Re: Another definition of Cardinality:
zuhair
Here I shall present another definition of Cardinality:
For any set X, if there exist a set Y that is equinumerous to X
such that there do not exist a set Z that is equinumerous to X
and TC(Z) strictly subnumerous to TC(Y), then Y is said to have a
minimal transitive closure, and
The class of all sets equinumerous to X with transitive closures
equinumerous to TC(Y) is said to be the Cardinality of X
Or more generally:
A Cardinal is an equivalence class of sets of minimal transitive
closures, under equivalence relation 'bijection'.
Cardinality(x) is the class of all sets equinumeorus to x, having
minimal transitive closures.
With Choice these cardinals are 'sets' and they are non empty.
Without Choice, I don't have any proof that for every set x,
cardinality(x) is a set; and even if they are sets I don't have a
proof that they will be non empty.
My guess is that assuming Regularity then those classes would be non
empty.
However the proof that those classes are sets, is something that
might need other assumptions independent of ZF,
or might actually be proved in ZF?
However without Regularity I highly doubt there can be minimal
transitive closures of equinumerous sets. However I really don't know
if this can be proved.
One further note, is that this definition as Scott's cardinals, and
as other definitions that I 've presented in this Usenet
of Cardinality, this definition requires
*Strong* Extensionality!
Axiom of Strong Extensionality:
For all z ( (z e x & ~ z=x) iff ( z e y & ~ z=y ) ) -> x=y
So Ur-elements are not premitted here at all, so all sets in these
cardinals do not have any Ur-element, or any set that would violate
strong extensionality in their transitive closures.
Intuitively speaking, if we assume Regularity it seems difficult for
me to imagine that a class of equinumerous sets can have
an infinitely decreasing chain of sizes of their transitive closures,
since we can only have finite membership chains in each set, it
appears to me that this process should come to an end, and we must
have a minimal transitive closure of equinuemrous sets. Thus with
Regularity it seems that this definition work.
However without Regularity I greatly doubt the matter.
However with or without Regularity the sethood of these classes
remains an open question.
Zuhair
zuhair
> Hi all,
>
> Here I shall present another definition of Cardinality:
>
> For any set X, if there exist a set Y that is equinumerous to X
> such that there do not exist a set Z that is equinumerous to X
> and TC(Z) strictly subnumerous to TC(Y), then Y is said to have a
> minimal transitive closure, and
>
> The class of all sets equinumerous to X with transitive closures
> equinumerous to TC(Y) is said to be the Cardinality of X
>
> Or more generally:
>
> A Cardinal is an equivalence class of sets of minimal transitive
> closures, under equivalence relation 'bijection'.
>
> Cardinality(x) is the class of all sets equinumeorus to x, having
> minimal transitive closures.
The strong thing about this definition is that it doesn't require full
choice,
it actually only requires the following axioms:
(1)For every set x there exist a set of all sets hereditarily
subnumerous to x.
were y is hereditarily subnumerous to x is defined as y having every
member
of its transitive closure subnumerous (i.e. injecitve) to x.
(2)The second axiom is the axiom of minimal transitive closure:
For every set x, there exist a set y such that y equinumerous to x
and every set z that is equinumerous to x has its transitive closure
not strictly subnumerous to the transitive closure of y.
With these two axioms added to ZF with strong Extensionality axiom,
then even without Regularity these cardinals can be defined for all
sets in the universe of discourse of that theory.
Zuhair
zuhair
Another idea might be to use the heirarchy of the sets of all
hereditarily sets subnumerous to an ordinal.
It is a theorem of ZF that for every ordinal there exist a set of all
sets hereditarily subnumerous (injective) to it.
So we will have a hierarchy of these sets, each set is transitive and
each contain the other as a subset.
Now I don't know if it is a theorem of ZF that every set must be a
subset of that hierarchy? if so, then we need to prove another
theorem
of ZF that for every set x there should exist a minimal ordinal that
contain a set that is equinumerous to x as a subset of the heirarchy
below it.
If so then we can really have a strong definition of cardinality as
the set of all equinumerous sets of the heirarchy below the minimal
ordinal.
Assuming strong extensionality, I don't know if Regularity would be
needed.
Zuhair
David Libert
> Hi all,
>
> Here I shall present another definition of Cardinality:
>
>
> For any set X, if there exist a set Y that is equinumerous to X
> such that there do not exist a set Z that is equinumerous to X
> and TC(Z) strictly subnumerous to TC(Y), then Y is said to have a
> minimal transitive closure, and
>
>
> The class of all sets equinumerous to X with transitive closures
> equinumerous to TC(Y) is said to be the Cardinality of X
>
>
> Or more generally:
>
>
> A Cardinal is an equivalence class of sets of minimal transitive
> closures, under equivalence relation 'bijection'.
>
>
>
>
> Cardinality(x) is the class of all sets equinumeorus to x, having
> minimal transitive closures.
>
>
> With Choice these cardinals are 'sets' and they are non empty.
Yes.
> Without Choice, I don't have any proof that for every set x,
> cardinality(x) is a set; and even if they are sets I don't have a
> proof that they will be non empty.
I don't know yet about these being sets. I don't have a proof and
I don't have a constructed counterexample model. I would guess it is
probably not a theorem of ZF, if ZF is consistent.
Regarding whether these can be empty. I previously posted something
similar to this, a proof outline, but it skipped lots of details.
That was in
[1] David Libert "Scott like trick? A question"
sci.logic, sci.math Dec 1, 2009
http://groups.google.com/group/sci.logic/msg/796730d70c793d7c
I still think that is probably ok. If so it would show this can
be empty.
> My guess is that assuming Regularity then those classes would be non
> empty.
My proof above, if it is ok.
> However the proof that those classes are sets, is something that
> might
> need other assumptions independent of ZF, or might actually be proved
> in ZF?
I haven't been able to settle this either, as I noted above. Whether
ZF proves it.
> However without Regularity I highly doubt there can be minimal
> transitive closures of equinumerous sets. However I really don't know
> if this can be proved.
My proof above if ok would give an example settling this weaker theory
also as not probving it. Or of yhst proof is ok it can probably be
modified to give models of ~regularity also with he set empty.
> One further note, is that this definition as Scott's cardinals, and
> as
> other definitions that I 've presented in this Usenet of Cardinality,
> this definition requires *Strong* Extensionality!
I think Scott cardinals work fine with the usual axoim of
extensionality. I will also note below, I think your axiom in spite
of its name is actually weaker than the usual axiom of extrensinality.
> Axiom of Strong Extensionality:
>
>
> For all z ( (z e x & ~ z=x) iff ( z e y & ~ z=y ) ) -> x=y
>
>
> So Ur-elements are not premitted here at all, so all sets in these
> cardinals do not have any Ur-element, or any set that would violate
> strong extensionality in their transitive closures.
I think this axiom is equivalent to usual extensionality if there
are no sets which are members of themselves. So in particular,
gregularoty -> this axiom is equvelent to the usual axiom of
extensionality.
If we drop regularity and extensionality from ZF, and so allow sets
to members of themselves, I think the following over that base is an
equivalent desciption of your axiom of strong exrensinality.
Given a set x, possibly with x member x, possibly with
x not member x, define an assymetric pair for x to
be a pair of sets <y, z> such that
x ~= y, x ~= z, y ~= z
and y has members exactly all the members of x and also y
and z has members exactly all the members of x and also y.
We do not claim that a set x must have any assymetric pairs
<y, z>.
Not do we claim if there are assymmetric pairs for a set x, then
they are unique.
We just say if there are one or more pairs <y, z> liie this
we call them an assymetric pair for x.
Similarly for a set x, possibly with x member x, possibly with
x not member x, we define a symmetric pair for x to be a
pair of sets <y, z> such that
x ~= y, x ~= z, y ~= z
and y has members exactly all the members of x and also y and z
and z has members exactly all the members of x and also y and z.
The ordinary axiom of extensionaliy rules our assymetric pairs and
symmetric pairs for x.
Your strong axiom of extensionality allows any sets x to have any combination
in any number of assymmetric and symmetric pairs for x. So some x may have
none. Some x may have only kind and not the other in either direction.
Some x may have both symmetric and assymetric pairs. Abd these can be
in any cardinality, and independly of each other.
Each such pair, each assymetric pair, and each symmetric pair, is
a coutnerexample to ordinary extensionality. But not a counterexample
to your strong extensionality. And these are the only counterexamples
to ordinary extensionality. In other words every pair of sets
violating ordinary extensionality can be realized as an assymetric
pair or a symmetric pair over some set x.
So your axiom of striong extensionality can be restated as
ordinary extensionality except for counterexaples such asymmetric or
symmetric pairs.
So in particular, ordinary extensionality -> your axiom of
so called strong extensionality, namely ordinary exrensionality
is just a special case whrre the counterexample class is empty.
And all these counterexamples involves sets members of themselves,
ie y in assymmetric pairs, and y and z in symmteric pairs.
So as I said above, if no sets are members of themselves,
extensionality ie equivalent to strong extensionality.
> Intuitively speaking, if we assume Regularity it seems difficult for
> me to imagine that a class of equinumerous sets can have
> an infinitely decreasing chain of sizes of their transitive closures,
> since we can only have finite membership chains in each set, it
> appears to me that this process should come to an end, and we must
> have a minimal transitive closure of equinuemrous sets. Thus with
> Regularity it seems that this definition work.
See my proof outline in [1].
> However without Regularity I greatly doubt the matter.
>
>
> However with or without Regularity the sethood of these classes
> remains an open question.
Yes, I still don't know that question either.
> Zuhair
--
David Libert ah...@FreeNet.Carleton.CA
David Libert
> On Dec 8, 4:52=A0pm, zuhair <zaljo...@gmail.com> wrote:
>> Hi all,
>>
>> Here I shall present another definition of Cardinality:
>>
>> For any set X, if there exist a set Y that is equinumerous to X
>> such that there do not exist a set Z that is equinumerous to X
>> and TC(Z) strictly subnumerous to TC(Y), then Y is said to have a
>> minimal transitive closure, and
>>
>> equinumerous to TC(Y) is said to be the Cardinality of X
>>
>> Or more generally:
>>
>> =A0closures, under equivalence relation 'bijection'.
>>
>> =A0Cardinality(x) is the class of all sets equinumeorus to x, having
>> minimal transitive closures.
>
> The strong thing about this definition is that it doesn't require full
> choice,
> it actually only requires the following axioms:
>
> (1)For every set x there exist a set of all sets hereditarily
> subnumerous to x.
>
> were y is hereditarily subnumerous to x is defined as y having every
> member
> of its transitive closure subnumerous (i.e. injecitve) to x.
I don't see why you need axiom 1. Axiom 1 would be relevant to your
previous definitions, involving hereditarily subnumerous to x.
In
[1] David Libert "Z+Size Limitation: Exposition"
sci.logic, sci.math Dev 6, 2009
http://groups.google.com/group/sci.math/msg/c4b2f05cc5a77fb7
I reviewed the first 5 definitions of cardinality.
This new defintion doesn't involve hereditarily subnumerous to x.
You just take Y equinumerous to X with no other cardinality restrictions
on members within TC(Y). Instead your cardinality restiction is the
on the overall #TC(Y), not its parts as the previous definitions.
Aciom 1 seems to relate to the previous definitions, not this new one
above.
> (2)The second axiom is the axiom of minimal transitive closure:
>
> For every set x, there exist a set y such that y equinumerous to x
> and every set z that is equinumerous to x has its transitive closure
> not strictly subnumerous to the transitive closure of y.
Earklier in the thread you raised the possibily that your new
defintion of cardinality could give some sets empty cardinailty.
I posted here that I thought I had a previous proof outline making
a model in which this happened.
So your axiom above is ruling out that case by an axiom. So you
have indeed handled a previously difficult case, and this new
axiom is acordingly relevant to this definition.
> With these two axioms added to ZF with strong Extensionality axiom,
> then even without Regularity these cardinals can be defined for all
> sets in the universe of discourse of that theory.
>
> Zuhair
Earlier in the thread you raised the question of whether the
cardinalities are proper classes. I noted this as an unresolved
question for me too.
I don't think the axioms above have done anything about this
difficult case. So that remains an incompleteness for this
definition, even with these new axioms.
Your article goes on to quote yourself from earlier in the
thread raising this very case I just noted, so I will leave
that here undeleted.
>> With Choice these cardinals are 'sets' and they are non empty.
>>
>> Without Choice, I don't have any proof that for every set x,
>> cardinality(x) is a set; and even if they are sets I don't have a
>> proof that they will be non empty.
--
David Libert ah...@FreeNet.Carleton.CA
zuhair
Unless axiom 1 is a theorem of ZF, which I don't think it is, axiom 1
seems to be indepedent of ZF, so I say unless that, then we must
axiomatize it here, unless you actually have a proof from ZF that
those cardinals are sets, even without this being a theorem of ZF,
then this is another matter.
Axiom 1 is surely related to the definition:
The definition states that the Cardinality of any set x is the class
of all equinumerous sets having minimal transitive closures.
Lemma: Every set x is hereditarily subnumerous to its transitive
closure!
This is a theorem of ZF.
Proof: all members of the transitive closure of x are subsets of the
transitive closure of x, so every one of them is injective(i.e
subnumerous) to the transitive closure of x, so x is hereditarily
subnumerous to its transitive closure.
Now with axiom (2) we can have a minimal transitive closure of
equinumerous sets.
So the Cardinality of x would be the class of all sets equinumerous to
x having minimal transitive closures.
In other words:
Cardinality of x is the class of all sets equinumerous to x, that are
hereditarily subnumerous to the minimal transitive closure.
So here you see that axiom 1 is important!
Because axiom 1 states that the class of all hereditarily subnumerous
to x sets is a set.
Since for every set x, TC(x) is a set.
So Cardinality(x) would be a subset of the class of all sets
hereditarily subnumerous to the minimal transitive closure of sets
equinumerous to x.
So Cardinality(x) is a set if x is a set.
> David Libert ah...@FreeNet.Carleton.CA- Hide quoted text -
>
> - Show quoted text -
zuhair
Correction:
Cardinality of x is a subset of the class of all sets equinumerous to
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>
> - Show quoted text -- Hide quoted text -
zuhair
So do strong Extensionality.
You said:
Given a set x, possibly with x member x, possibly with
x not member x, define an assymetric pair for x to
be a pair of sets <y, z> such that
x ~= y, x ~= z, y ~= z
and y has members exactly all the members of x and also y
If so then, then y=x according to strong extensionality,because both
have
the same proper members!
and z has members exactly all the members of x and also
y.
so z=y=x.
We do not claim that a set x must have any assymetric pairs
<y, z>.
there doesn't.
>
> Your strong axiom of extensionality allows any sets x to have any combination
> in any number of assymmetric and symmetric pairs for x.
How ??? it actually shuns them all.
So some x may have
> none. Some x may have only kind and not the other in either direction.
> Some x may have both symmetric and assymetric pairs. Abd these can be
> in any cardinality, and independly of each other.
All of these violate Stronge extensionality clearly.
>
> Each such pair, each assymetric pair, and each symmetric pair, is
> a coutnerexample to ordinary extensionality.
And stronge extensionality also.
But not a counterexample
> to your strong extensionality.
?????
And these are the only counterexamples
> to ordinary extensionality. In other words every pair of sets
> violating ordinary extensionality can be realized as an assymetric
> pair or a symmetric pair over some set x.
>
> So your axiom of striong extensionality can be restated as
> ordinary extensionality except for counterexaples such asymmetric or
> symmetric pairs.
>
> So in particular, ordinary extensionality -> your axiom of
> so called strong extensionality, namely ordinary exrensionality
> is just a special case whrre the counterexample class is empty.
>
> And all these counterexamples involves sets members of themselves,
> ie y in assymmetric pairs, and y and z in symmteric pairs.
>
> So as I said above, if no sets are members of themselves,
> extensionality ie equivalent to strong extensionality.
Yes, that is true.
What stronge Extensionality does, is that it forbides have two sets x
and y
such that ~ x = y and having the same proper members.
Suppose ~ x e x then all members of x are proper members
Now Suppose that y e y
Now if all proper members of y (i.e. members of y other than y itself)
are members of x, and there do not exist any proper member of y
that is not in x, then y=x according to Stonge Extensionality, because
y and x would have the same proper members.
Now suppose that ~ y e y, then we have the ordinary case
all members of y would be proper and we have ordinary extensionality
and x=y if they have the same members.
Suppose x e x
Suppose y e y
Now same thing happen x cannot be different from y and both x and y
having
the same proper members.
Suppose x e x
Suppose ~ y e y
we'll have the example above.
So the examples of pairs you gave is erronous.
Stronge Extensionality is stonge Than ordinary Extensionality.
With Ordinary Extensionality if we drop Regularity then we can have
two sets x and y that are not identical but still having the same
proper members
so we can have for example x={m,n,q,...,x} and y={m,n,q,...,y}
and still ~y=x, this is the case with ordinary extensionality if we
drop Regularity.
However with Stronge Extensionality we cannot have that, since x must
be identical to y if both have the same proper members.
Zuhair
>
> > Intuitively speaking, if we assume Regularity it seems difficult for
> > me to imagine that a class of equinumerous sets can have
> > an infinitely decreasing chain of sizes of their transitive closures,
> > since we can only have finite membership chains in each set, it
> > appears to me that this process should come to an end, and we must
> > have a minimal transitive closure of equinuemrous sets. Thus with
> > Regularity it seems that this definition work.
>
> See my proof outline in [1].
>
> > However without Regularity I greatly doubt the matter.
>
> > However with or without Regularity the sethood of these classes
> > remains an open question.
>
> Yes, I still don't know that question either.
>
> > Zuhair
>
> --
Don Stockbauer
1. Potential
2. Actualized.
zuhair
No that is incorrect.
>
> So in particular, ordinary extensionality -> your axiom of
> so called strong extensionality, namely ordinary exrensionality
> is just a special case whrre the counterexample class is empty.
IF we drop Regularity then with the ordinary axiom of Extensionality
we can
axiomatize the existence of a proper class of Quine atoms, this is
easy
Axiom: for all x ( not exist m ( m e TC(x) & m={m} ) ->
exist y (y equinumerous to x & for all z (z in y -> z ={z}))
I think this would be consistent with ZF minus Regularity.
anyhow, we actually can axiomatize for example
Axiom: Exist x,y (x={x} & y={y} & ~x=y)
This is definitely not contradictive with ZF minus Regularity.
So you see ordinary extensionality cannot shun the exitence of such
sets.
However with Stronge Extensionality you cannot have the above axiom,
since this would lead to a contradiction of ~x=x & ~y=y; since every
proper member of x is a proper member of y! isn't it?!
You can have at most one quine atom with stronge Extensionality, while
you can have a proper class of them with the ordinary Extensionaity.
Zuhair
>
> And all these counterexamples involves sets members of themselves,
> ie y in assymmetric pairs, and y and z in symmteric pairs.
>
> So as I said above, if no sets are members of themselves,
> extensionality ie equivalent to strong extensionality.
>
> > Intuitively speaking, if we assume Regularity it seems difficult for
> > me to imagine that a class of equinumerous sets can have
> > an infinitely decreasing chain of sizes of their transitive closures,
> > since we can only have finite membership chains in each set, it
> > appears to me that this process should come to an end, and we must
> > have a minimal transitive closure of equinuemrous sets. Thus with
> > Regularity it seems that this definition work.
>
> See my proof outline in [1].
>
> > However without Regularity I greatly doubt the matter.
>
> > However with or without Regularity the sethood of these classes
> > remains an open question.
>
> Yes, I still don't know that question either.
>
> > Zuhair
>
> --
Don Stockbauer
1. Potential
2. Actualized.
Oh, I'm sorry, ignore that, yall like infinite debates. Literally.
zuhair
Correction: if all proper members of y are members of x, and all
members of x (all are proper because ~ x e x here) are proper members
of y, then x=y.
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Marshall
>
> 1. Potential
>
> 2. Actualized.
1. You
2. Suck.
Marshall