Cardinality once more
in [Logic]
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From: zuhair on 2 Jan 2010 13:00 On Jan 1, 6:02 am, Rupert <rupertmccal...(a)yahoo.com> wrote: > On Jan 1, 8:37 pm, zuhair <zaljo...(a)gmail.com> wrote: > > > > > > > Hi all, > > > In the last one month I've posted many topics on cardinality. This one > > is also another trial to define cardinality in ZF, i.e. without > > Choice. > > > First from T. Jeck's paper, it appears that his proof can be > > generalized to every well orderable set x. > > > So for every well order-able set x, there exist the set of all > > sets hereditarily subnumerous to x. > > > This is a theorem of ZF. > > > Now this time the main idea of the trial is to associate a unique > > ordinal with each set in ZF such that all sets equinumerous to each > > other have a corresponding unique ordinal. > > > So for example lets take any set x, then all sets equinumerous to x > > would be associated with the same ordinal d(x), now every y that are > > strictly subnumerous to x would be associated with an ordinal > > d(y) that is also strictly subnumerous to d(x), same thing applies > > every set z that is strictly supernumerous to x would be associated > > with an ordinal d(z) that is strictly supernumerous to d(x). > > If ZF is consistent, then you cannot define such a function d(x) and > prove in ZF that it has the requisite properties. > > Because, if ZF is consistent, then it is consistent with ZF that there > exists a set S of real numbers which is Tarski infinite but Dedekind > finite. Then if k and l are two natural numbers with k<l, the set S > with l elements removed is subnumerous to the set S with k elements > removed but not equinumerous with it. If it were possible to define > your function d(x) then this would yield an infinite descending > sequence of ordinals and so a contradiction. > > So working in ZF alone it will not be possible to do this. Agreed. Yes that is correct. Thanks. Zuhair
From: zuhair on 2 Jan 2010 13:14 On Jan 2, 3:32 am, ah...(a)FreeNet.Carleton.CA (David Libert) wrote: > zuhair (zaljo...(a)gmail.com) writes: > > On Jan 1, 4:37=A0am, zuhair <zaljo...(a)gmail.com> wrote: > >> Hi all, > > >> In the last one month I've posted many topics on cardinality. This one > >> is also another trial to define cardinality in ZF, i.e. without > >> Choice. > > >> First from T. Jeck's paper, it appears that his proof can be > >> generalized to every well orderable set x. > > >> So for every well order-able set x, there exist the set of all > >> sets hereditarily subnumerous to x. > > >> This is a theorem of ZF. > > >> Now this time the main idea of the trial is to associate a unique > >> ordinal with each set in ZF such that all sets equinumerous to each > >> other have a corresponding unique ordinal. > > >> So for example lets take any set x, then all sets equinumerous to x > >> would be associated with the same ordinal d(x), now every y that are > >> strictly subnumerous to x would be associated with an ordinal > >> d(y) that is also strictly subnumerous to d(x), same thing applies > >> every set z that is strictly supernumerous to x would be associated > >> with an ordinal d(z) that is strictly supernumerous to d(x). > > Rupert already reponded on this point, for a Dedekind set the > strictly subnumerous sets below are not well-founded so this can't > be done all levels down in that case. > > >> 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. Because we must define d(x) to exist for every x, that's the idea, so Cardinality(x) would also have d(x) as a member so it cannot be empty. > > 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? > > 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.. > > >> The point is how to define d(x). > > >> This is a trial: > > >> Define(d(x)): > > Somehow a system along the way has replaced spaces by other characters > which makes the original quote hard to read. I will manually edit those > back to spaces, but I might make mistakes and change your original > spacing. > > >> A= d(x) <-> for all y ( y e A <-> > >> Exist u,z( u strictly subnumerous to x > > & z=d(u) & > >> & y is ordinal & y equinumerous to d > >> (u))). > > >> This is a trick really, > > The form of your defintion is a transfinite induction over strictly > subnumerous, since you invoke d(u). Rupert already gave examples > where this is not well-founded. NO Rupert gave an example when this is contradictive, since d(x) would be a set of ordinals, and thus must be well-founded, and at the same time we'll have a non well founded set of them with the case of non Dedekinds sets, so it would be contradictive. We may resolve this contradiction of I remove the condition " d(x) is ordinal" and replace it by another condition, like d(x) is hereditarily hereditary etc.. this would solve it, but as far as d(x) is ordinal then this would lead to the contradiction that Rupert referred to. > > We could interpret your definition as just defining d for sets with > well founded strictly subnumerous relation below them. So maybe it > is only defining cardinality for a subclass of the universe. Yea, that my be case indeed. > > This raises another interesting question. Is it possible to have in > a ZF model a set x which cannot be well-ordered, with well-founded > strictly subnumerous partial ordering below it? > > I don't know. > > Anyway, if it is and some cases of x get d(x) defined this way, > then my previous question for these of why Cardinality(x) is > non-empty remains. see above. > > Below you discuss example of familar small x. I will delete that > because it doesn't touch on the 2 trickier questions I just mentioned. > > [Deletion} > > >> Lets take the Cardinality of Power(omega) and lets assume that power > >> omega is not comparable to any of the uncountable ordinals. > > >> Now we have H(Aleph_1) as a set in ZF. > > By the way, you are also using the generalized Jech result as you > mentioned, but that used regualrity. Below you will consider what > happens with definitions like this without regularity. > > My first model of those discussed recently got H(= 1) a proper > class. So this is another difficulty for the version dropping > regularity. > > [Deletion] > > >> two sets that are strictly non equinumerous will also have different > >> cardinals also since their members would not be equinumerous. > > Except the emptiness issue above if it goes the wrong way. > > > > >> so: > > >> (1)Card(x)=3DCard(y) iff x and y have the same members(Extensionality). > > >> (2)Card(x) > Card (y) iff [(every member of Card(x) is strictly > >> supernumerous to every member of Card (y)) and =A0(every > >> subset of d(x) that is equinumerous to d(x) is a member of Card(x) iff > >> every subset of d(y) that is equinumerous to d(y) is a member of Card > >> (y)) ]. > > >> (3)Card(x) < Card(y) <-> Card(y) > Card(x) > > >> (4)Card(x) incomparable with Card(y) iff non of the above. > > >> Although very complex definition but I think it works. > > >> Of course this definition requires Regularity, so it works in ZF, it > >> doesn't require choice. > > Yes, about regularity as I mentioned above and you say now. > > > > > > >> However I have the guess, that this definition can be modified to > >> exclude Regularity also. > > > An example of such modification is: > > > Card(x) is the set of d(x) and all subsets of d(x) that are > > equinumerous to x. > > > However the problem with this definition would be sets that are not > > well orderable that are not comparable at the same time. > > > If we stipulate that every two non well orderable sets are comparable > > (having an injection from one to the other) then this definition seems > > to work, but I don't know if this is equivalent to choice really. > > >> Zuhair > > I think I have a proof over ZF that your last suggested property is > equivalent to AC. > > AC -> everything is well-orderable, so your clause about non-well-orderable > set is vacuous. So AC -> your stipulation is easy. > > For the other direction, your stipulation -> AC: > > Assume your stipulation. > > Suppose also A and B are arbitrary sets which cannot be well-ordered. > I will derive a property of them I will state after discussion. > > Let alpha be the usual Hartog's number for B, defined by injections. > Namely alpha is the least ordinal which cannot inject into B. ZF (no > AC needed) proves there is such an ordinal alpha and it is a set, not a > proper class. > > Without loss of generality I will assume A is disjoint from alpha. > (Else replace A by a suitable isomorphic copy). > > Consider the sets B and A union alpha. > > A union alpha cannot be well-ordered, otherwise this woud induce a > well-ordering on A, contrary to assumption on A. > > So your stipulation applies to B and A union alpha, so one of these > must inject into the other. > > But if A union alpha injects into B, this would induce an injection > of alpha into B, contrary to alpha being the injective Hartog numbwer > for B. > > So B must inject into A union alpha. > > Pulling back preimages of A and alpha along this injection. > we conclude the property I state now: B can be partitioned into > a subset isomorphic to a subset of A and a subset which is well-orderable. > > This for any pair A, B both non-well-orderable. > > Next, consider any A which cannot be well-ordered. > > Let beta be the surjective Hartog number for A. > > This is the least ordinal which cannot surject onto A. > > ZF (no AC needed) proves this ordinal exists and is a set > and not a proper class. > > I wrote abopit that in > > [1] David Libert "A new definition of Cardinality" > sci.logic, sci.math Nov 24, 2009 > http://groups.google.com/group/sci.logic/msg/23d5368fc03e61ca > > I quote that: > > > > > ZF proves for any set x there is a set of what I will call > >the surjective Hartog ordinal. The usual definition of the > >Hartog ordinal of a set x is the least ordinal which does > >not inject into x. ZF proves the so called Hartog ordinal > >exists and is a set. > > > I take instead a surjecive version: the least ordinal > >alpha such that x does not surject onto alpha. > > > ZF proves there is always such a set sized ordinal. Namely > >to see this, any surjection of x onto alpha induces an > >equivalence realtiion on x : x elements are equivalent > >if they are sent to the same ordinal. > > > So any ordinal surejcted by x is isomorphic to a well-ordering > >on some subset of P(x), ie pull the wellordering onn the ordinal > >back to the equivalence class preimages. > > > So all the ordinals less than the surjective Hartog ordinal of > >x are ismimorphic to various well-orderings on a single set > >P(x). > > > So the collection of all well-orderings on P(x) has natural > >well ordeing greater than all these, so its von Neuman ordinal > >can't be surjected by x, so there is a least such. > > That quote locally called that surjective Hartog ordinal > alpha, but I will resume calling the surjective Hartog ordinal > for A beta, to avoid confusion with the previous alpha I used > above in this proof. > > So A is an arbitrary non-well-orderable set, and > beta is A's surejctive Hartog number. > > Let B = A x beta. > > If B could be well ordered, it would nice. > ... > > read more »
From: zuhair on 2 Jan 2010 13:16 On Jan 2, 3:32 am, ah...(a)FreeNet.Carleton.CA (David Libert) wrote: > zuhair (zaljo...(a)gmail.com) writes: > > On Jan 1, 4:37=A0am, zuhair <zaljo...(a)gmail.com> wrote: > >> Hi all, > > >> In the last one month I've posted many topics on cardinality. This one > >> is also another trial to define cardinality in ZF, i.e. without > >> Choice. > > >> First from T. Jeck's paper, it appears that his proof can be > >> generalized to every well orderable set x. > > >> So for every well order-able set x, there exist the set of all > >> sets hereditarily subnumerous to x. > > >> This is a theorem of ZF. > > >> Now this time the main idea of the trial is to associate a unique > >> ordinal with each set in ZF such that all sets equinumerous to each > >> other have a corresponding unique ordinal. > > >> So for example lets take any set x, then all sets equinumerous to x > >> would be associated with the same ordinal d(x), now every y that are > >> strictly subnumerous to x would be associated with an ordinal > >> d(y) that is also strictly subnumerous to d(x), same thing applies > >> every set z that is strictly supernumerous to x would be associated > >> with an ordinal d(z) that is strictly supernumerous to d(x). > > Rupert already reponded on this point, for a Dedekind set the > strictly subnumerous sets below are not well-founded so this can't > be done all levels down in that case. > > >> 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? > > 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.. > > >> The point is how to define d(x). > > >> This is a trial: > > >> Define(d(x)): > > Somehow a system along the way has replaced spaces by other characters > which makes the original quote hard to read. I will manually edit those > back to spaces, but I might make mistakes and change your original > spacing. > > >> A= d(x) <-> for all y ( y e A <-> > >> Exist u,z( u strictly subnumerous to x > > & z=d(u) & > >> & y is ordinal & y equinumerous to d > >> (u))). > > >> This is a trick really, > > The form of your defintion is a transfinite induction over strictly > subnumerous, since you invoke d(u). Rupert already gave examples > where this is not well-founded. > > We could interpret your definition as just defining d for sets with > well founded strictly subnumerous relation below them. So maybe it > is only defining cardinality for a subclass of the universe. > > This raises another interesting question. Is it possible to have in > a ZF model a set x which cannot be well-ordered, with well-founded > strictly subnumerous partial ordering below it? > > I don't know. > > Anyway, if it is and some cases of x get d(x) defined this way, > then my previous question for these of why Cardinality(x) is > non-empty remains. > > Below you discuss example of familar small x. I will delete that > because it doesn't touch on the 2 trickier questions I just mentioned. > > [Deletion} > > >> Lets take the Cardinality of Power(omega) and lets assume that power > >> omega is not comparable to any of the uncountable ordinals. > > >> Now we have H(Aleph_1) as a set in ZF. > > By the way, you are also using the generalized Jech result as you > mentioned, but that used regualrity. Below you will consider what > happens with definitions like this without regularity. > > My first model of those discussed recently got H(= 1) a proper > class. So this is another difficulty for the version dropping > regularity. > > [Deletion] > > >> two sets that are strictly non equinumerous will also have different > >> cardinals also since their members would not be equinumerous. > > Except the emptiness issue above if it goes the wrong way. > > >> so: > > >> (1)Card(x)=3DCard(y) iff x and y have the same members(Extensionality). > > >> (2)Card(x) > Card (y) iff [(every member of Card(x) is strictly > >> supernumerous to every member of Card (y)) and =A0(every > >> subset of d(x) that is equinumerous to d(x) is a member of Card(x) iff > >> every subset of d(y) that is equinumerous to d(y) is a member of Card > >> (y)) ]. > > >> (3)Card(x) < Card(y) <-> Card(y) > Card(x) > > >> (4)Card(x) incomparable with Card(y) iff non of the above. > > >> Although very complex definition but I think it works. > > >> Of course this definition requires Regularity, so it works in ZF, it > >> doesn't require choice. > > Yes, about regularity as I mentioned above and you say now. > > >> However I have the guess, that this definition can be modified to > >> exclude Regularity also. > > > An example of such modification is: > > > Card(x) is the set of d(x) and all subsets of d(x) that are > > equinumerous to x. > > > However the problem with this definition would be sets that are not > > well orderable that are not comparable at the same time. > > > If we stipulate that every two non well orderable sets are comparable > > (having an injection from one to the other) then this definition seems > > to work, but I don't know if this is equivalent to choice really. > > >> Zuhair > > I think I have a proof over ZF that your last suggested property is > equivalent to AC. > > AC -> everything is well-orderable, so your clause about non-well-orderable > set is vacuous. So AC -> your stipulation is easy. > > For the other direction, your stipulation -> AC: > > Assume your stipulation. > > Suppose also A and B are arbitrary sets which cannot be well-ordered. > I will derive a property of them I will state after discussion. > > Let alpha be the usual Hartog's number for B, defined by injections. > Namely alpha is the least ordinal which cannot inject into B. ZF (no > AC needed) proves there is such an ordinal alpha and it is a set, not a > proper class. > > Without loss of generality I will assume A is disjoint from alpha. > (Else replace A by a suitable isomorphic copy). > > Consider the sets B and A union alpha. > > A union alpha cannot be well-ordered, otherwise this woud induce a > well-ordering on A, contrary to assumption on A. > > So your stipulation applies to B and A union alpha, so one of these > must inject into the other. > > But if A union alpha injects into B, this would induce an injection > of alpha into B, contrary to alpha being the injective Hartog numbwer > for B. > > So B must inject into A union alpha. > > Pulling back preimages of A and alpha along this injection. > we conclude the property I state now: B can be partitioned into > a subset isomorphic to a subset of A and a subset which is well-orderable. > > This for any pair A, B both non-well-orderable. > > Next, consider any A which cannot be well-ordered. > > Let beta be the surjective Hartog number for A. > > This is the least ordinal which cannot surject onto A. > > ZF (no AC needed) proves this ordinal exists and is a set > and not a proper class. > > I wrote abopit that in > > [1] David Libert "A new definition of Cardinality" > sci.logic, sci.math Nov 24, 2009 > http://groups.google.com/group/sci.logic/msg/23d5368fc03e61ca > > I quote that: > > > ZF proves for any set x there is a set of what I will call > >the surjective Hartog ordinal. The usual definition of the > >Hartog ordinal of a set x is the least ordinal which does > >not inject into x. ZF proves the so called Hartog ordinal > >exists and is a set. > > > I take instead a surjecive version: the least ordinal > >alpha such that x does not surject onto alpha. > > > ZF proves there is always such a set sized ordinal. Namely > >to see this, any surjection of x onto alpha induces an > >equivalence realtiion on x : x elements are equivalent > >if they are sent to the same ordinal. > > > So any ordinal surejcted by x is isomorphic to a well-ordering > >on some subset of P(x), ie pull the wellordering onn the ordinal > >back to the equivalence class preimages. > > > So all the ordinals less than the surjective Hartog ordinal of > >x are ismimorphic to various well-orderings on a single set > >P(x). > > > So the collection of all well-orderings on P(x) has natural > >well ordeing greater than all these, so its von Neuman ordinal > >can't be surjected by x, so there is a least such. > > That quote locally called that surjective Hartog ordinal > alpha, but I will resume calling the surjective Hartog ordinal > for A beta, to avoid confusion with the previous alpha I used > above in this proof. > > So A is an arbitrary non-well-orderable set, and > beta is A's surejctive Hartog number. > > Let B = A x beta. > > If B could be well ordered, it would induce a well-ordering > on A by an isomporphic copy of A sitting as a slice in B. > This contrary to A not being well-orderable. > > So apply my last statement to this A, B. > > Conclude B has a subset isomorphic to a subset of > A and it compliment in B well-orderable. > > Let A' be that subset of B, isomorphic to a subset > of A. > > First suppose every A copy slice inside B has members > in A'. Ie for each gamma < beta, A x {gamma} > as a subset of B has nonempty intersection with A'. > > A' is isomorphic to a subset of A. So for the A members > in that subset, map to A' and then map to the gamma > where they live. > > For A members outside the subset if any just map to > ordinal 0. > > Sonce we are for the moment assuming every slice gamma > has some A' members, this map is surjective onto > beta. > > We just surjected A onto beta, contrary to beta being > the surjective Hartog ordinal for A. > > We conclude some gamma slice of B is disjoint from > A'. > > So the entire A copy on slice gamma was instead sent > to B - A', the set which is well-orderable. > > This well ordering induces a well-ordering on its > subset the gamma'th slice, which being isomprphic to > A induces a well-ordering on A. > > Contrary to A not being well-orderable. > > We got this contradiction by assuming your stipulation > and also assuming some A cannot be well-ordered. > > So every wset can be well-ordered, hence AC, > from the assumption of your stipulation. Yea, that's nice. Thanks. > > Another point. I did my models of ZF - regualrity with > cardinality undefinable. For the momeent those seem to be > ok. > > So what is this about seeking a definition of cardinality > in ZF - regularity? Do you mean to add some axioms to that? > Or do you mean really just ZF - regularity, and you are > still questioning those models? > > -- > David Libert ah...(a)FreeNet.Carleton.CA
From: zuhair on 2 Jan 2010 16:20 On Jan 2, 1:14 pm, zuhair <zaljo...(a)gmail.com> wrote: > On Jan 2, 3:32 am, ah...(a)FreeNet.Carleton.CA (David Libert) wrote: > > > > > > > zuhair (zaljo...(a)gmail.com) writes: > > > On Jan 1, 4:37=A0am, zuhair <zaljo...(a)gmail.com> wrote: > > >> Hi all, > > > >> In the last one month I've posted many topics on cardinality. This one > > >> is also another trial to define cardinality in ZF, i.e. without > > >> Choice. > > > >> First from T. Jeck's paper, it appears that his proof can be > > >> generalized to every well orderable set x. > > > >> So for every well order-able set x, there exist the set of all > > >> sets hereditarily subnumerous to x. > > > >> This is a theorem of ZF. > > > >> Now this time the main idea of the trial is to associate a unique > > >> ordinal with each set in ZF such that all sets equinumerous to each > > >> other have a corresponding unique ordinal. > > > >> So for example lets take any set x, then all sets equinumerous to x > > >> would be associated with the same ordinal d(x), now every y that are > > >> strictly subnumerous to x would be associated with an ordinal > > >> d(y) that is also strictly subnumerous to d(x), same thing applies > > >> every set z that is strictly supernumerous to x would be associated > > >> with an ordinal d(z) that is strictly supernumerous to d(x). > > > Rupert already reponded on this point, for a Dedekind set the > > strictly subnumerous sets below are not well-founded so this can't > > be done all levels down in that case. > > > >> 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. > > Because we must define d(x) to exist for every x, that's the idea, so > Cardinality(x) would also have d(x) as a member so it cannot be empty. > > > > > > > > > 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? > > > 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. > > > >> The point is how to define d(x). > > > >> This is a trial: > > > >> Define(d(x)): > > > Somehow a system along the way has replaced spaces by other characters > > which makes the original quote hard to read. I will manually edit those > > back to spaces, but I might make mistakes and change your original > > spacing. > > > >> A= d(x) <-> for all y ( y e A <-> > > >> Exist u,z( u strictly subnumerous to x > > > & z=d(u) & > > >> & y is ordinal & y equinumerous to d > > >> (u))). > > > >> This is a trick really, > > > The form of your defintion is a transfinite induction over strictly > > subnumerous, since you invoke d(u). Rupert already gave examples > > where this is not well-founded. > > NO Rupert gave an example when this is contradictive, since d(x) would > be a set of ordinals, and thus must be well-founded, and at the same > time we'll have a non well founded set of them with the case of non > Dedekinds sets, so it would be contradictive. Actually I don't see why d(x) would be non well founded, d(x) is contradictive because it is a set of ordinals so it must be well orderable, and at the same time it would be NON WELL Orderable, so the conflict is all about "well orderability" and not about well foundedeness. > > We may resolve this contradiction of I remove the condition " d(x) is > ordinal" > and replace it by another condition, like d(x) is hereditarily > hereditary etc.. > this would solve it, but as far as d(x) is ordinal then this would > lead to the contradiction that Rupert referred to. > > > > > We could interpret your definition as just defining d for sets with > > well founded strictly subnumerous relation below them. So maybe it > > is only defining cardinality for a subclass of the universe. > > Yea, that my be case indeed. > > > > > This raises another interesting question. Is it possible to have in > > a ZF model a set x which cannot be well-ordered, with well-founded > > strictly subnumerous partial ordering below it? > > > I don't know. > > > Anyway, if it is and some cases of x get d(x) defined this way, > > then my previous question for these of why Cardinality(x) is > > non-empty remains. > > see above. > > > > > > > Below you discuss example of familar small x. I will delete that > > because it doesn't touch on the 2 trickier questions I just mentioned. > > > [Deletion} > > > >> Lets take the Cardinality of Power(omega) and lets assume that power > > >> omega is not comparable to any of the uncountable ordinals. > > > >> Now we have H(Aleph_1) as a set in ZF. > > > By the way, you are also using the generalized Jech result as you > > mentioned, but that used regualrity. Below you will consider what > > happens with definitions like this without regularity. > > > My first model of those discussed recently got H(= 1) a proper > > class. So this is another difficulty for the version dropping > > regularity. > > > [Deletion] > > > >> two sets that are strictly non equinumerous will also have different > > >> cardinals also since their members would not be equinumerous. > > > Except the emptiness issue above if it goes the wrong way. > > > >> so: > > > >> (1)Card(x)=3DCard(y) iff x and y have the same members(Extensionality). > > > >> (2)Card(x) > Card (y) iff [(every member of Card(x) is strictly > > >> supernumerous to every member of Card (y)) and =A0(every > > >> subset of d(x) that is equinumerous to d(x) is a member of Card(x) iff > > >> every subset of d(y) that is equinumerous to d(y) is a member of Card > > >> (y)) ]. > > > >> (3)Card(x) < Card(y) <-> Card(y) > Card(x) > > > >> (4)Card(x) incomparable with Card(y) iff non of the above. > > > >> Although very complex definition but I think it works. > > > >> Of course this definition requires Regularity, so it works in ZF, it > > >> doesn't require choice. > > > Yes, about regularity as I mentioned above and you say now. > > > >> However I have the guess, that this definition can be modified to > > >> exclude Regularity also. > > > > An example of such modification is: > > > > Card(x) is the set of d(x) and all subsets of d(x) that are > > > equinumerous to x. > > > > However the problem with this definition would be sets that are not > > > well orderable that are not comparable at the same time. > > > > If we stipulate that every two non well orderable sets are comparable > > > (having an injection from one to the other) then this definition seems > > > to work, but I don't know if this is equivalent to choice really. > > > >> Zuhair > > > I think I have a proof over ZF that your last suggested property is > > equivalent to AC. > > > AC -> everything is well-orderable, so your clause about non-well-orderable > > set is vacuous. So AC -> your stipulation is easy. > > > For the other direction, your stipulation -> AC: > > > Assume your stipulation. > > > Suppose also A and B are arbitrary sets which cannot be well-ordered. > > I will derive a property of them I will state after discussion. > > > Let alpha be the usual Hartog's number for B, defined by injections. > > Namely alpha is the least ordinal which cannot inject into B. ZF (no > > AC needed) proves there is such an ordinal alpha and it is a set, not a > > proper class. > > > Without loss of generality I will assume A is disjoint from alpha. > > (Else replace A by a suitable isomorphic copy). > > > Consider the sets B and A union alpha. > > > A union alpha cannot be well-ordered, otherwise this woud induce a > > well-ordering on A, contrary to assumption on A. > > > So your stipulation applies to B and A union alpha, so one of these > > must inject into the other. > > > But if A union alpha injects into B, this would induce an injection > > of alpha into B, contrary to alpha being the injective Hartog numbwer > > for B. > > > So B must inject into A union alpha. > > > Pulling back preimages of A and alpha along this injection. > > we conclude the property I state now: B can be partitioned into > > a subset isomorphic to a subset of A and a subset which is well-orderable. > > > This for any pair A, B both non-well-orderable. > > > Next, consider any A which cannot be well-ordered. > > > Let beta be the surjective Hartog number for A. > > > This is the least ordinal which cannot surject onto A. > > > ZF (no AC needed) proves this ordinal exists and is a set > > and not a proper class. > > > I wrote abopit that in > > > [1] David Libert "A new definition of Cardinality" > > sci.logic, sci.math Nov 24, 2009 > > http://groups.google.com/group/sci.logic/msg/23d5368fc03e61ca > > > I quote that: > > > > ZF proves for any set x there is a set of what I will call > > >the surjective Hartog > > ... > > read more »- Hide quoted text - > > - Show quoted text -- Hide quoted text - > > - Show quoted text -- Hide quoted text - > > - Show quoted text -
From: MoeBlee on 2 Jan 2010 16:44
On Jan 2, 8:30 am, zuhair <zaljo...(a)gmail.com> wrote: > Suppose we add Cardinality symbolized by "| |" as a primitive to the > language of ZF minus Reg., so"| |" would be a primitive one place > function symbol. Let's > name this theory ZF minus Reg.(=,e,| |) to differentiate it from > ZF minus Reg. > > And suppose we add the following axiom to the list of axioms of > ZF minus Reg.(=,e,| |). > > Axiom of Cardinality: > > For all x,y : |x|=|y| <-> Exist f ( f:x-->y , f is bijective ) > > So now we have the theory ZF minus Reg.+Axiom of Cardinality(=,e,| |). Okay, so far. You can find this (Tarskian, if I recall) approach (even whether or not we have regularity) in certain textbooks as a temporary way of dealing with cardinality without using choice (or even regularity): card(x) = card(y) <-> x equinumerous with y It may be adopted as a temporary axiom (let's call it "Tarski's axiom") until we add choice to prove a corollary of the numeration theorem (consider all formulas closed as appropriate): E!y(y is an ordinal & y equinumerous with x & Az((z ordinal & z equinumerous with x) -> ~zey)) (i.e, there is a unique y that is the least ordinal equinumerous with x) or it may be added as a temporary axiom until we add regularity (if we didnt' already have regularity) to prove: E!yAt(tey <-> (t equinumerous with x & Az(z equinumerous with x -> ~ rank(z) e rank(t)))) (i.e. there exists the set of all sets that are equinumerous with x and are of lowest rank among sets equinumerous with x) I gather that you pretty much already know that, but I'm including it for context, along with mentioning, that with choice and regularity combined, the two approaches above pick out the same set to serve as card(x). > So for example I want to come with a "definition of Cardinality" in ZF > minus Reg. > suppose that was, for a specific formula Phi(x): > > Card(x) <-> Phi(x) That is not a definition of a function symbol. It is not even a well formed formula. On the left of the biconditional you have a TERM but the left of a biconditional must be a FORMULA. One form for definining a function symbol is: card(x) = y <-> Phi(x y) where we have the thereom already proven: E!y Phi(x y) In case you don't have that theorem already proven, then (in a theory with '0' as a (primitive or defined) 0-place function symbol) you can use "the Fregean method": card(x) = y <-> ((AxE!y Phi(x y) & Phi(x y)) or (~AxE!y Phi(x y) & y=0)) So let's assume you've adopted such an approach. > Then I will add the axiom: > > Axiom of Cardinality: For all x Exist y ( y=Card(x) ). If 'card' (primitive or defined) is now in the language (primitive or extended by definitions), then the above formula is a theorem of logic alone anyway; you don't need it as an additional axiom. This has been mentioned many many times in discusssions on sci.logic that I think you probably have read. > So we'll have ZF minus Reg. + Axiom of Cardinality. > > Now what I want is that > > ZF minus Reg. + Axiom of Cardinality Equi-interpretable with > ZF minus Reg. + Axiom of Cardinality (e,=,| | ). > > Can that be done? If we correct your formulation, then, if I correctly understand your intent, your question becomes: Is there a formula Phi (in the language of set theory extended to include 'card') 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 ? MoeBlee |