Recursive Cardinals.
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From: zuhair on 18 Dec 2009 20:57 Hi all, Define(circular): x is circular <-> x e TC(x) Define(hereditarily non circular): x hereditarily non circular <-> for all y ( y e TC(x) -> ~ y is circular) Define(x-recursive): S is x-recursive <-> S equinumerous to x & for all y ( y e TC(S) -> y equinumerous to x )& S hereditarily non circular. in other words S is x-recursive <-> S equinumerous to x & S hereditarily equinumerous to x & S hereditarily non circular. Now we can define Cardinality in ZF minus Regularity as: ----------------------------------------------------------------------- Define(Cardinality(x)): Cardinality(x) is the class of all x-recursive sets. ----------------------------------------------------------------------- So for example the Cardinality of the empty set { } is the class of all sets equinumerous to { } that are hereditarily equinumerous to { } and since we have Extensionality then this would be the class { { } } So Card({}) = { {} } Now the cardinality of any singleton set {y} would be the class of all recursive singletons, i.e the class of all hereditarily non circular singleton hereditarily singleton sets. The cardinality of any doubleton sets {y,z} were ~y=z is the class of all recursive doubletons, i.e the class of all hereditarily non circular doubleton hereditarily doubleton sets. Same thing is applied for every set x. Now we may call these cardinals as : "Recursive Cardinals". Now what is the proof in ZF minus Regularity that these cardinals would be *set*s and not proper classes? The proof is the following Lemma in ZF minus Regularity: For all x , for all y y e TC(x) if and only if there exist a finite sequence <x0,x1,x2,...,xn> were x0 e x and xi+1 e xi for every i=0,1,2,...,n-1 and y=xn. The most important thing is that because of the Lemma above the class of all x-recursive sets for a particular x would be a set because the process is exhaustive!, we cannot go on having increasing numbers of these x-recursive sets in such a manner as to have a proper class of these x-recursive sets for a particular x. While with the Ur-elements and with Quine atoms, we can have proper classes of them, in such a manner that for each x there would exist a proper class of sets of Quine atoms or of Ur-elements that are equinumerous to x. The reason is that with Ur-elements and Quine atoms there is nothing to check on their identity, i.e. their identity is not checked by their membership (the identity of the members they contain) they are self-determined identity objects so one can proliferate them at will to the size of proper classes. For example we can have the axiom For all x Exist u (for all y (y e u -> y is a Ur-element) & u equinumerous to x) we can have the same axiom with Quine atoms. For all x Exist u (for all y (y e u -> y is a Quine atom) & u equinumerous to x) BUT, we cannot have the same axiom with recursive sets. i.e. the following is refuted in ZF minus Regularity: For all m For all x Exist u (for all y (y e u -> y is m-recursive) & u equinumerous to x) That's why we can use these recursive sets to define cardinals. That's why these recursive cardinals are sets! Of course all of the above needs the axiom: Axiom of Anti-Foundation: For all x Exist y ( y is x-recursive ). I don't think there is a way in ZF minus Regularity to construct these sets without the above axiom. The above is an example of actually defining Cardinality without Regularity and Choice, and without Coret's assumption also. Of course these recursive cardinals are shunned in ZF. However to define cardinality that works both in ZF and ZF minus Regularity one can actually have a joint definition of Scott cardinals and the above. -------------------------------------------------------------------------------------------- Define(Cardinality(x)): Cardinality(x) is the class of all pure well founded sets Equinumerous to x of the least possible rank, and of all x-recursive sets. --------------------------------------------------------------------------------------------- "A set is said to be pure iff no Ur-element exist in its transitive closure". In ZF this would be reduced to the Scott cardinals. In ZF minus Regularity + Anti-Foundation then outside Coret's assumption this will be reduced to the recursive cardinals, While when satisfying Coret's assumption the cardinal would be the union of a Scott cardinal and a recursive cardinal. Zuhair --------------------------------------------------------------- " And treat the sickness by what caused it in the first place" An ancient saying!
From: zuhair on 20 Dec 2009 10:33 On Dec 18, 8:57 pm, zuhair <zaljo...(a)gmail.com> wrote: > Hi all, > > Define(circular): x is circular <-> x e TC(x) > > Define(hereditarily non circular): > > x hereditarily non circular <-> > for all y ( y e TC(x) -> ~ y is circular) > > Define(x-recursive): > > S is x-recursive <-> > S equinumerous to x & > for all y ( y e TC(S) -> y equinumerous to x )& > S hereditarily non circular. > > in other words > > S is x-recursive <-> > S equinumerous to x & > S hereditarily equinumerous to x & > S hereditarily non circular. > > Now we can define Cardinality in ZF minus Regularity as: > > ----------------------------------------------------------------------- > Define(Cardinality(x)): > > Cardinality(x) is the class of all x-recursive sets. > ----------------------------------------------------------------------- > > So for example the Cardinality of the empty set { } > is the class of all sets equinumerous to { } > that are hereditarily equinumerous to { } > and since we have Extensionality then this would be > the class { { } } > > So Card({}) = { {} } > > Now the cardinality of any singleton set {y} would be > the class of all recursive singletons, i.e > the class of all hereditarily non circular > singleton hereditarily singleton sets. > > The cardinality of any doubleton sets {y,z} were ~y=z > is the class of all recursive doubletons, i.e > the class of all hereditarily non circular > doubleton hereditarily doubleton sets. > > Same thing is applied for every set x. > > Now we may call these cardinals as : > > "Recursive Cardinals". > > Now what is the proof in ZF minus Regularity > that these cardinals would be *set*s > and not proper classes? > > The proof is the following Lemma in ZF minus Regularity: > > For all x , for all y > > y e TC(x) if and only if there exist a finite sequence > <x0,x1,x2,...,xn> were x0 e x and > xi+1 e xi for every i=0,1,2,...,n-1 and y=xn. > > The most important thing is that because of the Lemma > above the class of all x-recursive sets for a particular x would be > a set because the process is exhaustive!, we cannot go > on having increasing numbers of these x-recursive sets > in such a manner as to have a proper class of these > x-recursive sets for a particular x. > > While with the Ur-elements and with Quine atoms, we can > have proper classes of them, in such a manner that > for each x there would exist a proper class of sets of Quine atoms > or of Ur-elements that are equinumerous to x. > > The reason is that with Ur-elements and Quine atoms there is > nothing to check on their identity, i.e. their identity is not > checked by their membership > (the identity of the members they contain) > they are self-determined identity objects so one can proliferate > them at will to the size of proper classes. > > For example we can have the axiom > > For all x Exist u > (for all y (y e u -> y is a Ur-element) & u equinumerous to x) > > we can have the same axiom with Quine atoms. > > For all x Exist u > (for all y (y e u -> y is a Quine atom) & u equinumerous to x) > > BUT, we cannot have the same axiom with recursive sets. > > i.e. the following is refuted in ZF minus Regularity: > > For all m For all x Exist u > (for all y (y e u -> y is m-recursive) & u equinumerous to x) > > That's why we can use these recursive sets to define cardinals. > > That's why these recursive cardinals are sets! > > Of course all of the above needs the axiom: > > Axiom of Anti-Foundation: > For all x Exist y ( y is x-recursive ). > > I don't think there is a way in ZF minus Regularity to construct > these sets without the above axiom. > > The above is an example of actually defining Cardinality without > Regularity and Choice, and without Coret's assumption also. > > Of course these recursive cardinals are shunned in ZF. > > However to define cardinality that works both in ZF and > ZF minus Regularity one can actually have a joint definition of Scott > cardinals and the above. > > -------------------------------------------------------------------------------------------- > Define(Cardinality(x)): > > Cardinality(x) is the class of all pure well founded sets Equinumerous > to x of the least possible rank, and of all x-recursive sets. > --------------------------------------------------------------------------------------------- > > "A set is said to be pure iff no Ur-element exist in its transitive > closure". > > In ZF this would be reduced to the Scott cardinals. > In ZF minus Regularity + Anti-Foundation > then outside Coret's assumption > this will be reduced to the recursive cardinals, > While when satisfying Coret's assumption > the cardinal would be the union of a Scott cardinal > and a recursive cardinal. > > Zuhair > --------------------------------------------------------------- > " And treat the sickness by what caused it in the first place" > > An ancient saying! How to construct these cardinals in ZF minus Regularity+Anti- Foundation axiom above? Take any set x. According to the Anti-Foundation axiom, there must exist at least one set y such that y is x-recursive. Now we have for every set y, there exist the Transitive closure of y "TC(y)" which is a set of course, as a theorem of ZF minus Regularity. Now we take the P_x(TC(y)) were P_x(z) generally stands for the set of all subsets of z that are equinumerous to x. Since y is equinumerous to x and y is a subset of TC(y), then we have x subnumerous to TC(y), then P_x(TC(y)) is non empty. Now we continuing the P_x powering omega times i.e. we continue powering like that P_x(TC(y)), P_x(P_x(TC(y))), P_x(P_x(P_x(TC(y)))),.... Let's define these inductively as: (P_x)0(TC(y)) = TC(y) (P_x)1(TC(y)) = P_x(TC(y)) (P_x)2(TC(y)) = P_x(P_x(TC(y))) generally (P_x)i+1(TC(y)) = P_x((P_x)i(TC(y))) for every finite ordinal i. Now (P_x)w(TC(y)) = Union(P_x)i(TC(y)) for i=1,2,3,..... Now (P_x)w (TC(y)) would be the Cardinality of x, because it would be the set of all x-recursive sets. (see the lemma in the first post). So we can define Recursive Cardinals for *ALL* sets in ZF minus Regularity + Anti-Foundation. QED. Actually this would prove that it is possible to define Cardinality for every set in ZF minus Regularity! Now I don't know how can that be reconciled with Gauntt's proof which proves the opposite(although Gauntt's models violates Extensionality,and comparable models has Quine atoms and violate "strong Extensionality"). A second issue is that it would a theorem that For all x , Cardinality(x) not subnumerous to x. Otherwise we'll arrive at a paradox of Cardinality(x) e Cardinality(x) while at the same time Cardinality(x) is non circular. A contradiction. Thus it is a theorem of ZF-Regularity+Anti-Foundation that Card(x) not subnumerous to x, for all x. (of course I am speaking about Cardinality defined in the recursive manner as detailed in this post, and the Anti-Foundation axiom refer to the one written here in this post). Zuhair
From: zuhair on 20 Dec 2009 10:36 On Dec 20, 10:33 am, zuhair <zaljo...(a)gmail.com> wrote: > On Dec 18, 8:57 pm, zuhair <zaljo...(a)gmail.com> wrote: > > > > > > > Hi all, > > > Define(circular): x is circular <-> x e TC(x) > > > Define(hereditarily non circular): > > > x hereditarily non circular <-> > > for all y ( y e TC(x) -> ~ y is circular) > > > Define(x-recursive): > > > S is x-recursive <-> > > S equinumerous to x & > > for all y ( y e TC(S) -> y equinumerous to x )& > > S hereditarily non circular. > > > in other words > > > S is x-recursive <-> > > S equinumerous to x & > > S hereditarily equinumerous to x & > > S hereditarily non circular. > > > Now we can define Cardinality in ZF minus Regularity as: > > > ----------------------------------------------------------------------- > > Define(Cardinality(x)): > > > Cardinality(x) is the class of all x-recursive sets. > > ----------------------------------------------------------------------- > > > So for example the Cardinality of the empty set { } > > is the class of all sets equinumerous to { } > > that are hereditarily equinumerous to { } > > and since we have Extensionality then this would be > > the class { { } } > > > So Card({}) = { {} } > > > Now the cardinality of any singleton set {y} would be > > the class of all recursive singletons, i.e > > the class of all hereditarily non circular > > singleton hereditarily singleton sets. > > > The cardinality of any doubleton sets {y,z} were ~y=z > > is the class of all recursive doubletons, i.e > > the class of all hereditarily non circular > > doubleton hereditarily doubleton sets. > > > Same thing is applied for every set x. > > > Now we may call these cardinals as : > > > "Recursive Cardinals". > > > Now what is the proof in ZF minus Regularity > > that these cardinals would be *set*s > > and not proper classes? > > > The proof is the following Lemma in ZF minus Regularity: > > > For all x , for all y > > > y e TC(x) if and only if there exist a finite sequence > > <x0,x1,x2,...,xn> were x0 e x and > > xi+1 e xi for every i=0,1,2,...,n-1 and y=xn. > > > The most important thing is that because of the Lemma > > above the class of all x-recursive sets for a particular x would be > > a set because the process is exhaustive!, we cannot go > > on having increasing numbers of these x-recursive sets > > in such a manner as to have a proper class of these > > x-recursive sets for a particular x. > > > While with the Ur-elements and with Quine atoms, we can > > have proper classes of them, in such a manner that > > for each x there would exist a proper class of sets of Quine atoms > > or of Ur-elements that are equinumerous to x. > > > The reason is that with Ur-elements and Quine atoms there is > > nothing to check on their identity, i.e. their identity is not > > checked by their membership > > (the identity of the members they contain) > > they are self-determined identity objects so one can proliferate > > them at will to the size of proper classes. > > > For example we can have the axiom > > > For all x Exist u > > (for all y (y e u -> y is a Ur-element) & u equinumerous to x) > > > we can have the same axiom with Quine atoms. > > > For all x Exist u > > (for all y (y e u -> y is a Quine atom) & u equinumerous to x) > > > BUT, we cannot have the same axiom with recursive sets. > > > i.e. the following is refuted in ZF minus Regularity: > > > For all m For all x Exist u > > (for all y (y e u -> y is m-recursive) & u equinumerous to x) > > > That's why we can use these recursive sets to define cardinals. > > > That's why these recursive cardinals are sets! > > > Of course all of the above needs the axiom: > > > Axiom of Anti-Foundation: > > For all x Exist y ( y is x-recursive ). > > > I don't think there is a way in ZF minus Regularity to construct > > these sets without the above axiom. > > > The above is an example of actually defining Cardinality without > > Regularity and Choice, and without Coret's assumption also. > > > Of course these recursive cardinals are shunned in ZF. > > > However to define cardinality that works both in ZF and > > ZF minus Regularity one can actually have a joint definition of Scott > > cardinals and the above. > > > -------------------------------------------------------------------------------------------- > > Define(Cardinality(x)): > > > Cardinality(x) is the class of all pure well founded sets Equinumerous > > to x of the least possible rank, and of all x-recursive sets. > > --------------------------------------------------------------------------------------------- > > > "A set is said to be pure iff no Ur-element exist in its transitive > > closure". > > > In ZF this would be reduced to the Scott cardinals. > > In ZF minus Regularity + Anti-Foundation > > then outside Coret's assumption > > this will be reduced to the recursive cardinals, > > While when satisfying Coret's assumption > > the cardinal would be the union of a Scott cardinal > > and a recursive cardinal. > > > Zuhair > > --------------------------------------------------------------- > > " And treat the sickness by what caused it in the first place" > > > An ancient saying! How to construct these cardinals in ZF minus Regularity+Anti- Foundation axiom above? Take any set x. According to the Anti-Foundation axiom, there must exist at least one set y such that y is x-recursive. Now we have for every set y, there exist the Transitive closure of y "TC(y)" which is a set of course, as a theorem of ZF minus Regularity. Now we take the P_x(TC(y)) were P_x(z) generally stands for the set of all subsets of z that are equinumerous to x. Since y is equinumerous to x and y is a subset of TC(y), then we have x subnumerous to TC(y), then P_x(TC(y)) is non empty. Now we continuing the P_x powering omega times i.e. we continue powering like that P_x(TC(y)), P_x(P_x(TC(y))), P_x(P_x(P_x(TC(y)))),.... Let's define these inductively as: (P_x)0(TC(y)) = TC(y) (P_x)1(TC(y)) = P_x(TC(y)) (P_x)2(TC(y)) = P_x(P_x(TC(y))) generally (P_x)i+1(TC(y)) = P_x((P_x)i(TC(y))) for every finite ordinal i. Now (P_x)w(TC(y)) = Union(P_x)i(TC(y)) for i=0,1,2,3,..... Now (P_x)w (TC(y)) would be the Cardinality of x, because it would be the set of all x-recursive sets. (see the lemma in the first post). So we can define Recursive Cardinals for *ALL* sets in ZF minus Regularity + Anti-Foundation. QED. A second issue is that it would a theorem that For all x , Cardinality(x) not subnumerous to x. Otherwise we'll arrive at a paradox of Cardinality(x) e Cardinality (x) while at the same time Cardinality(x) is non circular. A contradiction. Thus it is a theorem of ZF-Regularity+Anti-Foundation that Card(x) not subnumerous to x, for all x. (of course I am speaking about Cardinality defined in the recursive manner as detailed in this post, and the Anti-Foundation axiom refer to the one written here in this post). Zuhair
From: zuhair on 25 Dec 2009 09:28 The following is a modified quote from another subject in reply to David Libert, but I think it is better to be mentioned here. Here I have presented an example of these x-Recursive Cardinals were x is singleton, It shows that we cannot have uncountably many recursive singletons, at least at informal level, here these recursive singletons shall be called "singleton towers": First, a singleton tower, or what I call a "recursive singleton" can be defined in the following manner: x is a singleton tower iff x is singleton & For all y ( y e TC(x) -> y is singleton ) & For all y ( y e TC(x) -> ~ y e TC(y) ) So singleton towers are: hereditarily non circular singleton hereditarily singletons. we can symbolize that as x={{{... ...}}} is a singleton tower. However for the sake of simplicity lets attach a natural number to each bracket in these singleton towers, i.e. let's number the brackets in these towers, so let's say that x=0{1{2{... ...}}} i.e. the bracket number sequence of x is <0,1,2,3,.....> so the outer bracket has the number 0, the one inside it has the number 1, and the one inside it has the number 2, etc... Now we can see that the object y defined by the bracket sequence <1,2,3,....> will be inside x which is defined by the bracket sequence <0,1,2,3,....> So in general the singleton tower xi+1 with bracket sequence of <i+1,i+2,i+3,....> will be the member of the singleton tower xi with bracket sequence of <i,i+1,i+2,i+3,...> I think this is clear, so we have xi+1 e xi for all singleton towers defined above. Now lets take the tower were i=0, i.e. x0 with bracket sequence of <0,1,2,3,....> Now what is the transitive closure of x0 This would be TC(x0)= {x1,x2,x3,..........} right! Now obviously TC(x0) is countable! Now let's perform iterative singleton operations on x0, and let's use the negative integers for that purpose, so we'll have x(-1) = {x0} ,i.e. the bracket sequence of x(-1) is <-1,0,1,2,3,....> x(-2) = { x(-1) } , i.e. the bracket sequence x(-2) is <-2,-1,0,1,2,3,...> .. .. .. (x(-i)) = {x(-i+1)} for all i=1,2,3,..... Now how many x(-i) we have? of course we have countably many of them! that is clear. Now the set of all singleton towers would be ST={...,x(-2),x(-1),x0,x1,x2,......} Which is countable! We cannot have more than those! If we take the singleton power of ST denoted as P1(ST) (i.e. the set of all singleton subsets of ST) , we'll only have ST itself ST=P1(ST) Now because of the following lemma in ZF minus Regularity: Lemma: For all x , for all y y e TC(x) if and only if there exist a finite sequence <x0,x1,x2,...,xn> were x0 e x and xi+1 e xi for every i=0,1,2,...,n-1 and y=xn. Then we cannot go down further, nor we can go up further, i.e. we cannot have for example a singleton tower with a bracket sequence of <0,1,2,.....,w,w+1,w+2,...> , this is forbidden by the lemma above, also we cannot have a singleton tower with a bracket sequence of <-w,.....,-2,-1,0,1,2,....>, or even <...,-2,-1,0,1,2,....>, all of these cases are forbidden because of the lemma above. Also we cannot have a set with a bracket sequence of for example <0,2,3,4,...>, because this would be equal to the x1 i.e. the singleton tower with bracket sequence of <1,2,3,....>, also we cannot have sets with bracket sequences of for example <1,3,5,7,...>, or <1,2,3,5,7,11,...>, these have gaps in their sequences which is not compatible with the definition of singleton powers. Actually I don't see how we can ever have an uncountable number of these singleton towers?! Actually even if we suppose that we can have these gap sequences above, and by them one might conclude that we can have an uncountable number of these singleton towers, but still the issue remains, how can we have a proper class of them??? Because of the lemma above it is clear that Singleton powering of ST will stop giving different sets at some point, i.e. the process is exhaustive! So the class of all singleton towers, which is the cardinal number one would be a set! I might be mistaken though, but I would like to know how can we have an uncountable number of these singleton towers? and even more how can we have a proper class of them? I want a proof from within the model, and not from outside it. Zuhair
From: zuhair on 26 Dec 2009 02:26
On Dec 25, 9:28 am, zuhair <zaljo...(a)gmail.com> wrote: > The following is a modified quote from another subject in reply to > David Libert, but I think it is better to be mentioned here. > > Here I have presented an example of these x-Recursive Cardinals > were x is singleton, It shows that we cannot have > uncountably many recursive singletons, at least at informal level, > here these recursive singletons shall be called "singleton towers": > > First, a singleton tower, or what I call a "recursive singleton" can > be defined in the following manner: > > x is a singleton tower iff > x is singleton & > For all y ( y e TC(x) -> y is singleton ) & > For all y ( y e TC(x) -> ~ y e TC(y) ) > > So singleton towers are: hereditarily non circular singleton > hereditarily singletons. > > we can symbolize that as x={{{... ...}}} is a singleton tower. > > However for the sake of simplicity lets attach a natural number to > each > bracket in these singleton towers, i.e. let's number the brackets in > these towers, > > so let's say that x=0{1{2{... ...}}} > > i.e. the bracket number sequence of x is <0,1,2,3,.....> > so the outer bracket has the number 0, the one inside it has the > number 1, and the one inside it has the number 2, etc... > > Now we can see that the object y defined by the bracket sequence > <1,2,3,....> > will be inside x which is defined by the bracket sequence > <0,1,2,3,....> > > So in general the singleton tower xi+1 with bracket sequence of > <i+1,i+2,i+3,....> will be the member of the singleton tower xi with > bracket sequence of <i,i+1,i+2,i+3,...> > > I think this is clear, so we have xi+1 e xi for all singleton towers > defined above. > > Now lets take the tower were i=0, i.e. x0 with bracket sequence of > <0,1,2,3,....> > > Now what is the transitive closure of x0 > > This would be > > TC(x0)= {x1,x2,x3,..........} > > right! > > Now obviously TC(x0) is countable! > > Now let's perform iterative singleton operations on x0, and let's use > the negative integers for that purpose, so we'll have > > x(-1) = {x0} ,i.e. the bracket sequence of x(-1) is <-1,0,1,2,3,....> > x(-2) = { x(-1) } , i.e. the bracket sequence x(-2) is > <-2,-1,0,1,2,3,...> > . > . > . > (x(-i)) = {x(-i+1)} for all i=1,2,3,..... > > Now how many x(-i) we have? > > of course we have countably many of them! that is clear. > > Now the set of all singleton towers would be > > ST={...,x(-2),x(-1),x0,x1,x2,......} > > Which is countable! > > We cannot have more than those! > > If we take the singleton power of ST > denoted as P1(ST) (i.e. the set of all singleton subsets of ST) > , we'll only have ST itself > > ST=P1(ST) > > Now because of the following lemma in ZF minus Regularity: > > Lemma: For all x , for all y > > y e TC(x) if and only if there exist a finite sequence > <x0,x1,x2,...,xn> were x0 e x and > xi+1 e xi for every i=0,1,2,...,n-1 and y=xn. > > Then we cannot go down further, nor we can go up further, i.e. we > cannot have for example a singleton tower with a bracket sequence of > <0,1,2,.....,w,w+1,w+2,...> , this is forbidden by the lemma above, > also we cannot have a singleton tower with a bracket sequence of > <-w,.....,-2,-1,0,1,2,....>, or even <...,-2,-1,0,1,2,....>, all of > these cases are forbidden because of the lemma above. > > Also we cannot have a set with a bracket sequence of for example > <0,2,3,4,...>, because this would be equal to the x1 i.e. the > singleton > tower with bracket sequence of <1,2,3,....>, also we cannot > have sets with bracket sequences of for example > <1,3,5,7,...>, or <1,2,3,5,7,11,...>, these have gaps in their > sequences > which is not compatible with the definition of singleton powers. > > Actually I don't see how we can ever have an uncountable number of > these singleton towers?! > > Actually even if we suppose that we can have these gap sequences > above, and by them one might conclude that we can have > an uncountable number of these singleton towers, but still the issue > remains, > how can we have a proper class of them??? > > Because of the lemma above it is clear that Singleton powering of ST > will stop giving different sets at some point, i.e. the process is > exhaustive! > So the class of all singleton towers, which is the cardinal number one > would be a set! > > I might be mistaken though, but I would like to know how can we have > an uncountable number of these singleton towers? and even more > how can we have a proper class of them? I want a proof from within > the model, and not from outside it. > > Zuhair I am coming to think that these recursive cardinals cannot do the job really. For example what would be the recursive cardinal of Aleph_1, this would need 2^Aleph_1 of bracket sequences, But we only have maximally Aleph_1 of such sequences. Actually I am coming to be that only singleton recursives are possible, one cannot larger recursive sets. The reason I am saying that is because I think we actually cannot have more than countable number of bracket sequences for any recursive set,but if we say for example we have a recursive doubleton then it seems that we'll need an uncountable number of them, which is not possible, therefore we would have only singleton recursive sets. IF this is correct, then matters would be much easier really. One would easily define Cardinality if the axiom of strong Extensionality (see:http://groups.google.com.jm/group/sci.logic/browse_thread/thread/ 4bf8472f33458c84?hl=en) is adopted. Zuhair |