailment'). MoeBlee
From: MoeBlee on 3 Jan 2010 17:11 On Jan 2, 5:44 pm, zuhair <zaljo...(a)gmail.com> wrote: > Yes, but let us be clear here, this theory is in FOL(e,=), I can > dispense with > the symbol card(x) altogether. Whether that is the case depends on your exact reformulation now. MoeBlee
From: MoeBlee on 3 Jan 2010 17:16
On Jan 2, 6:51 pm, zuhair <zaljo...(a)gmail.com> wrote: > 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. Yes, NOW you've got a precise mathematical question and it is indeed equi-interpretability that it s appropriate to ask about. MoeBlee |