Axiom of infinity and the set of all hereditary finite sets.
Zalj...@gmail.com
I have two questions:
1) Can every set have a transitive closure in ZFC-Infinity?
The transitive closure of a set x is defined as
Tc(x)= U{x,Ux,UUx,UUUx,........}
I ask this question because from Regularity the transitive closure of
any set would always be finite. i.e we have
Ax: Tc(x) is finite
as a theorem of ZFC.
So it appears as if the theorem
AxEy( y=Tc(x) )
doesn't depend on axiom of infinity .
I know that AxEy( y=Tc(x) ) is a theorem of ZFC, but I don't know if
the proof of this theorem requires axiom of infinity.
2) Is ExAy ( yex<-> y is hereditary finite ) is a theorem of ZFC-
infinity?
In a previous topic I defined Hyperfinite sets as
Define: x is hyperfinite<->(x is finite & Ay( yeTc(x) -> y is
finite)).
and I came to know that this is equivalent to the concept of
hereditary finite sets.
and I also came to know that the set of all hereditary finite sets is
a set, i.e ExAy( yex<-> y is hereditary finite ) is a theorem of ZFC.
So my question is: does the proof of this theorem requires axiom of
infinity.
The reason beyond these two questions is that:
if ExAy ( yex<-> y is hereditary finite ) is a theorem of ZFC-
infinity.
Then there is no need for axiom of infinity.
Because if so then from Separation we can prove the following schema
to be a theorem schema of ZFC-infinity
Schema of hereditary finiteness:If F is a formula in which x is not
free, then all closures of
ExAy ( yex<->( y is hereditary finite & F(y) ) )
are theorems.
Now we can easily prove the existence of an infinite set, because the
set of all hereditary finite sets should be infinite, otherwise
regularity would be violated. The existence of Omega can simply follow
from the above schema by having F as
F(y)<-> y is a natural number.
And thus axiom of infinity would be a theorem of ZFC-Infinity, and
thus redundant.
Zuhair
Aatu Koskensilta
> 1) Can every set have a transitive closure in ZFC-Infinity?
It is provable in ZFC without infinity that every set has a transitive
closure.
> 2) Is ExAy ( yex<-> y is hereditary finite ) is a theorem of ZFC-
> infinity?
No.
> And thus axiom of infinity would be a theorem of ZFC-Infinity, and
> thus redundant.
The existence of the set of hereditarily finite sets is not provable
in ZFC without infinity.
--
Aatu Koskensilta (aatu.kos...@xortec.fi)
"Wovon man nicht sprechen kann, daruber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
Zalj...@gmail.com
> Hi all,
>
> I have two questions:
>
> 1) Can every set have a transitive closure in ZFC-Infinity?
>
> The transitive closure of a set x is defined as
>
> Tc(x)= U{x,Ux,UUx,UUUx,........}
>
> I ask this question because from Regularity the transitive closure of
> any set would always be finite. i.e we have
>
> Ax: Tc(x) is finite
>
> as a theorem of ZFC.
Correction : the above is not a theorem of ZFC
What I was actually talking about is the set
{x,Ux,UUx,UUUx,...........} is finite for every x , i.e
Ax:{x,Ux,UUx,UUUx,...........} is finite
Aatu Koskensilta
> What I was actually talking about is the set
>
> {x,Ux,UUx,UUUx,...........} is finite for every x , i.e
>
> Ax:{x,Ux,UUx,UUUx,...........} is finite
>
> is a theorem of ZFC.
It's not.
Zalj...@gmail.com
wrote:
> Zaljo...@gmail.com writes:
> > What I was actually talking about is the set
>
> > {x,Ux,UUx,UUUx,...........} is finite for every x , i.e
>
> > Ax:{x,Ux,UUx,UUUx,...........} is finite
>
> > is a theorem of ZFC.
>
> It's not.
>
How come? if it is not a theorem of ZFC, then this mean that we can
have infinite descending membership of x which is impossible according
to Regularity.
So the set {x,Ux,UUx,UUUx,.........} MUST be finite.
and it should be a theorem of ZFC.
Can you give me an example of a set x such that the set
{x,Ux,UUx,UUUx,.....} is infinite.
Zuhair
> --
> Aatu Koskensilta (aatu.koskensi...@xortec.fi)
Aatu Koskensilta
> Can you give me an example of a set x such that the set
> {x,Ux,UUx,UUUx,.....} is infinite.
Yes: the set of finite von Neumann ordinals, omega. Incidentally,
{x,Ux,UUx,UUUx,.....} is not the transitive closure of x -- its union
is.
Why do you think the existence of sets with infinite transitive
closure, or, equivalently, the existence of infinite transitive sets,
contradicts regularity?
--
Aatu Koskensilta (aatu.kos...@xortec.fi)
Ross A. Finlayson
wrote:
> Zaljo...@gmail.com writes:
> > Can you give me an example of a set x such that the set
> > {x,Ux,UUx,UUUx,.....} is infinite.
>
> Yes: the set of finite von Neumann ordinals, omega. Incidentally,
> {x,Ux,UUx,UUUx,.....} is not the transitive closure of x -- its union
> is.
>
> Why do you think the existence of sets with infinite transitive
> closure, or, equivalently, the existence of infinite transitive sets,
> contradicts regularity?
>
> --
>
> "Wovon man nicht sprechen kann, daruber muss man schweigen"
> - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
I think it does, for a variety of reasons.
The set of all hereditarily finite sets, comprised only of
hereditarily finite sets, would be a hereditarily finite set, and thus
contain itself, similarly to how the set of all ordinals would be an
ordinal and contain itself. (?)
(Consider Goedelian incompleteness interpreted as that there are no
absolute truths about the natural integers. Is that not an absolute
truth about the natural integers?)
In any theory containing all the sets of ZF(C) and any set containing
itself, eg itself, the universe of ZF would be the Russell set and
contain itself.
If there's a countable model (universe) of ZF then even the "least"
infinite set, one corresponding to the natural integers, would be non-
well-founded.
Infinite sets are irregular.
Ross
--
Finlayson Consulting
Aatu Koskensilta
> The set of all hereditarily finite sets, comprised only of
> hereditarily finite sets, would be a hereditarily finite set, and thus
> contain itself, similarly to how the set of all ordinals would be an
> ordinal and contain itself. (?)
The set of hereditarily finite sets is not a hereditarily finite set.
> (Consider Goedelian incompleteness interpreted as that there are no
> absolute truths about the natural integers. Is that not an absolute
> truth about the natural integers?)
One might just as well consider Gödelian incompleteness interpreted as
the claim that bananas are tastier than broccoli.
--
Aatu Koskensilta (aatu.kos...@xortec.fi)
Zalj...@gmail.com
wrote:
> Zaljo...@gmail.com writes:
> > Can you give me an example of a set x such that the set
> > {x,Ux,UUx,UUUx,.....} is infinite.
>
> Yes: the set of finite von Neumann ordinals, omega.
How come?
I am not understanding your thinking.
You are saying that Omega 'w' has the set {w, Uw, UUw,UUUw,....}
as an infinite set, how is that, I am really losing you here.
>From what I knew is that Uw=w, and thus UUw=w and UUUw=w
and so for the case of w we have { w,Uw,UUw,UUUw,....} ={w} which is a
singlton set
and it is finite.
Incidentally,
> {x,Ux,UUx,UUUx,.....} is not the transitive closure of x -- its union
> is.
I know that {x,Ux,UUx,UUUx,...} is not the transitive closure,
and I am refering to that set and not the transitive closure, I have
already mentioned
that in my correction.
>
> Why do you think the existence of sets with infinite transitive
> closure, or, equivalently, the existence of infinite transitive sets,
> contradicts regularity?
I didn't say that. You are not concentrating on what I am saying
What I am saying is that
Ax: {x,Ux,UUx,UUUx,...} is finite
is a theorem of ZFC, and it is.
I am not saying that Ax: Tc(x) is finite.
I already mentioned that.
>
> --
> Aatu Koskensilta (aatu.koskensi...@xortec.fi)
Aatu Koskensilta
> I am not understanding your thinking.
It seems I misunderstood your question.
--
Aatu Koskensilta (aatu.kos...@xortec.fi)
Zalj...@gmail.com
wrote:
> Zaljo...@gmail.com writes:
> > Can you give me an example of a set x such that the set
> > {x,Ux,UUx,UUUx,.....} is infinite.
>
> Yes: the set of finite von Neumann ordinals, omega. Incidentally,
> {x,Ux,UUx,UUUx,.....} is not the transitive closure of x -- its union
> is.
>
> Why do you think the existence of sets with infinite transitive
> closure, or, equivalently, the existence of infinite transitive sets,
> contradicts regularity?
As I mentioned. what I am saying is that for every set x
the set {x,Ux,UUx,UUUx,...} is finite.
The reason is because if the later set is not finite, then
this mean that for some x there is an infinite descending membership
of x
and this clearly violates Regularity.
Zuhair
>
> --
> Aatu Koskensilta (aatu.koskensi...@xortec.fi)
>
Ross A. Finlayson
wrote:
> "Ross A. Finlayson" <r...@tiki-lounge.com> writes:
>
> > The set of all hereditarily finite sets, comprised only of
> > hereditarily finite sets, would be a hereditarily finite set, and thus
> > contain itself, similarly to how the set of all ordinals would be an
> > ordinal and contain itself. (?)
>
> The set of hereditarily finite sets is not a hereditarily finite set.
>
It's comprised only of hereditarily finite sets, else it's not....
> > (Consider Goedelian incompleteness interpreted as that there are no
> > absolute truths about the natural integers. Is that not an absolute
> > truth about the natural integers?)
>
> One might just as well consider Gödelian incompleteness interpreted as
> the claim that bananas are tastier than broccoli.
>
Opinion?
Ross
Zalj...@gmail.com
>
> > The set of all hereditarily finite sets, comprised only of
> > hereditarily finite sets, would be a hereditarily finite set, and thus
> > contain itself, similarly to how the set of all ordinals would be an
> > ordinal and contain itself. (?)
>
> The set of hereditarily finite sets is not a hereditarily finite set.
I agree.
The set of all hereditarily finite sets is an infinite set.
Since if it is finite, then it would be hereditarily finite
and it would contain itself as a member and this would
violate regularity, and thus by negation it is not finite.
Now the set which has the set of all hereditarily finite sets
as its member is finite but not hereditarily finite, thus
it is not a member of the set of all hereditarily finite sets
thus it doesn't suffer the same problem as that of the class of all
finite sets sufferes in ZFC ( this later class is not a set in ZFC).
For more information see my new alternative axiomatization of ZFC
that I have posted in anther topic in this usenet.
http://groups.google.com/group/sci.logic/browse_frm/thread/7ec7ea91363316bf
This system of axiomatization proves the existence of the set of all
hereditary finite sets, as an infinite set, and it replaces the axiom
of infinity, as well as the axioms of pairing, union and power
also it replaces the schemas of separation and replacement.
Zuhair
>
> > (Consider Goedelian incompleteness interpreted as that there are no
> > absolute truths about the natural integers. Is that not an absolute
> > truth about the natural integers?)
>
> One might just as well consider Gödelian incompleteness interpreted as
> the claim that bananas are tastier than broccoli.
>
> --
> Aatu Koskensilta (aatu.koskensi...@xortec.fi)
Jesse F. Hughes
> On Sep 29, 12:58 pm, Aatu Koskensilta <aatu.koskensi...@xortec.fi>
> wrote:
>> "Ross A. Finlayson" <r...@tiki-lounge.com> writes:
>>
>> > The set of all hereditarily finite sets, comprised only of
>> > hereditarily finite sets, would be a hereditarily finite set, and thus
>> > contain itself, similarly to how the set of all ordinals would be an
>> > ordinal and contain itself. (?)
>>
>> The set of hereditarily finite sets is not a hereditarily finite set.
>>
>
> It's comprised only of hereditarily finite sets, else it's not....
It's not a finite set, and hence it is not a hereditarily finite set.
--
"The math doesn't care about their mortgages. It doesn't care about
their political needs. [...] Today's mathematicians have to hate
mathematics because mathematics doesn't look out for them. It doesn't
pay attention to their needs." --- JSH analyzes mathematicians
Ross A. Finlayson
> "Ross A. Finlayson" <r...@tiki-lounge.com> writes:
>
> > On Sep 29, 12:58 pm, Aatu Koskensilta <aatu.koskensi...@xortec.fi>
> > wrote:
> >> "Ross A. Finlayson" <r...@tiki-lounge.com> writes:
>
> >> > The set of all hereditarily finite sets, comprised only of
> >> > hereditarily finite sets, would be a hereditarily finite set, and thus
> >> > contain itself, similarly to how the set of all ordinals would be an
> >> > ordinal and contain itself. (?)
>
> >> The set of hereditarily finite sets is not a hereditarily finite set.
>
> > It's comprised only of hereditarily finite sets, else it's not....
>
> It's not a finite set, and hence it is not a hereditarily finite set.
>
> --
> "The math doesn't care about their mortgages. It doesn't care about
> their political needs. [...] Today's mathematicians have to hate
> mathematics because mathematics doesn't look out for them. It doesn't
> pay attention to their needs." --- JSH analyzes mathematicians
You see, there's a difference between "for each", "for any", "for
every" and "for all."
Jesse, I agree with you and Zuhair and Aatu. The set of all
hereditarily finite sets is an infinite set.
Ross
--
Finlayson Consulting
Nam D. Nguyen
> "Ross A. Finlayson" <r...@tiki-lounge.com> writes:
>> (Consider Goedelian incompleteness interpreted as that there are no
>> absolute truths about the natural integers. Is that not an absolute
>> truth about the natural integers?)
>
> One might just as well consider Gödelian incompleteness interpreted as
> the claim that bananas are tastier than broccoli.
Are you implying that truths about the natural integers are absolute?
David C. Ullrich
>On Sep 29, 10:57 am, Aatu Koskensilta <aatu.koskensi...@xortec.fi>
>wrote:
>> Zaljo...@gmail.com writes:
>> > Can you give me an example of a set x such that the set
>> > {x,Ux,UUx,UUUx,.....} is infinite.
>>[...]
For example, let x be the set that one
might informally define by
x = {{1}, {{2}}, {{{3}}}, ...}.
For each natural number n, n is an element of U^n x but
not an element of U^k x for any k < n.
(for example, 3 is in UUUx but not in x, Ux, or UUx. Hence
UUUx is not equal to any of x, Ux or UUx. Etc.)
>The reason is because if the later set is not finite, then
>this mean that for some x there is an infinite descending membership
>of x
Oh? How do you prove that?
>and this clearly violates Regularity.
>
>Zuhair
>>
>> --
>> Aatu Koskensilta (aatu.koskensi...@xortec.fi)
>>
>> "Wovon man nicht sprechen kann, daruber muss man schweigen"
>> - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
>
************************
David C. Ullrich
Aatu Koskensilta
> Are you implying that truths about the natural integers are absolute?
What does it mean for truths about the naturals to be absolute or fail
to be absolute?
Zalj...@gmail.com
> On Sat, 29 Sep 2007 13:16:43 -0700, Zaljo...@gmail.com wrote:
> >On Sep 29, 10:57 am, Aatu Koskensilta <aatu.koskensi...@xortec.fi>
> >wrote:
> >> Zaljo...@gmail.com writes:
> >> > Can you give me an example of a set x such that the set
> >> > {x,Ux,UUx,UUUx,.....} is infinite.
> >>[...]
>
> For example, let x be the set that one
> might informally define by
>
> x = {{1}, {{2}}, {{{3}}}, ...}.
>
> For each natural number n, n is an element of U^n x but
> not an element of U^k x for any k < n.
>
> (for example, 3 is in UUUx but not in x, Ux, or UUx. Hence
> UUUx is not equal to any of x, Ux or UUx. Etc.)
Let's see:
U1x= {1,{2},{{3}},{{{4}}},...........}
U2x= {2,{3},{{4}},.....}
U3x= {3,{4},{{5}},.....}
So the i-th Union of x always starts with member i.
so the i-th Union of x is never empty. Yes, I agree
this would be an example of a set x such that
{x,Ux,UUx,UUUx,....} is infinite.
Thank you.
Zuhair
Nam D. Nguyen
> "Nam D. Nguyen" <namduc...@shaw.ca> writes:
>
>> Are you implying that truths about the natural integers are absolute?
>
> What does it mean for truths about the naturals to be absolute or fail
> to be absolute?
>
Within the context of FOL=, a modal truth is called an absolute truth one
if its formula is a logical theorem (i.e. a theorem whose axioms
of the shortest proof are logical). In the same context, a modal truth
that's not absolute is called relative. For example, in the context of
the naturals, 0=0 is an absolute truth, while Ax[~(S(x)=0)] is relative.
The intuition of this definition of absoluteness/relativity is that
some truths are invariant *over the existences* of the individuals in the
set comprising models of theories of the same language. Those invariant
truths are absolute. Otherwise, the truths would be called relative.
Jesse F. Hughes
Then you agree that it is not hereditarily finite. By definition, a
set is hereditarily finite iff *it is finite* and each set in it is
hereditarily finite.
--
Jesse F. Hughes
"To all Leaders of the World, buy or rent the movie 'The Day
After'[...] I assure you will have a new perspective on WMDs."
-- practical advice from online petitions
aatu.kos...@xortec.fi
> Within the context of FOL=, a modal truth is called an absolute truth one
> if its formula is a logical theorem (i.e. a theorem whose axioms
> of the shortest proof are logical).
In this somewhat odd sense truths about naturals obviously aren't in
general absolute. How is this an interpretation of Gödelian
incompleteness? On the face of it it has as much to do with the
incompleteness theorems as the tastiness of bananas.
Zalj...@gmail.com
> "Ross A. Finlayson" <r...@tiki-lounge.com> writes:
>
>
>
>
>
> > On Sep 29, 6:23 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> >> "Ross A. Finlayson" <r...@tiki-lounge.com> writes:
>
> >> > It's comprised only of hereditarily finite sets, else it's not....
>
> >> It's not a finite set, and hence it is not a hereditarily finite set.
>
> >> --
> >> "The math doesn't care about their mortgages. It doesn't care about
> >> their political needs. [...] Today's mathematicians have to hate
> >> mathematics because mathematics doesn't look out for them. It doesn't
> >> pay attention to their needs." --- JSH analyzes mathematicians
>
> > You see, there's a difference between "for each", "for any", "for
> > every" and "for all."
>
> > Jesse, I agree with you and Zuhair and Aatu. The set of all
> > hereditarily finite sets is an infinite set.
>
> Then you agree that it is not hereditarily finite. By definition, a
> set is hereditarily finite iff *it is finite* and each set in it is
> hereditarily finite.
This is a clear circular (cyclical) definition! since it defines
hereditarily infinite in terms of itself.
The correct definition is that:
A set is hereditarily finite if *it is finite* and each set in its
transitive closure is finite.
x is hereditarily finite <-> ( x is finite & Ay ( yeTc(x) -> y is
finite ).
were Tc(x) = U { x,Ux,UUx,UUUx,........ }
I also have another definition for Tc(x) that I've already mentioned
in another post ( but I am not so sure weather it is equivalent to the
standard one I've just mentioned above ) and this is
Definition: y=Tc(x) <-> ( Am(mex->mey) & y is transitive &
Am( (mey & ~mex) -> Ez( zey & mez ) ) ).
Definition: y is transitive <-> Anm( (nem&mey)->ney)
I would really like to know if this definition is equivalent to the
standard one according to ZFC.
The reason why the set of all hereditarily finite sets is not finite
is that
Any hereditarily finite set is a well founded set, so the set of all
these sets should be well founded, now if this set is finite, then it
would be hereditarily finite since every member in its transitive
closure is finite
and by then it should be in itself and thus not well founded: A
contradiction.
Thus the set of all hereditarily finite sets should be *infinite*.
If we accept regularity then a simpler proof is that:
if the set of all hereditarily finite sets is finite, then it is
hereditarily finite
then it would be in itself by definition, thus violating regularity.
Zuhair
>
> --
> Jesse F. Hughes
> "To all Leaders of the World, buy or rent the movie 'The Day
> After'[...] I assure you will have a new perspective on WMDs."
> -- practical advice from online petitions- Hide quoted text -
>
> - Show quoted text -
Nam D. Nguyen
> Nam D. Nguyen wrote:
>> Within the context of FOL=, a modal truth is called an absolute truth one
>> if its formula is a logical theorem (i.e. a theorem whose axioms
>> of the shortest proof are logical).
>
> In this somewhat odd sense truths about naturals obviously aren't in
> general absolute.
Would you care to discuss - in some details - why you think it's "odd"?
> How is this an interpretation of Gödelian
> incompleteness? On the face of it it has as much to do with the
> incompleteness theorems as the tastiness of bananas.
GIT is a meta theorem and I guess one could interpret a meta theorem
any which way one desires: it's "useful", "philosophically wrong", etc...
I wasn't really after any debate about interpretation of any meta statement,
per se. My original question to you:
"Are you implying that truths about the natural integers are absolute?"
is basically a question about what you might think the absoluteness of the
truths of the naturals be? If you believe truths about naturals are "absolute",
then what is your definition of "absolute" that everybody else _ought to_
stick with ?
Jesse F. Hughes
> On Sep 30, 9:44 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>>
>> Then you agree that it is not hereditarily finite. By definition, a
>> set is hereditarily finite iff *it is finite* and each set in it is
>> hereditarily finite.
>
> This is a clear circular (cyclical) definition! since it defines
> hereditarily infinite in terms of itself.
No, I was simply being too curt. What I wrote *follows* from the
definition, but it is not (usually taken as) the definition. That is
what I meant when I wrote "by definition", but obviously that phrase
is usually used to introduce the definition.
Sorry for the misleading text.
--
"Do you know why I'm tall?" "Why?"
"Because I eat apples." "Do you know why I'm short?"
"Why?" "Because I'm a kid."
--Quincy P. Hughes (age almost 4) bests his father.
David Libert
wrote:
> Zaljo...@gmail.com writes:
> > 1) Can every set have a transitive closure in ZFC-Infinity?
>
> It is provable in ZFC without infinity that every set has a transitive
> closure.
I don't think that is provable. Here is my proposed construction of
a counter model, if its ok:
Work in meta-theory ZFC. In there I will attempt to construct a
countable model of ZFC without infinity + there exists a set
without a transitive closure.
So basically, I am going to start with some atoms, and I will
declare explicit "membership" relations among them, so the constructed
structure will view them as sets. By using atoms with only
"membership" declared by fiat I can put strange structure into this
starting membership relation. Then I am going to close off the
structure by adding certain sets above the atoms, to get the other
axioms of ZFC - infiinity. I will add sets which do not correspond to
the previous ones created as atoms.
So to start constructing the model with the atoms. The starting
elements will be set of the form
<0, n> for n in Z = the usual integers (positive, negative
and 0). The <0, _> part is to keep track of atoms versus sets.
I will declare the membership relation E among these elements,
which will be the final interpretation of epsilon in the constructed
structure. Namely each <0, n+1> will become {<0, n>}.
So I define E among these elements by <0, n> E <0, m> <-
> m = n+1 for n,m in Z .
So this set of elements with its E is the 0th stage approximation to
the constructed structure.
I grow this structure in omega stages after the 0'th by at stage
k+1 for k in omega adding in all elements of the form <1, F>
where F is a finite subset of the k'th stage, and F is not of
the form {<0, n>} for any n in Z.
So each k+1 'st stage consists of the <0,n> 's and the <1, F>
's added in stages 1...k+1 .
At stage k+1 we are unioning some <1, F> that were added at
earlier positive stages for F small enough sets.
I extend the definition of E to membership for new elements;
<i, x> E <1, F> <-> <i,x> member (real working
universe) of F
for <i,x> member of the structure ( so i = 0 and x in Z
or i = 1 and x a finite non-singleton
atom subset of earlier stages.
The final structure is the union over omega of all finite stages
with the inherited E.
I claim this satisfies extensionality. Each atom has distinct
members from each other atom. There is no atom coextensive with and
<1, F>, since we kept singletons of atoms out of the F's. And the
F's are distinct among themselves.
This satisfies pairing separation, union, replacement. All sets are
finite so the replacement axiom style replacement of a finite set got
thrown in.
The tricky one is regularity, which might look dubious since I put
an epsilon descending chain of E membership into the structure.
So define a rank function on the structure (in model theory outside
the structure). Namely,
<0, n> has rank n (a member of Z). Abd define <1, F> has rank
1 + the maximum rank of the E members of <1, F> . (Finitely many E
members, so can take maximum).
So overall, these ranks are in Z and are not well-founded.
But given any element of the structure, it has only finitely many E
members. So even though ranks over the entire structure are not well-
founded, the ranks appearing inside any particular element are only
finitely many and so have a smallest rank. Then this element is
disjoint for the starting element, as regularity requires.
AC: all the sets are finite, so get choice functions.
So we have ZFC - inifinity.
If <0, 0> in the structure had a transitive closure every <0,
n> for n < 0 would have to be in it, so it would be an infinite
set, contrary to each set being finite.
So <0, 0> has no transitive closure.
So that completes the construction, if I haven't made a mistake.
basically, you could say we made a weird model by putting in non-
well-founded membership that seems odd for regularity, but because the
model also lacked infiniity it was not able to recognize its own
weirdness to negate regulariy.
Earlier, inspried some previous discussion in the math or logic
geroups, I thought of similar intersting constructions of models of
ZC (leave off replacement). Similar thing, put strange non-weel-
founded structure into the model. And it turns out you still get
regularity because the ways to try to contradict it depend on making
sets with replacement which you are missing.
Or you can also make models like this of ZF, have inifiinty and
replacement, but ~AC stops you from amalgating non-well-foundedness
into an actual set to violate regularity.
So in all these cases you can make unusal models and regularity is
sort of a cheat, because something else missing from the model makles
the model unable to see its own non-well-foundedness.
--
David Libert