Definition of finite.
zuhair
I think the best definition of 'finite' would be the following:
x is finite <-> Ayz((y subset_of x & ~Em(mey & mez))->
yUz supernumerous to y)).
r supernumerous to s <-> Af((f:s->r,f is injective)->~ f is
surjective).
This is better than both Dedekind's and Tarsky's definitions.
x is infinite <-> ~( x is finite).
Best,
Zuhair
zuhair
> Hi all,
>
> I think the best definition of 'finite' would be the following:
>
> x is finite <-> Ayz((y subset_of x & ~Em(mey & mez))->
> yUz supernumerous to y)).
Correction:
x is finite <-> Ayz((y subset_of x & ~Em(mey & mez))->
yUz supernumerous to y).
Jesse F. Hughes
> Hi all,
>
> I think the best definition of 'finite' would be the following:
>
> x is finite <-> Ayz((y subset_of x & ~Em(mey & mez))->
> yUz supernumerous to y)).
>
> r supernumerous to s <-> Af((f:s->r,f is injective)->~ f is
> surjective).
>
> This is better than both Dedekind's and Tarsky's definitions.
Right. No doubt. A *lot* clearer!
Congrats! This will get you in the textbooks if anything will!
> x is infinite <-> ~( x is finite).
>
> Best,
>
> Zuhair
>
--
Jesse F. Hughes
"Every country has its stupid people. It just so happens that
America's stupid people are louder." -- Ling Cheung, Sociologist
thero...@yahoo.com
> I think the best definition of 'finite' would be the following:
> This is better than both Dedekind's and Tarsky's definitions.
Step 1. Define the binary relation "is better than" applied to the set
FinDefs of all definitions of "finite".
Step 2. Prove that "is better than" is irreflexive, asymmetric,
transitive and satisfies the Trichotomy Law. In other words: prove
that it is a strict total order.
Step 3. Prove that "is better than" orders the set FinDefs in such a
way that there exists a greatest element.
Step 4. Define the best definition of "finite" to be the greatest
element from Step 3.
Step 5. Try to prove that "is better than" is unique - there is no
other strict total order on FinDefs.
Step 6. If you think you have a proof for Step 5, contact the local
asylum to check your proof attempt.
Perhaps you will need Step 0 to define "definition of X" in such a way
that it is a set and FinDefs exists. If you do this, maybe you can try
to prove that your definition of "definition of X" is the best using
the same step-by-step instructions but with the set of all definitions
of "definition of X" instead of FinDefs.
Theron
zuhair
.
>
> Right. No doubt. A *lot* clearer!
>
> Congrats! This will get you in the textbooks if anything will!
I don't care about this issue, neither did I think of it at all.
in addition the definition I gave was erronous I will correct it to:
x is finite <-> Ayz((y subset_of x & ~Em(mey & mez) &~z=y)->yUz
supernumerous to y)).
were
r supernumerous to s <->( Ef(f:s->r, f is injective) & Af((f:s->r,f is
injective)
-> ~ f is surjective) ).
This definition works with choice or without choice ( thus better than
Dedekinds' which only works with choice), and this definition is not
essentially circular (cyclical) , while Tarsky's definition is
essentially circular.
Zuhair
Virgil
thero...@yahoo.com wrote:
> On Apr 29, 5:40 am, zuhair <zaljo...@yahoo.com> wrote:
> > I think the best definition of 'finite' would be the following:
> > ...
> > This is better than both Dedekind's and Tarsky's definitions.
>
> Step 1. Define the binary relation "is better than" applied to the set
> FinDefs of all definitions of "finite".
>
> Step 2. Prove that "is better than" is irreflexive, asymmetric,
> transitive and satisfies the Trichotomy Law. In other words: prove
> that it is a strict total order.
Actually, it doesn't need to be a total order, a partial order will do
as long as every set of definitions has an upper bound in the set of all
such definitions.
zuhair
> On Apr 28, 10:41 pm, "Jesse F. Hughes" <j...@phiwumbda.org> wrote:
> .
>
>
>
> > Right. No doubt. A *lot* clearer!
Although I think the definition I made (and Corrected) is pretty
clear, yet I will try to clarify it.
Define x disjoint y <-> ~Em(mex & mey)
Define x subset_of y <-> Az(zex->zey)
x supernumerous to y is defined in the correction.
Define x is increamental <-> Az( (z disjoint x & ~z=x) -> zUx
supernumerous to x).
In words any set is increamental if and only if the set union of it
with any disjoint set from it that is different from it IS
supernumerous to it.
Now:
x is finite <-> Ay( y subset _of x -> y is increamental).
In words any set is said to be finite if and only if any subset of it
IS increamental.
Don't tell me that is not clear.
Not only that, we can proceed and define the set of all finite
ordinals.
x is transitive <-> Ayz( (zey & yex) -> zex ).
x is regular <-> ( Ez(zex) ->Ey(yex & y disjoint x) )
x is ordinal <-> (x is regular & x is transitive & Ay(yex -> y is
transitive)).
x is a finite ordinal <-> ( x is ordinal & x is finite ).
Now we can axiomatize existence of the set of all finite ordinals as
Axiom of infinity: ExAy(yex<-> y is a finite ordinal )
Theorem E!xAy(yex<-> y is a finite ordinal )
Proof: Extensionality.
Define x=N<->Ay(yex<-> y is a finite ordinal )
This way of defining N is better than the official one present
in ZF and others, although it is a longer way.
The peano-like way of defining N present in these set theories
makes no sense, though it defines N in a shorter way.
Zuhair
zuhair
I forgot to mention , that this axiom works in a set theory which has
separation as an axiom or a theorem in it.
otherwise there is no guarantee that x would be infinite.
>
> Theorem E!xAy(yex<-> y is a finite ordinal )
>
> Proof: Extensionality.
>
> Define x=N<->Ay(yex<-> y is a finite ordinal )
>
> This way of defining N is better than the official one present
> in ZF and others, although it is a longer way.
> The peano-like way of defining N present in these set theories
> makes no sense, though it defines N in a shorter way.
>
> Zuhair
>
>
>
>
>
> > > Congrats! This will get you in the textbooks if anything will!
>
> > I don't care about this issue, neither did I think of it at all.
>
> > in addition the definition I gave was erronous I will correct it to:
>
> > x is finite <-> Ayz((y subset_of x & ~Em(mey & mez) &~z=y)->yUz
> > supernumerous to y)).
>
> > were
>
> > r supernumerous to s <->( Ef(f:s->r, f is injective) & Af((f:s->r,f is
> > injective)
> > -> ~ f is surjective) ).
>
> > This definition works with choice or without choice ( thus better than
> > Dedekinds' which only works with choice), and this definition is not
> > essentially circular (cyclical) , while Tarsky's definition is
> > essentially circular.
>
> > Zuhair- Hide quoted text -
>
> - Show quoted text -
John Maurera
The defenition you proposed for "finite" is, in fact, equivalent to Dedekind's definition (and thus doesn't convey the usual meaning of finiteness).
To show the equivalence:
For the first direction, if A is not Dedeking finite, then there is some strict subset B of A which is equipotent with A. It is obvious, then, that the subset B is not incremental (because the union of A and B-A are disjoint sets whose union is equipotent with A).
For the other direction, suppose A is not finite in your sense, then there is a subset B which is not incremental. Suppose for simplicity that B is a strict subset of A (although the same is true if B=A), and let x be an element of A not in B, then the set B /union {x} is not Dedekind finite, and thus the original set A is not Dedekind finite.
Best Regards.
zuhair
> Hello, Zuhair.
>
> The defenition you proposed for "finite" is, in fact, equivalent to Dedekind's definition (and thus doesn't convey the usual meaning of finiteness).
>
> To show the equivalence:
>
> For the first direction, if A is not Dedeking finite, then there is some strict subset B of A which is equipotent with A. It is obvious, then, that the subset B is not incremental (because the union of A and B-A are disjoint sets whose union is equipotent with A).
Ok, this proves that :
x is Dedekindian infinite -> x is Z-infinite
>
> For the other direction, suppose A is not finite in your sense, then there is a subset B which is not incremental. Suppose for simplicity that B is a strict subset of A (although the same is true if B=A), and let x be an element of A not in B, then the set B /union {x} is not Dedekind finite,
You should prove that B union {x} is not Dedekindian finite.
You didn't prove that.
zuhair
You only managed to proved one direction: that is
Ax( x is Dedekindian infinite -> x is Z-infinite ).
You need to prove the opposite direction. Which you didn't.
Zuhair
Dave L. Renfro
[snipped]
On the off-hand chance that anyone getting to this thread
is interested in seriously pursuing various notions of
"finite", a very complete survey (but dated) is:
Alfred Tarski, "Sur les ensembles finis", Fundamenta
Mathematicae 6 (1924), 45-95.
http://matwbn.icm.edu.pl/tresc.php?wyd=1&tom=6
http://matwbn.icm.edu.pl/ksiazki/fm/fm6/fm619.pdf
Tarski introduces many notions of "finite" (over 20,
I think) and studies the logical relationships between
them.
Dave L. Renfro
zuhair
be the following:
x is finite <-> ER( Em( (mex&~En(nex&nRm)) ->
Ek(kex&~Ed(dex&kRd) & Ay((yex&~y=m&~y=k) -> EuEw(uRy & yRw & uex & wex
& ~Eb~Ec( bex & cex & uRb & bRy & yRc & cRw)))))).
In words: every set x is said to be finite if and only if there exist
a relation R that arrange the members of x in such a manner that the
existence of a first member m in x implies the existence of a last
member k in x and implies that every member y in x that is neither the
first nor the last member in x , should have two members u and w both
of which are in x, were uRy and yRw,such that no member b in x exist
such that uRb & bRy , nor their exist a member c in x such that yRc
and cRw.
So if such a relation R exist on any set x then x is finite, if
such a relation doesn't exist , then x is infinite.
This is the complete definition of x is finite, and it perfectly
parallel's the intuitive concept of finitude.
Zuhair
John Maurera
It is obvious that B U {x} is not Dedekind finite: B is not icremental, so there is a non-empty set C disjoint from B such that |B U C| = |B|. Let y be an element of C, then |B| <= |B U {y}| <= |B U C| = |B|, so |B U {y}| = |B|. Now let x be an element of A\B, then |B U {x}| = |B U {y}| = |B|, so B U {x} is not Dedekind finite.
zuhair
Ok, that's fine.
Now let x be an element of A\B, then |B U {x}| = |B U {y}| = |B|, so B
U {x} is not Dedekind finite.
You have a problem in proving that |B U {x}| = |B U {y}|
You didn't prove that.
You seem to say that xeC. But how do you know that?
B is not increamental means that there exist at least
one set C that is disjoint from B and different from B
such that |CUB|=|B|. But how do you prove that
there must exist xeA and xeC, for every B
were B is a proper subset of A.
In other words lets say that D is a subset of A
that is disjoint from B, then you should prove that
AA AB EC( (|CUB|=|B|, B proper subset_of A) -> ED(D proper subset_of A
& D disjoint from A & ~D.C={ }) )
Zuhair
zuhair
Sorry, that is not correct what you should prove is the following:
AA( A is Z-infinite ->EB EC( (|CUB|=|B|, B proper subset_of A) -> ED(D
proper subset_of A & D disjoint from A & ~D.C={ }))).
Zuhair
zuhair
>
> AA( A is Z-infinite ->EB EC( (|CUB|=|B|, B proper subset_of A) -> ED(D
> proper subset_of A & D disjoint from A & ~D.C={ }))).
Correction: Even better:
AA( A is Z-infinite ->EB EC( (|CUB|=|B|, B proper subset_of A,C.B={ },
~C=B) -> ED(D proper subset_of A & D disjoint from A & ~D.C={ }))).
Zuhair
John Maurera
You're forcing me to be very formal, but never mind, that's fine. We know that x is not in B (because it's an element of A\B) and y is not in B as well (because it's an element of C which is disjoint from B). Now let f:B U {x} --> B U {y} be the function defined as follows: f(b)=b for every b in B and f(x)=y, then f is a bijection between B U {x} and B {y} proving the equipotence of these two sets.
zuhair
Do I smell choice here. I don't know, but I think your argument is
based on choice.
f(x)=y, how do you know that you can get such a function. I doubt such
a function exist
in any set theory T; take NF for example, and take the case were xey,
then
f would be unstratified , and by then f would not be bijective as you
think.
Zuhair
>
>
>
> > You seem to say that xeC. But how do you know that?
> > B is not increamental means that there exist at least
> > one set C that is disjoint from B and different from
> > B
> > such that |CUB|=|B|. But how do you prove that
> > there must exist xeA and xeC, for every B
> > were B is a proper subset of A.
>
> > In other words lets say that D is a subset of A
> > that is disjoint from B, then you should prove that
>
> > AA AB EC( (|CUB|=|B|, B proper subset_of A) -> ED(D
> > proper subset_of A
> > & D disjoint from A & ~D.C={ }) )
>
> > Zuhair- Hide quoted text -
>
> - Show quoted text -- Hide quoted text -
zuhair
although I think your argument is not true in all set theories. But
yet
you need to prove the existence of B as a proper subset and non
increamental. You didn't do that.
You see John if you want to illustrate a perfect formal proof. you
should state all possibilities.
so x is Z-infinite <-> EB( B subset_of x & ~ B is increamental )
Now you should prove that this B is a proper subset also, i.e for your
proof to be complete then you should also prove that for every x that
is Z-infinite then there exist a Proper subset of x that is not
increamental.
Otherwise I can say that A itself is increamental and that there do
not exist a proper subset B of A that is not increamental. and
accordingly A would be Dedekindian finite but Z-infinite
Zuhair
Jesse F. Hughes
> On May 1, 3:07 pm, John Maurera <j...@mailinator.com> wrote:
>> You're forcing me to be very formal, but never mind, that's
>> fine. We know that x is not in B (because it's an element of A\B)
>> and y is not in B as well (because it's an element of C which is
>> disjoint from B). Now let f:B U {x} --> B U {y} be the function
>> defined as follows: f(b)=b for every b in B and f(x)=y, then f is a
>> bijection between B U {x} and B {y} proving the equipotence of
>> these two sets.
>
> Do I smell choice here. I don't know, but I think your argument is
> based on choice.
You seem to be having some olfactory issues. The axiom of choice has
nothing to do with the above trivial argument.
> f(x)=y, how do you know that you can get such a function. I doubt
> such a function exist in any set theory T; take NF for example, and
> take the case were xey, then f would be unstratified , and by then f
> would not be bijective as you think.
He's not arguing in NF. This is a very simple argument in ZF.
Personally, I have no idea about NF, but it would be strange if it is
any harder to prove the theorem there.
Theorem: Suppose x and y are not elements of B. Then there is a
bijection from B u {x} to B u {y}.
This *is* the theorem you find doubtful, yes?
--
Jesse F. Hughes
"Disney has now succeeded in preventing anyone from doing to Mickey
Mouse what Disney did to Quasimodo."
-- Randolph Rackovitz, on Eldred vs. Ashcroft
sg...@hotmail.co.uk
> zuhair wrote:
>
> [snipped]
>
> On the off-hand chance that anyone getting to this thread
> is interested in seriously pursuing various notions of
> "finite", a very complete survey (but dated) is:
>
> Alfred Tarski, "Sur les ensembles finis", Fundamenta
>
> Tarski introduces many notions of "finite" (over 20,
> I think) and studies the logical relationships between
> them.
I had no idea that there was more than one. Is it possible to explain
in terms that an amateur might understand what is inadequate about the
definition that a set x is finite iff there exists a bijection between
x and an element of |N?
-Rotwang
zuhair
> zuhair <zaljo...@yahoo.com> writes:
> > On May 1, 3:07 pm, John Maurera <j...@mailinator.com> wrote:
> >> You're forcing me to be very formal, but never mind, that's
> >> fine. We know that x is not in B (because it's an element of A\B)
> >> and y is not in B as well (because it's an element of C which is
> >> disjoint from B). Now let f:B U {x} --> B U {y} be the function
> >> defined as follows: f(b)=b for every b in B and f(x)=y, then f is a
> >> bijection between B U {x} and B {y} proving the equipotence of
> >> these two sets.
>
> > Do I smell choice here. I don't know, but I think your argument is
> > based on choice.
>
> You seem to be having some olfactory issues. The axiom of choice has
> nothing to do with the above trivial argument.
>
> > f(x)=y, how do you know that you can get such a function. I doubt
> > such a function exist in any set theory T; take NF for example, and
> > take the case were xey, then f would be unstratified , and by then f
> > would not be bijective as you think.
>
> He's not arguing in NF. This is a very simple argument in ZF.
> Personally, I have no idea about NF, but it would be strange if it is
> any harder to prove the theorem there.
It doesn't matter in what theory he is arguying about, I meant by
equivalence to be equivalence in any set theory that allows for the
definition of finite according to the two ways , the Z way and
Dedekinds.
>
> Theorem: Suppose x and y are not elements of B. Then there is a
> bijection from B u {x} to B u {y}.
>
> This *is* the theorem you find doubtful, yes?
Yes, I think it is doubtful in NF, since if xey
then no bijection would exist between Bu{x}
and Bu{y}.
But I am not sure of that. But the function he
put that is f(x)=y and f(b)=b for every b in B
is not a stratified formula , since f(x)=y is
not a stratified formula.
However as I said I am not so sure of that
I would rather need the help of one who has
more experience with NF about this situation.
Zuhair
MoeBlee
> r supernumerous to s <->( Ef(f:s->r, f is injective) & Af((f:s->r,f is injective) -> ~ f is surjective) ).
It seems to me that "r supernumerous to s" is equivalent to "s
strictly_dominated_by r":
s strictly_dominated_by r <-> (Ef f is an injection from s into r &
~Ef f is a bijection from r onto s).
> Define x disjoint y <-> ~Em(mex & mey)
So "x disjoint y" is equivalent to "x/\y=0".
> Define x subset_of y <-> Az(zex->zey)
> Define x is increamental <-> Az( (z disjoint x & ~z=x) -> zUx
> supernumerous to x).
>
> x is finite <-> Ay( y subset _of x -> y is increamental).
I call that 'zufinite'.
So in Z set theory we have:
Def: x strictly_dominated_by y <-> (Ef f is an injection from x into y
& ~Ef f is a bijection from y onto x)
Def: x incremental <-> Az((z/\x=0 & ~z=x) -> x strictly_dominated_by
zux)
Def: x zufinite <-> Ay(y subset_of x -> y incremental)
Def: x zuinfinite <-> ~ x is zufinite
Theorem Ax(~x=0 -> x is zuinfinite)
Proof:
Let pex.
So {p} subset_of x, and {p}/\0=0, and ~0={p}, and ~ {p}
strictly_dominated_by 0u{p}.
So, unless I'm missing something, in Z set theory, 0 is zufinite and
all other sets are zuinfinite.
MoeBlee
zuhair
> On 1 May, 15:28, "Dave L. Renfro" <renfr...@cmich.edu> wrote:
>
> > zuhair wrote:
>
> > [snipped]
>
> > On the off-hand chance that anyone getting to this thread
> > is interested in seriously pursuing various notions of
> > "finite", a very complete survey (but dated) is:
>
> > Alfred Tarski, "Sur les ensembles finis", Fundamenta
>
> > Tarski introduces many notions of "finite" (over 20,
> > I think) and studies the logical relationships between
> > them.
>
> I had no idea that there was more than one. Is it possible to explain
> in terms that an amateur might understand what is inadequate about the
> definition that a set x is finite iff there exists a bijection between
> x and an element of |N?
>
> -Rotwang
This last definition is Tarsky's, it is essentially a circular
definition , that doesn't tell us what 'finite' means, it is a
definition by reference, which is not adequate here.
I have presented two definitions here, in a previous post I have
presented a definition which looks similar to this:
x is finite <-> ER( R is asymmetric & Ay( y subset_of x -> ( Em(mey &
~En(ney&nRm)) -> Er(rey & ~Es(sey&rRs)))).
Em(mey & ~En(ney&nRm) <-> m is first member in y as arranged by R.
Er(rey & ~Es(sey&rRs)) <-> r is last member in y as arranged by R.
So the above definition can be simplified to
x is finite <->ER( R is asymmetric & Ay( y subset_of x ->( Em( m is
first member in y as arranged by R) -> Er( r is last member in y as
arranged by R ) ) ) ).
This is a simpler one than the one I have presented earlier with R in
it.
However it would be interesting to see the equivalence between
this definition and the increamental definition John Maurera trying to
prove that it is equivalent to Dedekind's.
Zuhair
zuhair
>
> This last definition is Tarsky's, it is essentially a circular
> definition , that doesn't tell us what 'finite' means, it is a
> definition by reference, which is not adequate here.
>
> I have presented two definitions here, in a previous post I have
> presented a definition which looks similar to this:
>
> x is finite <-> ER( R is asymmetric & Ay( y subset_of x -> ( Em(mey &
> ~En(ney&nRm)) -> Er(rey & ~Es(sey&rRs)))).
>
> Em(mey & ~En(ney&nRm) <-> m is first member in y as arranged by R.
> Er(rey & ~Es(sey&rRs)) <-> r is last member in y as arranged by R.
>
> So the above definition can be simplified to
>
> x is finite <->ER( R is asymmetric & Ay( y subset_of x ->( Em( m is
> first member in y as arranged by R) -> Er( r is last member in y as
> arranged by R ) ) ) ).
Oh shit, what is this.
Correction the definition should be:
x is finite <->ER( R is well ordering relation & Ay( y subset_of x -
>( Em( m is first member in y as arranged by R) -> Er( r is last
member in y as
arranged by R ) ) ) ).
Or in symboles:
x is finite <-> ER( R is well ordering relation & Ay( y subset_of x -
>
( Em(mey & ~En(ney&nRm)) -> Er(rey & ~Es(sey&rRs))))).
Zuhair
zuhair
Yes, you are right Moe. z in my definition shouldn't be allowed to be
0.
>
> MoeBlee
x incremental <-> Az((z/\x=0 & ~z=x & ~z=0) -> x strictly_dominated_by
zux).
Thanks Moe.
Zuhair
zuhair
Summary:
Just to nail things down:
We have three definitions of x is finite in this thread:
1) Definition by increament:
Define x disjoint y <-> ~Em(mex & mey)
Define x subset_of y <-> Az(zex->zey)
r supernumerous to s <->( Ef(f:s->r, f is injective) & Af((f:s->r,f
is
injective) -> ~ f is surjective) ).
Define x is increamental <-> Az( (z disjoint x & ~z=x &~z={}) -> zUx
supernumerous to x).
x is finite <-> Ay( y subset _of x -> y is increamental).
However John Maurera tried to prove that this definition is equivalent
to Dedekind's.
But he didn't complete his formal proof yet.
But it looks that he is right, i.e in all theories which allow for
both definitions , it looks that these two definitions of x is finite
are equivalent.
2) Definition by well ordering.
x is finite <-> ER( R is well ordering relation &
Ay( (y subset_of x & ~y={}) -> (Ez( zex & ~Eu(uex & zRu)) &
Em(mex &~En(nex & nRm))))).
In words: x is said to be finite if and only if there exist a well
ordering relation R that arranges the members of x such that every non
empty subset of x has a first and a last member.
3) The complete definition.
x is finite <-> ER( R is asymmetric relation &
Em( (mex&~En(nex&nRm)) ->
Ek(kex&~Ed(dex&kRd) & Ay((yex&~y=m&~y=k) -> EuEw(uRy & yRw & uex & wex
& ~Eb~Ec( bex & cex & uRb & bRy & yRc & cRw))))) &
(Ez(zex) -> Em(mex&~En(nex&nRm))) ).
The essence of this definition is that a set x is said to be finite if
and only if
There exist an asymmetric relation R that arrange x such that the
existence of a member z in x implies the existence of a first member m
in x as arranged by R , and the existence of m implies the existence
of a last member k in x as arranged by R, and implies also that for
every member y in x other than m and k, there should exist TWO sets u
and w both of which are in x such that uRy and yRw, and such that
there cannot exist a member b in x such that uRb and bRy, and that
there cannot exist a member c in x such that yRc and cRw.
This is the complete definition of x is finite, it is a long one, but
it is the only definition that I am sure of.
According to this definition one can see why {} is finite, and why
all members of Omega are finite. All of the members of Omega
are finite because they have such a relation R that arranges them,
and this relation R is x R xU{x} , which fulfills all the conditions
(outlined above)of R that defines a finite set .
Now I don't know if 3) can be reduced to 2) , I personally doubt that.
And what I doubt most is that 2) or 3) can be reduced to 1).
It would be nice to see if all of these definitions are equivalent to
Dedekind's, but I doubt that largely.
I personally cannot do this work, since it is above my knoweldge base.
Zuhair
John Maurera
Hello Zuhair.
To tell the truth, your attitude begins to discourage me, so unless something new appears on this thread, these will be my final words:
(1) My argument does not use choice. The bijection I used is a simple straight-forward function explicitly defined which exists in every reasonable axiomatic set theory.
(2) None of the above is relevant for our discussion. Say I did use choice, so what ? People nowadays use choice, and if the use of choice makes your definition unusable, then that's a problem.
(3) As I said in my first reply, just for simplicity I assume that B is a proper subset of A. The same argument with minor technical changes would show that the conclusion holds for this case as well.
Good luck,
John
Daryl McCullough
I'm pretty sure that (3) is implied by (2).
If S is finite by definition (2), then there is
a total ordering R on S such that every nonempty
subset S' of S has a first and last element.
So let a be an arbitrary element of S that is
neither the first nor last element of S. Then
let S' = { x in S | x R a }. This has a smallest
element, b. So b R a, but there is no element
between b and a. Similiarly, let S'' = { x in S | a R x }.
This has a largest element, c . So a R c but there is
no element between a and c.
--
Daryl McCullough
Ithaca, NY
zuhair
>
> - Show quoted text -
Ok John, thank you. I think you are right, this increamental
definition is equivalent to Dedekind's.
Zuhair
zuhair
>
> - Show quoted text -
Thank you again. but I already finished with this increamental
definition. you were right regarding its equivalence with Dedekind's.
Zuhair
zuhair
how do you prove the non existence of these in between
elements, since a well ordering relation is transtive,
so these in between elements are their, except when
S has cardinality 3 or below.
Zuhair
>
> --
> Daryl McCullough
> Ithaca, NY- Hide quoted text -
Daryl McCullough
>stevendaryl3...@yahoo.com (Daryl McCullough) wrote:
>> If S is finite by definition (2), then there is
>> a total ordering R on S such that every nonempty
>> subset S' of S has a first and last element.
>>
>> So let a be an arbitrary element of S that is
>> neither the first nor last element of S. Then
>> let S' = { x in S | x R a }. This has a smallest
>> element, b.
I meant "largest element" here.
>> So b R a, but there is no element
>> between b and a. Similiarly, let S'' = { x in S | a R x }.
>> This has a largest element, c.
And I meant smallest element here.
>> So a R c but there is
>> no element between a and c.
>how do you prove the non existence of these in between
>elements, since a well ordering relation is transtive,
>so these in between elements are their, except when
>S has cardinality 3 or below.
We are assuming the following facts about R:
1. Forall x,y in S : x R y or y R x or x=y
2. Forall x,y,z in S: if x R y and y R z then x R z.
3. Forall x,y in S: x R y -> x~=y
4. If S' is a nonempty subset of S, then S' has
a smallest element and a largest element, according
to relation R.
The "smallest element" of S' is an element x such that
forall y in S': x R y or x=y
The "largest element" of S' is an element x such that
forall y in S': y R x or x=y
Okay, so now let a be an element of S that is neither
the largest nor smallest element of S. Let b be the
largest element of { x in S | x R a }. Let c be the
smallest element of { x in S | a R x }.
Now you are asking how do we know that there is no
element between a and c. Well, if d is between a and
c, that means a R d and d R c. But by definition of
c, c is the smallest element in { x in S | a R x }.
That means that forall x in S: a R x -> c R x or c = x.
Since a R d, it follows that c R d or c=d. So d is
*not* beween a and c.
MoeBlee
> Yes, you are right Moe. z in my definition shouldn't be allowed to be 0.
Then in Z set theory (without choice), x is zufinite iff x is Dedekind
finite (maybe this is basically the same proof as John Maurera gave?):
Theorem: x is Dedekind finite <-> x is zufinite
Proof:
Suppose x is Dedekind infinite.
So let y proper subset of x and y equinumerous with x.
So ~ x\y = 0 and ~ x\y = y and (x\y)/\y = 0.
But y equinumerous with (x\y)uy. So ~ y is incremental.
So x is zuinfinite.
Suppose x is zuinfinite.
So let y subset of x and ~z = 0 and ~z = y and z/\y = 0 and ~ y is
strictly dominated by zuy.
So either ~ y is dominated by zuy or y is equinumerous with zuy.
But y is dominated by zuy. So y is equinumerous with zuy.
Let pez. So y proper subset of zuy.
So y is Dedekind infinite.
So x is Dedekind infinite.
MoeBlee
MoeBlee
> This last definition is Tarsky's, it is essentially a circular
> definition , that doesn't tell us what 'finite' means, it is a
> definition by reference, which is not adequate here.
What specific definition do you claim to be 'essentially circular' and
'defiinition by reference' and what do you mean by that?
MoeBlee
Herman Rubin
zuhair <zalj...@yahoo.com> wrote:
>Hi all,
>I think the best definition of 'finite' would be the following:
>x is finite <-> Ayz((y subset_of x & ~Em(mey & mez))->
>yUz supernumerous to y)).
>r supernumerous to s <-> Af((f:s->r,f is injective)->~ f is
>surjective).
>This is better than both Dedekind's and Tarsky's definitions.
>x is infinite <-> ~( x is finite).
If one has a Dedekind finite set which is not equinumerous
to a finite ordinal, it will satisfy this criterion. Your
criterion is equivalent to Dedekind's.
The Tarsky definition is one which is equivalent to being
equinumerous to a finite ordinal. A simpler definition is
that there is a relation R on the set so that every
non-empty subset has a unique R-maximal and a unique
R-minimal element.
--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hru...@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
zuhair
> In article <1177814427.629019.200...@y80g2000hsf.googlegroups.com>,
>
> zuhair <zaljo...@yahoo.com> wrote:
> >Hi all,
> >I think the best definition of 'finite' would be the following:
> >x is finite <-> Ayz((y subset_of x & ~Em(mey & mez))->
> >yUz supernumerous to y)).
> >r supernumerous to s <-> Af((f:s->r,f is injective)->~ f is
> >surjective).
> >This is better than both Dedekind's and Tarsky's definitions.
> >x is infinite <-> ~( x is finite).
>
> If one has a Dedekind finite set which is not equinumerous
> to a finite ordinal, it will satisfy this criterion. Your
> criterion is equivalent to Dedekind's.
>
> The Tarsky definition is one which is equivalent to being
> equinumerous to a finite ordinal. A simpler definition is
> that there is a relation R on the set so that every
> non-empty subset has a unique R-maximal and a unique
> R-minimal element.
But this is exactly my second definition of x is finite. Apparantly
you only saw my first post, but I already have posted this definition
and asked if it is equivalent to the the third definition of mine of x
is finite.
Actually I guessed that this second definition of mine would be
a know definition of x is finite, since it is pretty clear one.
To save you time searching for the three definitions that I have
presented in this thread I will rewrite them again:
**********************************
Summary:
Just to nail things down:
We have three definitions of x is finite in this thread:
1) Definition by increament:
Define x disjoint y <-> ~Em(mex & mey)
Define x subset_of y <-> Az(zex->zey)
r supernumerous to s <->( Ef(f:s->r, f is injective) & Af((f:s->r,f
is
injective) -> ~ f is surjective) ).
Define x is increamental <-> Az( (z disjoint x & ~z=x &~z={}) -> zUx
supernumerous to x).
x is finite <-> Ay( y subset _of x -> y is increamental).
However John Maurera tried to prove that this definition is
equivalent
to Dedekind's.
But he didn't complete his formal proof yet.
But it looks that he is right, i.e in all theories which allow for
both definitions , it looks that these two definitions of x is finite
are equivalent.
3) The complete definition.
According to this definition one can see why {} is finite, and why
all members of Omega are finite. All of the members of Omega
are finite because they have such a relation R that arranges them,
and this relation R is x R xU{x} , which fulfills all the conditions
(outlined above)of R that defines a finite set .
Now I don't know if 3) can be reduced to 2) , I personally doubt
that.
And what I doubt most is that 2) or 3) can be reduced to 1).
It would be nice to see if all of these definitions are equivalent to
Dedekind's, but I doubt that largely.
I personally cannot do this work, since it is above my knoweldge
base.
***********************************************
Daryl McChullough already said that 3) is equivalent to 2).
Do any of the discussors agree with him on this issue.
and most important is 3) or 2) equivalent to 1).
Mind you that you only responded to definition 1) of mine, which as
you, John Maurera and Moe Blee have proved it to be equivalent to
Dedekind's with or without choice.
But what about definitions 2) and 3).
Zuhair
zuhair
Ok, Moe I will elaborate on this issue, since it is a fine one, it is
not easy to see.
I meant Tarsky's definition that a set is finite if it is
equivnumerous with a finite ordinal.
So Tarsky's definition is essentially:
x is finite <-> Ey( x equinumerous to y & y is a finite ordinal ).
This is the circular definition that I am talking about.
I think my definition number 2, which Herman Rubin has already
mentioned that it is a known definition of x is finite, and he
mentioned that it is a simpler definition than Tarsky's but yet
equivalent to it, I will add that it is a non essentially circular
definition, and thus it is a more meaningful definition.
Of course you might defend Tarsky's definition by saying that
the set of all natural numbers N is axiomatized by axiom of infinity
and
proved to be unique from Extensionality, and non of its definition
mention anything about it being the set of all 'finite ordinals'.
The set of all naturals is what Bertrand Russell's described as
the posterity of 0 with respect to relation ' immediate successor'
Review
Introduction to mathematical philosophy:Relations.
So in this manner, the set of all naturals is not defined in terms of
'finite ordinal',
So you can say that the definition is no circular.
My answer is that: Of course such a definition when given by a giant
in logic wouldn't be formaly( technically) circular one.
But 'essentially' it is circular, i.e. it doesn't tell us what
'finite' mean,
it tells us that 'finite' is some inherit property of the members of
N, but it doesn't tell us why.
What Tarsky actually has done is that he picked a special kind of
finite sets, which is essentially the kind of finite sets when they
are ordinals, and he seemed to be quite confident of their finitness,
and then said that any set , would be finite by being equinumerous
with them.
Actually if you want to go to a more deeper level, this definition of
Tarsky's has no difference from Peano's appraoch which states that
'finite' is an
un-defined property, and that 0 is finite and that any set that belong
to the posterity of 0 inherits this undefined property 'finite' from
0 by induction over relation ' immediate successor'.
All of this makes Tarsky's definition meaningless as regards what the
term 'finite' means.
However the second defintion I have mentioned which is equivalent to
Tarsky's is simpler, and more meaningful. It really defines what
'finite' means.
You will more clearly see the emptyness of Tarsky's definition of x is
finite, if you ask yourself why any member of N is finite.
Of course the answer is that because any member of N is equinumerous
with itself, i.e 'equinumerousity' is a relation that is NOT
ALIORELATIVE ( a relation is said to be aliorelative if it cannot
exist between an object and itself( see: Introducation to mathematical
philosophy:Russell)). and since every member of N is a member of N,
then being equinumerous with itself means that it is equinumerous to a
member of N, and thus it is finite.
Although this definition is not circular formally, but yet it is an
empty definition.
It doesn't explain to us in a meaningful manner why members of N are
finite.
I hope I was clear about that.
In nutshell the definition of x is finite number 2) that I have given
in the Summary is the best one.
I always thought that definition number 3) is the most complete
definition of x is finite, But Daryl McCullough have presented a proof
that 3) is equivalent to 2). If so then since 2) is simpler, then 2)
is the best definition of x is finite.
I tried to clarify this as much as I can.
Zuhair
MoeBlee
> On May 2, 11:12 am, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > On May 1, 5:17 pm, zuhair <zaljo...@yahoo.com> wrote:
>
> > > This last definition is Tarsky's, it is essentially a circular
> > > definition , that doesn't tell us what 'finite' means, it is a
> > > definition by reference, which is not adequate here.
>
> > What specific definition do you claim to be 'essentially circular' and
> > 'defiinition by reference' and what do you mean by that?
>
> > MoeBlee
>
> Ok, Moe I will elaborate on this issue, since it is a fine one, it is
> not easy to see.
>
> I meant Tarsky's definition that a set is finite if it is
> equivnumerous with a finite ordinal.
>
> So Tarsky's definition is essentially:
>
> x is finite <-> Ey( x equinumerous to y & y is a finite ordinal ).
>
> This is the circular definition that I am talking about.
Where did you see such a definition? I don't know of any such
defiinition.
There are different paths to take among the definitions of 'finite',
'ordinal', 'natural number'. 'member of omega', etc. So, depending on
which path you take, you will be able to use certain terms already
defined in the path. And, then, finally, no matter which of those
paths we take (not including the Dedekind defintions), we'll prove the
equivalence of the different definitions. But in no path will there be
the circularity you just mentioned.
This is yet another example of your being confused due to not now
studying this subject systematically but instead grabbing bits and
pieces from different sources, none of which is a systematic treatment
onto itself. And the rest of your post below is yet more of your
confusion on this matter as you think you are summarizing from among
different approaches, including those of Tarski and Russell, but you
are instead just getting the whole thing terribly mixed up. This is
just garbage what you write below; it is insulting even that you are
so blithe to post such ignorance and confusion:
You are terribly mixed up about this. I think it's a good thing that
you found an equivalent to Dedekind's definition. That shows some
creativity of yours. But, unfortunately, that is offset by the mess of
confusion you just posted above.
You need a book; and that's all there is to it. Suppes's 'Axiomatic
Set Theory' has an excellent chapter in which he gives Tarksi's
definition and proves the equivalence of it with other definitions.
MoeBlee
zuhair
I produced two other pretty good definitions of 'x is finite' in
addition to the one that is equivalent to Dedekind's, and actually the
other two definitions are better than the one equivalent to
Dedekind's.
It is definition 2) and definition 3) that really deserve to be
disucssed especially after definition 1) has been proved to be
equivalent to Dedekind's.
Instead of discussing definitions 2) and 3) you simply were botherd
with what you claim it to be a confusional record of mine.
Even if you were right about my confusions, then they are non
important things, you should concentrate on the important aspects, not
the confusional aspects.
Definitions 2) and 3) has nothing to do with last confusional post of
mine.
And I am pretty sure that these last two definitions are pretty sane
and actually much better than definition 1).
And I personally don't care if the last opinion of mine (about
Tarski's definition as being 'essentially' circular) is confusional or
not. It is not an important aspect of my ideation about this subject.
Perhaps if I read this chapter in Supps's about Tarski that you've
mentioned, then my confusions would resolve about that subject.
What is really left to be disucssed in this thread is Definitions 2)
and 3)
and as far as I see only Daryl McCullough have made a positive
approach to them when he proved that Definition 3) can be reduced to
Definition 2)
Also Herman Rubin mentioned a definition of x is finite, that appears
similar to Definition 2 of mine.
The outstanding question to be answered here is:
Without choice, is definition 2 equivalent to definition 1 ???
Zuhair
MoeBlee
> Even if you were right about my confusions, then they are non
> important things, you should concentrate on the important aspects, not
> the confusional aspects.
No, it is important not to incorrectly claim that Tarski proposed a
circular definition. Also, what is important to you (e.g., whether
certain formulations are equivalent) might not be important to other
people and other people may wish to comment on other matters that you
have mentioned or are related to your posts.
So, for example, here's a proof of an equivalence that is pertinent to
this discussion, whether or not it is important to you:
Def: conv(R) = {<c b> | <b c> e R}.
Theorem: x is finite <-> ER(R is a well ordering on x & conv(R) is a
well ordering on x).
Proof:
Suppose x is finite. So let f be a bijection from x onto a natural
number n.
Let R = {<b c> | bex & cex & f(b) e f(c)}.
So R is a well ordering on x and conv(R) is a well ordering on x.
Suppose ER(R is a well ordering on x & conv(R) is a well ordering on
x).
For all p in x, there exists a natural number k such that {y | <y p> e
R} is equinumerous with k, per proof by contradiction as follows:
Let q be the R-least member of x such that there is no natural number
k such that {y | <y q> e R} is equinumerous with k.
So q is not the least member of x, lest {y | <y q> e R} = 0 (thus {y |
<y q> e R} equinumerous with a natural number).
So q has R-predecessors.
So, since conv(R) is a well ordering on x, let g be the greatest R-
predecessor of q.
So there is a natural number j such that {y | <y g> e R} equinumerous
with j.
So let h be a bijection from {y | <y g> e R} onto j.
So hu{<g j>} is a bijection from {y | <y g> e R}u{g} onto j+.
But {y | <y g> e R}u{g} = {y | <y q> e R}, so we have a contradiction.
So, for all p in x, we have that {y | <y p> e R} is finite.
And. since every nonempty subset of x has a greatest member, every
subset of x is, for some p, a subset of {y | <y p> e R}u{p}.
So every subset of x is finite, so, since x itself is a subset of x,
we have that x is finite.
MoeBlee
zuhair
> On May 2, 6:34 pm, zuhair <zaljo...@yahoo.com> wrote:
>
> > Even if you were right about my confusions, then they are non
> > important things, you should concentrate on the important aspects, not
> > the confusional aspects.
>
> No, it is important not to incorrectly claim that Tarski proposed a
> circular definition. Also, what is important to you (e.g., whether
> certain formulations are equivalent) might not be important to other
> people and other people may wish to comment on other matters that you
> have mentioned or are related to your posts.
hmmm............, i c.
Well if you want a fruitful disscussion about this issue, then you
should
at least show me or others that might be interested in this issue How
Tarski's definition is not circular?
Tarski's definition is
x is finite <-> Ey( y is a natural number & x equinumerous to y).
Now Let y be a natural number, How do I use Tarski's definition
to show that y is finite, without being involved in some circularity.
The proof that is in my head ( which might be erronous ) is the
following:
Since y equinumerous to y
Since y is a natural number
Then y is a finite set.
But this proves the finitude of y
in terms of y itself, and it is that's why
I claim this definition to be 'essentially'
circulr.
Nice.
Ok, this mean that 2) is equivalent to Tarski's definition.
I have a couple of questions.
Is 2) equivalent to Dedekind's ( in a set theory without choice )?
You mentioned 'Suppose x is finite. So let f be a bijection from x
onto a natural number n Let R = {<b c> | bex & cex & f(b) e f(c)}'.
my question is : in your proof you mentioned 'f(b)ef(c)' , how do you
know that such an R exist. I mean based on what axioms and theorems in
what set theory you are stating your prove.
I think you are working in ZF.
Though I am not sure of this but it looks as if 'f(b)ef(c)' is not a
stratified formula, so you might have problems with your proof in NFU.
Also I want to make sure of this: Is this proof of yours choice
dependant or not?
MoeBlee
> Well if you want a fruitful disscussion about this issue, then you
> should
> at least show me or others that might be interested in this issue How
> Tarski's definition is not circular?
And that discussion starts with the questin I already asked: Where di
Tarski ever give such a definition as you mentioned?
> Tarski's definition is
>
> x is finite <-> Ey( y is a natural number & x equinumerous to y).
Hang on there for just a dagnabbit minute, Sparky. That is NOT the
definition you claimed to have been Tarkski's.
Now, as to the particular definition you just mentioned (whether it's
Tarski's or not), it's not circular as long as 'is a natural number'
and 'equinumerous' have been previously defined without use of
'finite'. And that is accomplished easily. First, 'equinumerous'
doesn't need 'finite' as you already know. And 'n is a natural number'
can be defined as 'new' ('n is a member of omega'), where 'w' is
defined without having to have first defined 'finite', as you SHOULD
already know by now since we've already talked about that so many
times over the last year or so.
> Now Let y be a natural number, How do I use Tarski's definition
> to show that y is finite, without being involved in some circularity.
With this definition you prove a natural number is finite by citing
the fact that it is equinumerous with itself. That is a PROOF of a
theorem and APPLIES a definition but is not ITSELF a DEFINITION, so
there's no circularity.
There are axioms, and definitions (which are a special kind of axiom
called a 'definitional axioms'), and the definitions are not circular
and the proofs use only axioms (indluding the definitional axioms, if
desired) and rules of inference, and there is no circularity in that.
> The proof that is in my head ( which might be erronous ) is the
> following:
>
> Since y equinumerous to y
> Since y is a natural number
> Then y is a finite set.
That is correct.
> But this proves the finitude of y
> in terms of y itself, and it is that's why
> I claim this definition to be 'essentially'
> circulr.
It's not circular. The reason you think it is circular is that you
still haven't studied the basic logic you need to study.
We will prove the finiitude of ANY set y in terms of properties of y
itself. If y is a natural number, then it just happens that the
property we is IMMEDIATE in the proof. Immediacy is not circularity.
> Nice.
It might be done more simply (and maybe I made a mistake somewhere?),
but it's just rountine technique of proving things in basic set theory
and requires very little mathematical imagination.
> Ok, this mean that 2) is equivalent to Tarski's definition.
> I have a couple of questions.
>
> Is 2) equivalent to Dedekind's ( in a set theory without choice )?
Not in ZF.
> You mentioned 'Suppose x is finite. So let f be a bijection from x
> onto a natural number n Let R = {<b c> | bex & cex & f(b) e f(c)}'.
>
> my question is : in your proof you mentioned 'f(b)ef(c)' , how do you
> know that such an R exist.
It's a subset of xXx.
> I mean based on what axioms and theorems in
> what set theory you are stating your prove.
Previous proof of existence of Cartesian products and axiom schema of
separation.
> I think you are working in ZF.
I am. But more particulary, just Z (without regularity). Usually,
whatever I do in set theory, I do it in Z (without regularity), then
mention regularity, replacement, or choice (or a weaker version of
choice) if I use any them also.
> Though I am not sure of this but it looks as if 'f(b)ef(c)' is not a
> stratified formula, so you might have problems with your proof in NFU.
I don't know. I haven't studied enough about NF to say.
> Also I want to make sure of this: Is this proof of yours choice
> dependant or not?
No choice, dependent or otherwise. Where do you see even a hint of
choice? Anyway, if I used any form of choice, I'd mention it. (By the
way, you do know that choice on a finite set doesn't require any
axioms other than those of Z, right? Though, I didn't even use finite
choice here.)
MoeBlee
MoeBlee
to my theorem list. I think you might do some interesting work
eventually; but it's really too bad that you're holding up your
progress by not getting the solid foundation in your studies that you
so very much need.
G. Frege
>>
>> [the] definition is
>>
>> x is finite <-> Ey(y is a natural number & x equinumerous to y).
>>
More formally:
finite x :<-> En(n e w & x ~ n)
>
> Now, as to the particular definition you just mentioned [...],
> it's not circular as long as 'is a natural number' and 'equi-
> numerous' have been previously defined without use of 'finite'.
>
Right.
>>
>> Now Let y be a natural number, How do I use [this] definition
>> to show that y is finite, without being involved in some circu-
>> larity?
>>
I guess, he wanted to stress the word "show" in this question.
That's indeed an interesting "problem", imho. Of course, you should
not interpret "show" here as "prove" (as one might be tempted to
do), but rather with: "show that this is _really_ the case".
I guess, his point is the following: Just assume _for the sake of
the argument_ that there were infinite sets in w (here I'll use an
intuitive notion of infinite!) - since we don't have a definition
of finite/infinite so far, this might very well be the case, at the
present stage of the development of our formal theory (i.e. we
don't have a proof that this is not the case). If we _define_ now
/finite/ (in our theory) the way we did, we could _prove_ (i.e. we
would get), that all sets in w are _finite_ (which _actually_ would
not be the case).
So proving that all n in w are finite by "assuming" that they are
finite is "circular" (from this point of view).
Actually, *I* personally would consider this as asome sort of
argument for NOT using the mentioned definition as our "basic"
definition for "finite" in set theory (say ZFC) - despite its
simplicity (and hence its attractiveness).
So one might claim (if you are to follow the argumentation from
above) that our "basic" definition of /finite sets/ should be via
_Dedekind's_ definition of /infinite sets/.
Of course, then we can show (at least in ZF_C_) that these two
definitions of "finite" are equivalent. So we might claim that the
natural numbers are _really_ finite. (And this might be considered
as a justification for using of the definition mentioned above.)
F.
--
E-mail: info<at>simple-line<dot>de
MoeBlee
> I guess, his point is the following: Just assume _for the sake of
> the argument_ that there were infinite sets in w (here I'll use an
> intuitive notion of infinite!) - since we don't have a definition
> of finite/infinite so far, this might very well be the case, at the
> present stage of the development of our formal theory (i.e. we
> don't have a proof that this is not the case).
The question doesn't make sense UNLESS we have a definition of
'finite'. So suppose, we have a definition of 'finite' BEFORE having
proven the existence of w. Then, after we've proven the existence of
the set that we name 'w', we prove:
new <-> n is finite.
> If we _define_ now
> /finite/ (in our theory) the way we did, we could _prove_ (i.e. we
> would get), that all sets in w are _finite_ (which _actually_ would
> not be the case).
We'd be using 'finite' in a sense that is not equivalent to the
definitions we could give (and not Dedekind finite, either) prior to
proving the existence of the set we call 'w'. And we'd KNOW that.
> So proving that all n in w are finite by "assuming" that they are
> finite is "circular" (from this point of view).
And that is a confused point of view, which is cleared up by
considering what definitions are.
> Actually, *I* personally would consider this as asome sort of
> argument for NOT using the mentioned definition as our "basic"
> definition for "finite" in set theory (say ZFC) - despite its
> simplicity (and hence its attractiveness).
>
> So one might claim (if you are to follow the argumentation from
> above) that our "basic" definition of /finite sets/ should be via
> _Dedekind's_ definition of /infinite sets/.
No, that doesn't follow even under the terms of the argument you
mentioned, because we can define an equivalent of the 'member of w'
definition even before proving the existence of the set we call 'w'.
Specifically, even before introducing the axiom of infinity and thus
even before proving the existence of a least successor-inductive set,
we can define 'finite' so that, after we do prove the existence of a
least successor-inductive set and name it 'w', our previous definition
will be equivalent to 'is a member of w'.
> Of course, then we can show (at least in ZF_C_) that these two
> definitions of "finite" are equivalent. So we might claim that the
> natural numbers are _really_ finite. (And this might be considered
> as a justification for using of the definition mentioned above.)
But we can also show in Z that 'member of w' is equivalent to certain
definitions of 'finite' that are prior to definiing 'w'.
MoeBlee
zuhair
>
> > The proof that is in my head ( which might be erronous ) is the
> > following:
>
> > Since y equinumerous to y
> > Since y is a natural number
> > Then y is a finite set.
>
> That is correct.
>
> > But this proves the finitude of y
> > in terms of y itself, and it is that's why
> > I claim this definition to be 'essentially'
> > circulr.
>
> It's not circular. The reason you think it is circular is that you
> still haven't studied the basic logic you need to study.
>
> We will prove the finiitude of ANY set y in terms of properties of y
> itself. If y is a natural number, then it just happens that the
> property we is IMMEDIATE in the proof. Immediacy is not circularity.
Hey Moe, you know that I know all of that.
My point is a little bit deeper than this.
This immediancy you call, to me it is
a form of circularity, it is a kind of subtle
circularity. It is not desirable to define
such an important concept like 'finite'
in this way.
Remember I said that Tarski's definition
is 'essentially' circular. I didn't say
it is formally circular.
This definition of finite , I mean the one
with the R and conv(R) that you've mentioned
is much better than Tarski's quasy-circular
definition. actually this definition using R
and conv(R) is my definition of x is finite
number 2.
Anyhow I think we'll discuss this subject later.
I need to define 'essentially' circular in a clear
manner so that we can have a frutiful clear
discussion about it.
MoeBlee
> Remember I said that Tarski's definition
> is 'essentially' circular. I didn't say
> it is formally circular.
>
> This definition of finite , I mean the one
> with the R and conv(R) that you've mentioned
> is much better than Tarski's quasy-circular
> definition. actually this definition using R
> and conv(R) is my definition of x is finite
> number 2.
In Z without the axiom of infinity, the 'R and conv(R)' definition is
equivalent to Tarski's 'subset maximal' definition, which is
equivalent to other definitions, and, in Z, all of these are
equivalent to 'member of w'.
Meanwhile your 'essentially circular' and 'quasi-circular' are your
own self-distractions.
MoeBlee
MoeBlee
> all of these are equivalent to 'member of w'.
I meant 'equinumerous to a member of w'; and in my post. G. Frege too.
MoeBlee
zuhair
Nup.
>
> MoeBlee
MoeBlee
> new <-> n is finite.
I meant:
n equinumerous to a member of w <-> n is finite
> ''member of w'
Again, I meant 'equinumerous with a member of w'.
And throughout my post.
MoeBlee
MoeBlee
Check your keyboard. It seems to have crossed 'N' for 'Y'.
MoeBlee
Herman Rubin
>> I guess, his point is the following: Just assume _for the sake of
>> the argument_ that there were infinite sets in w (here I'll use an
>> intuitive notion of infinite!) - since we don't have a definition
>> of finite/infinite so far, this might very well be the case, at the
>> present stage of the development of our formal theory (i.e. we
>> don't have a proof that this is not the case).
>The question doesn't make sense UNLESS we have a definition of
>'finite'. So suppose, we have a definition of 'finite' BEFORE having
>proven the existence of w. Then, after we've proven the existence of
>the set that we name 'w', we prove:
>new <-> n is finite.
All versions of set theory of which I am aware have an
axiom which guarantees an infinite set. One of the
"standard" ones is that there is a set x such that 0 (the
empty set) is an element of x and if y \in x then y u {y}
\in x. It is easily seen that the intersection of all such
sets is such a set, and the smallest of that type. This
set must be well ordered, i.e., any non-empty subset has a
smallest element under inclusion. Use induction to prove
this. We do not have to worry about the elements being
regular; this follow by induction.
Now we take an easy characterization of finite; there is an
ordering such that every non-empty subset has a unique largest
and smallest element. It can be shown that if y has this
property, so does y u {y}. Tnere cannot be a largest y
with this property, as y u {y} is larger.
zuhair
Ok, Moe let me put matters in this way.
Which constitutes a more plausable appraoch?
To define 'natural number' in terms in finitude
Or
To define 'finitude' in terms of 'natural number'.
Lets say that definition D1 is simpler than definition D2
if D1 requires less number of axioms then D2 requires.
Lets say that a definition D1 is more stable than definition D2
if D1 require more number of axioms to change before it changes
than do D2 require.
Let me clarify.
you see lets take the definition of 'x is a natural number'.
How do you define this in Z-I-R in an independent manner
from finitude.
I think the definition would be
x is a natural number <-> x is an ordinal.
were 'ordinal' is defined in the standard manner.
Now how do you define it in Z-R
you see the definition will CHANGE to the following:
x is a natural number <-> xew
While lets take the opposite approach
x is a natural number <-> x is a finite ordinal.
were
x is finite <-> EREconvR( R well order x & convR well order x).
and ordinal defined in the standard manner.
Now this definition is the same in Z-I-R and in Z-R
So the later definition is more stable than the first.
Not only this, the later definition is Simpler than the first
since it requires Z-I-R only
while the other is more complex it is actually two definitions
depending on the presence of infinity or not.
>From that I see that it is easier to define
x is finite , in a manner that doesn't depend on 'natural number'
and THEN we define 'natural number' as a finite ordinal.
And not the opposite manner of Tarski's.
Tarski is defining the Hen after the egg it layed,
While my approach define the Hen after the egg it hatched from.
I don't know, it seems common sense for me, to
define 'finite set' first in a manner that is independent
of 'natural number'.
And after that we proceed to define 'natural number'
as a special case of finite set like natural number is a finite
ordinal.
This constitutes a simpler, more plausable and more stable
approach than the opposite one.
I don't know, it seems a matter of preference.
Zuhair
Dave L. Renfro
>> On the off-hand chance that anyone getting to this thread
>> is interested in seriously pursuing various notions of
>> "finite", a very complete survey (but dated) is:
>>
>> Alfred Tarski, "Sur les ensembles finis", Fundamenta
>> Mathematicae 6 (1924), 45-95.
>> http://matwbn.icm.edu.pl/tresc.php?wyd=1&tom=6
>> http://matwbn.icm.edu.pl/ksiazki/fm/fm6/fm619.pdf
>>
>> Tarski introduces many notions of "finite" (over 20,
>> I think) and studies the logical relationships between
>> them.
Rotwang wrote:
> I had no idea that there was more than one. Is it possible
> to explain in terms that an amateur might understand what
> is inadequate about the definition that a set x is finite
> iff there exists a bijection between x and an element of |N?
I might have been mistaken about the "over 20" part.
I looked in several places for where I might have seen
a statement like this and the closest I found was Gregory
H. Moore's book "Zermelo's Axiom of Choice: Its Origins,
Development, and Influence" (Springer-Verlag, 1982). On
page 211, Moore gives the following five (5) notions of
"F is finite".
1. Every nonempty family of subsets of F has
a subset-maximal element.
2. Every nonempty family of subsets of F,
ordered by inclusion, has a subset-maximal
element.
3. No proper subset of the power set of F
has the same cardinality as the power
set of F.
4. No proper subset of F has the same
cardinality as F.
5. F is not the union of two disjoint
nonempty sets, each of which has the
same cardinality as F.
Tarski gives these in his 1924 paper, according to
Moore, and Tarski proves 1 implies 2 implies 3
implies 4 implies 5, where neither the axiom of
choice nor the axiom of infinity is assumed.
Tarski wasn't able to show that the notions are
distinct (in the absence of the axiom of choice
and the axiom of infinity), but Moore says that
mathematicians later managed to show that all
5 of these are distinct notions of "finite".
I'm sure Tarski must have given some other notions
of "finite" as well, but they might have been trivial
modifications of some of the five notions above, which
he likely showed were equivalent to one of the five notions
above. Thus, it might very well be the case that Tarski
introduced over 20 notions of "finite", but if he did,
I think it's likely that he was able to show the
equivalence of each one of them to one of the five
notions above.
Maybe someone who can actually read French could
spend a few moments glancing over Tarski's paper
to see if what I'm saying is essentially correct.
Dave L. Renfro
MoeBlee
> In article <1178229756.514840.212...@q75g2000hsh.googlegroups.com>,
I understand and, of course, agree with all of that.
MoeBlee
MoeBlee
> Which constitutes a more plausable appraoch?
>
> To define 'natural number' in terms in finitude
> Or
> To define 'finitude' in terms of 'natural number'.
I don't know what would be implausbile about either.
Anyway, personally, I prefer to define, separately, 'finite' and
'ordinal', then define 'natural number' as 'finite ordinal' then
define 'w' as 'the unique set that is both closed under successors and
is a subset of any other set closed under successors', then prove
that w is the set whose members are all and only the natural numbers.
But other paths come to the same set of theorems eventually anyway.
> Lets say that definition D1 is simpler than definition D2
> if D1 requires less number of axioms then D2 requires.
In a very strict sense, definitions, in the sense we're dealing with
here, don't require axioms. However, in this example (and, throughout
all my remarks, using the method of setting t=0 for improperly
referring terms 't'), with the 'member of w' definition, the axiom of
infinity would be used to prove there exist natural numbers (since
without the axiom of infinity, it is undetermined whether or not w=0).
> Lets say that a definition D1 is more stable than definition D2
> if D1 require more number of axioms to change before it changes
> than do D2 require.
>
> Let me clarify.
>
> you see lets take the definition of 'x is a natural number'.
>
> How do you define this in Z-I-R in an independent manner
> from finitude.
>
> I think the definition would be
>
> x is a natural number <-> x is an ordinal.
(1) You have not given a defintion, in mathematical logic, of
'stable'.
(2) In Z-R-I, we would never give the definition: x is a natural
number <-> x is an ordinal.
> were 'ordinal' is defined in the standard manner.
>
> Now how do you define it in Z-R
>
> you see the definition will CHANGE to the following:
>
> x is a natural number <-> xew
That is only one possiblity. We can also define, in Z-R-I or in Z-R,
'natural number' without mentioning 'w'.
Again, these definitions don't depend on the axioms in the way you
think they do.
Your analysis of this whole matter is based on your lack of
familiarity with how mathematical logic treats the subject of
definitions of predicate symbols.
> While lets take the opposite approach
>
> x is a natural number <-> x is a finite ordinal.
>
> were
>
> x is finite <-> EREconvR( R well order x & convR well order x).
>
> and ordinal defined in the standard manner.
>
> Now this definition is the same in Z-I-R and in Z-R
Definitions, in the sense we're working with here, don't change on
account of axioms. What changes is the set of formulas (particularly,
in this context, those formulas using defined symbols) that are
derivable.
But, anyway, personally, I do prefer the 'finite ordinal' definition
for 'natural number' over the 'member of w' definition, because the
latter does require the axiom of infinity to prove that there exist
natural numbers, even though, strictly speaking, we could still
formulate the definition without the axiom of infinity.
> So the later definition is more stable than the first.
I do understand your general informal sense, and I too use the 'finite
ordinal' defintion rather than the 'member of w' definition. But,
techncially, just formulating the definitions does not require axioms
in the way you are analysing.
> Not only this, the later definition is Simpler than the first
> since it requires Z-I-R only
> while the other is more complex it is actually two definitions
> depending on the presence of infinity or not.
> >From that I see that it is easier to define
>
> x is finite , in a manner that doesn't depend on 'natural number'
>
> and THEN we define 'natural number' as a finite ordinal.
>
> And not the opposite manner of Tarski's.
(1) I too prefer the 'finite ordinal' definition, for the reason I
gave, but you are unclear, at best, as to the PRECISE technical
implications of this.
(2) Tarski DOES give a defintion of 'finite' that does not refer to
'w' and which you can see presented in, e.g., Suppes, BEFORE the axiom
of infinity is introduced.
> Tarski is defining the Hen after the egg it layed,
> While my approach define the Hen after the egg it hatched from.
You keep talking about what Tarski did, even though you are not
correctly apprised of the historical facts about his work with respect
to these definitions.
> I don't know, it seems common sense for me, to
> define 'finite set' first in a manner that is independent
> of 'natural number'.
Both approaches make good sense in their own respective ways.
Please inform yourself about these matters before spouting categorical
pronouncements about them. You have a misunderstanding of the
technicalities of definitions, axioms, and theories. And you are not
informed as the historical matter of Tarski.
MoeBlee
MoeBlee
> Maybe someone who can actually read French could
> spend a few moments glancing over Tarski's paper
> to see if what I'm saying is essentially correct.
I would love to have a translation of that paper into English. Much of
it is conveyed in Suppes's set theory text, but it would be nice to
have the whole thing.
MoeBlee
zuhair
Yes of course I am ignorant regarding this.
However my argument is perfectly clear.
I prefer defining the class of finite sets ( which is a proper class
by the way) first, and after that I go define the set of natural
numbers as a proper subset of the class of finite sets using
separation with property_ ordinal.
i.e x is a natural number<-> x is a finite ordinal.
It is easier to define the class first and then you go define a
subclass of it using separation.
To do the opposite i.e to define the proper class of finite sets after
a proper subset of it that is the set of the natural numbers, even if
the last set is defined independently of finitude as to avoid
circularity; looks to me as an upside down approach.
I know that both approaches are equivalent, but I prefer defining
'finite'
first , then we define ' ordinal ' , and after that we define natural
number as
a finite ordinal.
Zuhair
zuhair
Hey Moe, tell me in Z-I-R, how x is a natural number is defined?
Zuhair
zuhair
How a subset-maximal element is defined in first order logic?
Can you write this definition in first order logic, please.
>
> 2. Every nonempty family of subsets of F,
> ordered by inclusion, has a subset-maximal
> element.
The formula please.
>
> 3. No proper subset of the power set of F
> has the same cardinality as the power
> set of F.
>
> 4. No proper subset of F has the same
> cardinality as F.
This is Dedekind's finite.
>
> 5. F is not the union of two disjoint
> nonempty sets, each of which has the
> same cardinality as F.
5 is actually my definition that I have presented in this thread and
that many people in this group say that they prove that it is
equivalent to 4 ( i.e. Dedekind's ).
>
> Tarski gives these in his 1924 paper, according to
> Moore, and Tarski proves 1 implies 2 implies 3
> implies 4 implies 5, where neither the axiom of
> choice nor the axiom of infinity is assumed.
>
> Tarski wasn't able to show that the notions are
> distinct (in the absence of the axiom of choice
> and the axiom of infinity), but Moore says that
> mathematicians later managed to show that all
> 5 of these are distinct notions of "finite".
You say they are distinct, then the definition I gave at this thread
which was said to be equivalent to Dedekind's is in reality not!
I have always been suspecious regarding this 'equivalence'.
>
> I'm sure Tarski must have given some other notions
> of "finite" as well, but they might have been trivial
> modifications of some of the five notions above, which
> he likely showed were equivalent to one of the five notions
> above. Thus, it might very well be the case that Tarski
> introduced over 20 notions of "finite", but if he did,
> I think it's likely that he was able to show the
> equivalence of each one of them to one of the five
> notions above.
>
> Maybe someone who can actually read French could
> spend a few moments glancing over Tarski's paper
> to see if what I'm saying is essentially correct.
>
> Dave L. Renfro- Hide quoted text -
>
> - Show quoted text -
zuhair
All versions of set theory of which I am aware have an
axiom which guarantees an infinite set.
.................
Zuhair wrote:
Not all set theories has such an axiom, take for example
Kripke-Platek set theory with urelements.
Now it would be interesting to see how one can define
natural number in Z-I-R
I propose the following definition of x is a natural number in Z-I-R
x is a natural number <-> x is ordinal.
I think this would do the job.
However in this theory there is no need to define x is finite, since
every x in Z-I-R is finite.
Zuhair
zuhair
>
> I know that both approaches are equivalent, but I prefer defining
> 'finite'
> first , then we define ' ordinal ' , and after that we define natural
> number as
> a finite ordinal.
>
>
> - Show quoted text -
Any definition of a class in terms of a proper subset of it, would
give an impression that the defining subset has some kind of
importance over all other sets in this class. I think one shouldn't
use such an approach unless he outlines why this defining set is so
important.
In a thoery which defines finite cardinals and finite ordinals , as
natural numbers , it is clear that Tarski's approach of defining x is
finite <-> x equinumerous to y & y is a natural number , is justified,
since the finitude of all sets is meansured after that of these
natural numbers.
However I still don't like it.
The easier way is to define x is finite , then define x is ordinal and
then go define x is natural number.
Zuhair
MoeBlee
> However my argument is perfectly clear.
It is clear what you prefer in one definition over another; and I
SHARE with you that preference. But what was wrong wa your
mathematical and historical characterizations of the matter.
> I prefer defining the class offinitesets ( which is a proper class
> by the way) first, and after that I go define the set of natural
> numbers as a proper subset of the class offinitesets using
> separation with property_ ordinal.
> i.e x is a natural number<-> x is afiniteordinal.
>
> It is easier to define the class first and then you go define a
> subclass of it using separation.
>
> To do the opposite i.e to define the proper class offinitesets after
> a proper subset of it that is the set of the natural numbers, even if
> the last set is defined independently of finitude as to avoid
> circularity; looks to me as an upside down approach.
Now you're in a class theory. That's okay too. Either in, say, Z or in
NBG, yes, we know that we can first define 'finite' and 'ordinal'
without mentioning the axiom of infinity or omega. I already
recommended Suppes's set theory textbook to see how he does that and
how he proves equivalences along the way (actually, some of it is in
exercises that I even worked out with the help of other posters in
sci.logic and sci.math).
> I know that both approaches are equivalent, but I prefer defining
> 'finite'
> first , then we define ' ordinal ' , and after that we define natural
> number as
> afiniteordinal.
So do I. But you need to be more careful about what you say about the
mathematics, metamathematics and the history of it.
MoeBlee
MoeBlee
> Hey Moe, tell me in Z-I-R, how x is a natural number is defined?
I ALREADY mentioned:
n is a natural number <-> n is a finite ordinal
And there are other equivalent definitions. I strongly suggest you get
Suppes's set theory book, which as a Dover paperback, and especially
as a used copy, can be purchased for literally just a few dollars.
MoeBlee
MoeBlee
> On May 4, 9:11 am, "Dave L. Renfro" <renfr...@cmich.edu> wrote:
> > 1. Every nonempty family of subsets of F has
> > a subset-maximal element.
>
> How a subset-maximal element is defined in first order logic?
> Can you write thisdefinitionin first order logic, please.
m is a subset-maximal element of S <-> (meS & Axes ~ m proper subset
of x)
Get Suppes's set theory book.
> > 2. Every nonempty family of subsets of F,
> > ordered by inclusion, has a subset-maximal
> > element.
>
> The formula please.
Here's Tarksi's definition ['P' stands for 'power set of']:
x is finite <-> AS((S subset of Px & ~S=0) -> Em m is a subset-maximal
element of S)
Get Suppes's set theory book. There's a bunch of this stuff in there.
> You say they are distinct, then thedefinitionI gave at this thread
> which was said to be equivalent to Dedekind's is in reality not!
> I have always been suspecious regarding this 'equivalence'.
Just get a good set theory textbook.
In Z set theories:
The Tarski definition I just gave is equivalent to the 'well ordered
by an R and its converse' definition.
The Dedekind definition is equivalent to the 'incremental'
definition.
The Tarski condition entails the Dedekind condition ('not equinumerous
with a proper subset of itself') (indeed, this is called the
'pigeonhole principle') but the Dedekind condition does not entail the
Tarski condition.
Adding the axiom of choice (even just the axiom of countable choice),
the Dedekind condition entails the Tarski condition so that with the
axiom of choice (or even just the axiom of countable choice), the
Tarski and Dedekind definitions are equivalent.
MoeBlee
MoeBlee
> On May 3, 7:38 pm, hru...@odds.stat.purdue.edu (Herman Rubin) wrote:
> Now it would be interesting to see how one can define
> natural number in Z-I-R
We don't need axioms to make these definitions. I keep explaining that
to you and you keep ignoring what I wrote. And I've already said that
we don't even have to mention omega to define 'natural number'.
n is a natual number <-> (n is finite & n is an ordinal)
where 'finite' is defined as any of the equivalents of 'well ordered
by an R and its converse', and 'ordinal' is defined as 'well ordered
by epsilon and is epsilon-transitive'.
> I propose the followingdefinitionof x is a natural number in Z-I-R
>
> x is a natural number <-> x is ordinal.
>
> I think this would do the job.
That's IDIOTIC.
And I ALREADY addressed that. You just skip what I wrote.
Are you thinking that because Z-R-I can't prove the existence of an
infinite set (thus, a fortiori, of an infinite ordinal) it follows
that we might as well take the naturals to be the ordinals?
If that's what you're thinking, it's IDIOTIC and you are RELAPSING
into illogical thinking that we ALREADY have been through with you.
What I thought you finally understood by this time is that even though
a theory might not prove the existence of objects having a certain
property, it doesn't follow that the theory RULES OUT that there are
such objects. Even though Z-R-I doesn't prove the existence of any
infinite ordinals, it still doesn't RULE OUT that they exist. So the
definition of 'natural number' still must be stated so as to rule out
that a natural number could be an infinite ordinal, otherwise it is
left UNdetermined whether there are natural numbers that are infinite
when what we WANT to do is make it DETERMINED that there are NOT
natural numbers that infinite.
> However in this theory there is no need to define x isfinite, since
> every x in Z-I-R isfinite.
NO!!! NO!!! NO!!!
We went through that with you about a year ago!!!
In Z-I-R it is UNDETERMINED whether every set is finite. In Z-I-R it
is NOT determined that every set is finite.
A theory that DOES determine that every set is finite is Z-R-I + ~I.
That is, Z-R-I plus the NEGATION of the axiom of infinity. There, yes,
all sets are finite. But in just Z-R-I, it is UNDETERMINED whether
there are infinite sets, thus UNDETERMINED whether all sets are
finite.
Come on, zuhair, we spent a lot of work a year ago to finally get you
to understand that but now you're repeating your old misconceptions
again.
MoeBlee
zuhair
Oops, yes that is true, what you said is true, actually I know that,
but it is due to my poor practice in this field that I sometimes keep
doing the same mistake again and again?
Thanks for the correction.
But Moe I have an argument behind all of that. You didn't see it.
What I want to do is to see weather Tarski's definition of
x is finite in terms of x is natural, is workable in Z-I-R.
I mean the definition of: x is finite <-> Ey( y is a natural number &
y equinumerous to x)
My point is that If I want to use the above definition of x is finite
then I cannot say that in Z-I-R a natural number is a finite ordinal.
Why?
because if I say: in Z-R-I : x is a natural number <-> ( x is ordinal
& x is finite)
then I can't use Tarski's definition without circularity, since if I
use it, the definition would be:
x is finite <-> Ey( y is an ordinal & y is finite & y equinumerous to
x)
which is clearly circular.
However there is a way around that!
I can't get Supp's now so I will make a primitive try, and I am
looking forwards for you assisting me in that.
In Z-I-R :
x is a natural number <-> ( x is an ordinal & Ay( (yex & ~y=0) ->
Ez( zex & zu{z}=y) ) ).
In words, in Z without infinity and without regularity: any x is said
to be a natural number if and only if x is an ordinal in which every
member y other than 0, has an immediate predecessor z in x.
of course 'immediate predecessor' is the converse of ' immediate
successor'.
Now we can proceed and define x is finite in Z-I-R
x is finite <-> Ey ( x equinumerous to y & y is a natural number ).
Now these formulae that I gave, are really independable of Infinity,
they are the same formulae used in Z-R and in Z-R-I, and are not
circular.
And they are better than the definition mentioned by Rubin which is
x is a natural number <-> xew
since this formula requires axiom of infinity, since the existence of
Omega 'w' is not garanteed without infinity, and so this formula
doesn't work in Z-R-I, since w is not defined to exist in Z-R-I.
So in summary:
In Z-I-R or Z-R.
x is a natural number <-> ( x is an ordinal & Ay( (yex & ~y=0) ->
Ez( zex & zu{z}=y) ) ).
x is finite <-> Ey ( x equinumerous to y & y is a natural number ).
Am I right this time?
Zuhair
MoeBlee
> On May 6, 3:40 am, MoeBlee <jazzm...@hotmail.com> wrote:
> What I want to do is to see weather Tarski's definition of
> x is finite in terms of x is natural, is workable in Z-I-R.
>
> I mean the definition of: x is finite <-> Ey( y is a natural number &
> y equinumerous to x)
Yes, I've answered that several times already. The answer is YES.
> My point is that If I want to use the above definition of x is finite
> then I cannot say that in Z-I-R a natural number is a finite ordinal.
I've alreay addressed this.
If we define
x is finite <-> Ey(y is a natural number & x equniumerous with y),
then of course we can't define 'natural number' as 'finite ordinal'.
But then we will define 'natural number' in some way so that we PROVE,
not define, that x is a natural number iff x is a finite ordinal.
Again, the formulas and all of that are in, e.g., Suppes's book.
> Why?
>
> because if I say: in Z-R-I : x is a natural number <-> ( x is ordinal
> & x is finite)
>
> then I can't use Tarski's definition without circularity, since if I
> use it, the definition would be:
>
> x is finite <-> Ey( y is an ordinal & y is finite & y equinumerous to
> x)
>
> which is clearly circular.
>
> However there is a way around that!
Of course there is. I've mentioned it already.
> I can't get Supp's now so I will make a primitive try, and I am
> looking forwards for you assisting me in that.
I hope I can help you with studying the book. I can't help you to get
the book. But it's widely available at a very cheap price.
> In Z-I-R :
>
> x is a natural number <-> ( x is an ordinal & Ay( (yex & ~y=0) ->
> Ez( zex & zu{z}=y) ) ).
That doesn't work, since you need to stipulate that z is also an
ordinal. This works:
x is a natural number <-> (x is an ordinal & Ay(yex -> (y is an
ordinal & (y=0 v Ez(z is an ordinal & y=zu{z}))))).
> In words, in Z without infinity and without regularity: any x is said
> to be a natural number if and only if x is an ordinal in which every
> member y other than 0, has an immediate predecessor z in x.
Change that to: x is an ordinal in which every member other than 0 is
a successor ordinal.
> of course 'immediate predecessor' is the converse of ' immediate
> successor'.
>
> Now we can proceed and define x is finite in Z-I-R
>
> x is finite <-> Ey ( x equinumerous to y & y is a natural number ).
Yes, we already knew that.
> Now these formulae that I gave, are really independable of Infinity,
Again, your notion of defintions being independent of axioms is not
precise. However,I do understand your point in a general sense, and I
too prefer to define 'finite' and 'natural number' without mentioning
'w'.
> they are the same formulae used in Z-R and in Z-R-I, and are not
> circular.
>
> And they are better than the definition mentioned by Rubin which is
>
> x is a natural number <-> xew
>
> since this formula requires axiom of infinity, since the existence of
> Omega 'w' is not garanteed without infinity, and so this formula
> doesn't work in Z-R-I, since w is not defined to exist in Z-R-I.
It's not as simple as you state it. I addressed this technical point
already.
MoeBlee
zuhair
>
> That doesn't work, since you need to stipulate that z is also an
> ordinal. This works:
>
> x is a natural number <-> (x is an ordinal & Ay(yex -> (y is an
> ordinal & (y=0 v Ez(z is an ordinal & y=zu{z}))))).
>
Sorry Moe, the formula that I have presented was actually incomplete:
What I meant was this:
x is a natural number <-> ( x is an ordinal & ( ~x=0 -> Uxex) &
Ay( (yex & ~y=0) ->Ez( zex & y= zu{z}) ) ).
This was my formula, but I missed mentioning the union part of it, I
am not concentrating these days , I have a lot of work, anyhow.
About z we don't need to mention that z is an ordinal, since every
member of an ordinal is an ordinal.
I am sure that this formula works.
To clarify this formula, ( x is an ordinal & ( ~x=0 -> Uxex) ; means
that any x that is ordinal and who's union is a member of it , then
and only then this x is a last membered ordinal ( which might be
finite or infinite of course ).
Now: Ay( (yex & ~y=0) ->Ez( zex & y= zu{z}) ) ; means that
every member in x other than 0 should have an immediate predecessor z
that is in x, and of course z would be an ordinal since every member
of an ordinal is an ordinal.
So in words this definition is saying.
A natural number x is a last membered ordinal in which every member y
has an immediate predecessor z in x.
I think that this definition works.
Zuhair
MoeBlee
> On May 6, 8:12 am, zuhair <zaljo...@yahoo.com> wrote:
> > x is a natural number <-> ( x is an ordinal & Ay( (yex & ~y=0) ->
> > Ez( zex & zu{z}=y) ) ).
>
> That doesn't work, since you need to stipulate that z is also an
> ordinal. This works:
>
> x is a natural number <-> (x is an ordinal & Ay(yex -> (y is an
> ordinal & (y=0 v Ez(z is an ordinal & y=zu{z}))))).
No wait, we're BOTH wrong.
You are right about not having to mention 'y is an ordinal' since yex
entails y is an ordinal (my only excuse for overlooking that is that
I'm posting while I'm very very hungry as I keep putting off going out
for a bite to eat and thus I can really only think about food, not
sets, right now!).
But the formula should be:
x is a natural number <-> (x is an ordinal & Ay(((yex v y=x) & ~y=0) -
> Ez(zex & zu{z}=y))).
The way we had it, omega would qualify to be a natural number. What I
added to correct that is the 'y=x' clause. I.e., x itself must also be
either 0 or a successor, and not just a set whose members are all
either 0 or a successor.
I think it's right now. (I'll check it again when I come back from
getting some food!) And I think it can be simplified too.
MoeBlee
zuhair
> On May 6, 4:24 pm, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > On May 6, 8:12 am, zuhair <zaljo...@yahoo.com> wrote:
> > > x is a natural number <-> ( x is an ordinal & Ay( (yex & ~y=0) ->
> > > Ez( zex & zu{z}=y) ) ).
>
> > That doesn't work, since you need to stipulate that z is also an
> > ordinal. This works:
>
> > x is a natural number <-> (x is an ordinal & Ay(yex -> (y is an
> > ordinal & (y=0 v Ez(z is an ordinal & y=zu{z}))))).
>
> No wait, we're BOTH wrong.
I was wrong, but I corrected my definition. The corrected definition
of mine is not wrong.
>
> You are right about not having to mention 'y is an ordinal' since yex
> entails y is an ordinal (my only excuse for overlooking that is that
> I'm posting while I'm very very hungry as I keep putting off going out
> for a bite to eat and thus I can really only think about food, not
> sets, right now!).
>
> But the formula should be:
>
> x is a natural number <-> (x is an ordinal & Ay(((yex v y=x) & ~y=0) -
>
> > Ez(zex & zu{z}=y))).
>
This is complex, my formula is simpler.
> The way we had it, omega would qualify to be a natural number.
I don't see how can that be???
I defined x is natural in the following manner ( the corrected
definition ):
x is a natural number <-> ( x is an ordinal & ( ~x=0 -> Uxex) &
Ay( (yex & ~y=0) ->Ez( zex & y= zu{z}) ) ).
Now with this definition , how can Omega satisfy it??
The union of Omega is not in Omega, because the union of Omega IS
Omega itself, and since every ordinal is not in itself, then we have
~Uwew, and so we have ~ w is a natural number.
This corrected formula of mine, is a simple formula, and it is simpler
than the one you gave.
Zuhair
MoeBlee
and see if it's correct:
x is a natural number <-> ((x=0 v x is a successor ordinal) & Anex(n=0
v Ez n=zu{z}))
That should be equivalent to:
x is a natural number <-> ((x=0 v x is a successor ordinal) & Anex(n=0
v n is a successor ordinal))
And both should be equivalent to the correction I offered in my
previous post.
MoeBlee
zuhair
I will check them. But both of these corrections are complex formulae.
The one I gave is much simpler than both of these.
zuhair
You should define clearly ' x is a successor ordinal'
I think you mean
x is a successor ordinal<-> Ey(y is an ordinal & x=yU{y}).
Yes, I think your correction works.
It is a nice definition.
Still I think my definition works to
x is a natural number <-> ( x is an ordinal & ( ~x=0 -> Uxex) &
Ay( (yex & ~y=0) ->Ez( zex & y= zu{z}) ) ).
I think we both corrected our definitions right!
You definition is equivalent to mine.
Zuhair
zuhair
Although both of your and my definitions are correct and equivalent,
but I still think that mine is better. Because your definition states
frankly that 0 is a natural number, while mine implies that, I mean in
your definition 0 is fixed to be a natural number, your definition
doesn't explain why 0 is a natural number, it orders zero to be a
natural number, while mine do explain why?
So though the two definitions are equivalent yet I think mine is more
meaningful.
Zuhair
MoeBlee
> On May 6, 8:52 pm, MoeBlee <jazzm...@hotmail.com> wrote:
> I was wrong, but I corrected my definition. The corrected definition
> of mine is not wrong.
Your correction:
x is a natural number <-> (x is an ordinal & (~x=0 -> Uxex) &
Ay((yex & ~y=0) -> Ez(zex & y=zu{z}))).
That can be simplified:
x is a natural number <-> (x is an ordinal & (~x=0 -> Ux e x) &
Ay((yex & ~y=0) -> Ez y=zu{z})).
I think that works as follows:
Suppose x is a finite ordinal. Then the only thing that it isn't
obvious (or that we haven't already established) as holding on the
right hand side of the biconditional in the definition is:
~x=0 -> Ux e x.
But we can show that as follows (though maybe there's an easier way?):
If x is a finite ordinal, then Ux is a finite ordinal.
If x is a finite ordinal and ~x=0, then ~ x subset of Ux as follows by
induction on x:
If x=1 then ~x subset of Ux.
Suppose ~x subset of Ux. Show ~ xu{x} subset of U(x u {x}).
U(x u {x}) = Ux u x. So we must show ~ xu{x} subset of Ux u x.
x e xu{x}. But if x e Ux u x, then x e Ux (since ~xex).
But if x e Ux, then, since Ux is an ordinal, we have x subset of Ux,
which contradicts the inductive hypothesis.
So ~x subset of Ux.
So, by trichotomy of ordinals, Ux is a proper subset of x.
So Ux e x.
To show the right hand side of the biconditional in the definition
entails that x is a finite ordinal, it suffices to show that x is not
a limit ordinal. And we have that since if x is a limit ordinal then
~x=0 and ~ x e Ux.
> > The way we had it, omega would qualify to be a natural number.
>
> I don't see how can that be???
Look at the old formulation, and you will see that w satisfies the
definition.
> I defined x is natural in the following manner ( the corrected
> definition ):
>
> x is a natural number <-> ( x is an ordinal & ( ~x=0 -> Uxex) &
> Ay( (yex & ~y=0) ->Ez( zex & y= zu{z}) ) ).
>
> Now with this definition , how can Omega satisfy it??
Right, with the CORRECTED version, w does not satisfy it. But with the
version BEFORE the corrected version (wihch is what I was talking
about), w satisfies it.
> This corrected formula of mine, is a simple formula, and it is simpler
> than the one you gave.
"Simpler" doesn't have a formal determination, but look at how we
simple it is to read one of the versions I gave:
x is 0 or (x is a sucessor ordinal and every member of x is either 0
or a successor).
That's surely no more complicated than your:
x is an ordinal and if x is not 0 then the union of x is a member of
x, and every member of x that is not 0 is equal to the successor of
some member of x.
MoeBlee
MoeBlee
Pshaw. Your definition has the clause:
~x=0 -> x e Ux.
which is equivalent to
x=0 v x e Ux
just as mine also treated x=0 as part of a disjunction.
There's no substantive difference there. It's just a matter of which
sentential equivalent we chose.
> So though the two definitions are equivalent yet I think mine is more
> meaningful.
Your continual claims of "better", "simpler", "more intuitive", etc.,
now "more meaningful" about your various proposals are irritating,
especially when they so often turn out not to be the case.
MoeBlee
zuhair
hahaha............, I was just jocking here.
I know that they are equivalent!
>
> MoeBlee
Herman Rubin
No problem.
(A X)(X subset P(F) -> (E Y)(Y \in X & (A Z) (Z \in X
-> ~(Y proper subset Z))))
>> 2. Every nonempty family of subsets of F,
>> ordered by inclusion, has a subset-maximal
>> element.
>The formula please.
Add to the condition on X the condition that
(A Y)(A Z)((Y \in X & Z \in X) -> (Y subset Z or Z subset Y))
>> 3. No proper subset of the power set of F
>> has the same cardinality as the power
>> set of F.
>> 4. No proper subset of F has the same
>> cardinality as F.
>This is Dedekind's finite.
>> 5. F is not the union of two disjoint
>> nonempty sets, each of which has the
>> same cardinality as F.
>5 is actually my definition that I have presented in this thread and
>that many people in this group say that they prove that it is
>equivalent to 4 ( i.e. Dedekind's ).
This is not so. If there is a Dedekind finite
set which is not finite in the strongest form,
the union of a transfinite (at least aleph_0)
set and that does not satisfy 5.
I have not yet had a chance to read Tarski's paper.
But I would suggest you read the book by Paul Howard
and Jean E. Rubin, and you might look up papers by
Norbert Brunner and others who have discussed these
problems. I know that Brunner discusses in one paper
seven types of Dedekind finite infinite sets.
--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hru...@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
zuhair
> In article <1178375530.054046.82...@n76g2000hsh.googlegroups.com>,
No please this is an important point. One should be clear about it.
The question is: do the following definition of 'x is ordinal' is
equivalent to 5, or equivalent to 4?
x is finite <-> Ayz((y subset_of x & y/\z=0 & ~z=0)->
yUz supernumerous to y)).
Both John Maurera and Moe Blee claim that it is equivalent to 4
( i.e.Dedekind's).
And they have presented proofs of their claim, even you yourself had
said that it is equivalent to 4.
While to me it looks equivalent to 5. Which is stated to be distinct
from 4.
And by then it should be distinct from 4.
Can you elaborate on this point please.
Zuhair
MoeBlee
> ( i.e.Dedekind's).
Just for the record, mine had some typos. Here it is cleaned up (as if
anyone needs such an obvious proof):
Theorem:
x is Dedekind infinite
<->
Eyz(y subset of x & ~ z=0 & z/\y=0 & ~ y strictly dominated by zuy)
Proof:
Suppose x is Dedekind infinite.
So let y proper subset of x and y equinumerous with x.
So x\y subset of x and ~ x\y = 0 and (x\y)/\y = 0.
But y equinumerous with (x\y)uy. So ~ y strictly dominated by (x\y)uy.
Let y subset of x and ~ z=0 and z/\y = 0 and ~ y strictly dominated by
zuy.
So either ~ y dominated by zuy or y equinumerous with zuy.
But y dominated by zuy. So y equinumerous with zuy.
Since ~ z=0 and z/\y = 0, we have y proper subset of zuy.
So zuy is Dedekind infinite.
So, since y equinumerous with zuy, we have y is Dedekind infinite.
So, since x subset of y, we have x is Dedekind infinite.
MoeBlee
MoeBlee
Oops, the last line should be:
So, since y subset of x, we have x is Dedekind infinite.
MoeBlee
zuhair
> On May 8, 3:47 pm, zuhair <zaljo...@yahoo.com> wrote:
>
> > Both John Maurera and Moe Blee claim that it is equivalent to 4
> > ( i.e.Dedekind's).
>
> Just for the record, mine had some typos. Here it is cleaned up (as if
> anyone needs such an obvious proof):
>
> Theorem:
> x is Dedekind infinite
> <->
> Eyz(y subset of x & ~ z=0 & z/\y=0 & ~ y strictly dominated by zuy)
>
> Proof:
> Suppose x is Dedekind infinite.
> So let y proper subset of x and y equinumerous with x.
> So x\y subset of x and ~ x\y = 0 and (x\y)/\y = 0.
> But y equinumerous with (x\y)uy. So ~ y strictly dominated by (x\y)uy.
So far so good. this proves that: x is Dedekind infinite -> x is z-
infinite.
that's correct.
>
> Let y subset of x and ~ z=0 and z/\y = 0 and ~ y strictly dominated by
> zuy.
> So either ~ y dominated by zuy or y equinumerous with zuy.
> But y dominated by zuy. So y equinumerous with zuy.
> Since ~ z=0 and z/\y = 0, we have y proper subset of zuy.
> So zuy is Dedekind infinite.
Yes.
> So, since y equinumerous with zuy, we have y is Dedekind infinite.
Strange??? that's what we want to prove isn't it?
Why y is Dedekind infinite?? I don't understand?
Your argument is vague, you should show that there exist a proper
subset of y that is equinumerous with y in order to prove that y is
Dedekind infinite, did you show that??? Perhpas, I don't know.
MoeBlee
> On May 8, 7:25 pm, MoeBlee <jazzm...@hotmail.com> wrote:
> > Theorem:
> > x is Dedekind infinite
> > <->
> > Eyz(y subset of x & ~ z=0 & z/\y=0 & ~ y strictly dominated by zuy)
>
> > Proof:
> > Suppose x is Dedekind infinite.
> > So let y proper subset of x and y equinumerous with x.
> > So x\y subset of x and ~ x\y = 0 and (x\y)/\y = 0.
> > But y equinumerous with (x\y)uy. So ~ y strictly dominated by (x\y)uy.
>
> So far so good. this proves that: x is Dedekind infinite -> x is z-
> infinite.
> that's correct.
>
> > Let y subset of x and ~ z=0 and z/\y = 0 and ~ y strictly dominated by
> > zuy.
> > So either ~ y dominated by zuy or y equinumerous with zuy.
> > But y dominated by zuy. So y equinumerous with zuy.
> > Since ~ z=0 and z/\y = 0, we have y proper subset of zuy.
> > So zuy is Dedekind infinite.
>
> Yes.
>
> > So, since y equinumerous with zuy, we have y is Dedekind infinite.
>
> Strange???
What is strange???
> that's what we want to prove isn't it?
It's not what we want to prove finally but you can see by the end of
the proof that it's a step toward what we want to prove, can't you?
> Why y is Dedekind infinite??
What don't you understand about a set being Dedekind infinite if it is
equinumerous with a Dedekind infinite set??
> I don't understand?
You don't?
> Your argument is vague, you should show that there exist a proper
> subset of y that is equinumerous with y in order to prove that y is
> Dedekind infinite, did you show that???
What don't you understand about a set being Dedekind infinite if it is
equinumerous with a Dedekind infinite set???
> Perhpas, I don't know.
Obviously, you don't.
> > So, since x subset of y, we have x is Dedekind infinite.
So why don't you just read the proof and ask whatever legitimate
questions you have about it instead of asking unnecessary questions as
a childishly petulant way of acting out your umbrage regarding an
exchange we had in another thread??? [Should that have been three
question marks or just two???]
MoeBlee
MoeBlee
Theorem:
x is Dedekind infinite
<->
Eyz(y subset of x & ~ z=0 & z/\y=0 & ~ y strictly dominated by zuy)
Theorem:
Ebc(~ b=0 & ~ c=0 & b/\c=0 & b equinumerous with x equninumerous with
c & x = buc)
->
x is Dedekind infinite
Theorem:
x is Dedekind infinite
->
x is infinite
Theorem (with axiom of denumerable choice):
x is infinite
->
x is Dedekind infinite
Theorem (with axiom of choice):
x is infinite
->
Ebc(~ b=0 & ~ c=0 & b/\c=0 & b equinumerous with x equninumerous with
c & x = buc)
But, without the axiom of choice, why do you think there is a proof of
the following?:
x Dedekind infinite
->
Ebc(~ b=0 & ~ c=0 & b/\c=0 & b equinumerous with x equninumerous with
c & x = buc)
MoeBlee
MoeBlee
By the way, just to be clear, I had corrected that last line
"So, since x subset of y, we have x is Dedekind infinite."
to:
"So, since y subset of x, we have x is Dedekind infinite."
MoeBlee
zuhair
zuhair
>
>
>
>
>
> > On May 8, 7:25 pm, MoeBlee <jazzm...@hotmail.com> wrote:
>
> So why don't you just read the proof
I read your proof.
You maintain that:
Axy ( (x equinumerous y & y is Dedekind infinite) -> x is Dedekind
infinite ).
Mind you I am not talking in Z, I am talking
about something in all set theories that allow for it.
This is clear in Z , but I don't know if it is the case in NFU,
or any other set theory that doesn't have separation.
That's why I asked you about the proof of this statement,
to see if it is workable in other theories.
Zuhair
zuhair
What I said is that I believe that:
Eyz(y subset of x & ~ z=0 & z/\y=0 & ~ y strictly dominated by zuy)
<->
Ebc(~ b=0 & ~ c=0 & b/\c=0 & b equinumerous with x equninumerous with
c & x = buc).
They look equivalent to me.
Zuhair
zuhair
>
> What I said is that I believe that:
>
> Eyz(y subset of x & ~ z=0 & z/\y=0 & ~ y strictly dominated by zuy)
>
> <->
>
> Ebc(~ b=0 & ~ c=0 & b/\c=0 & b equinumerous with x equninumerous with
> c & x = buc).
>
> They look equivalent to me.
No, they are not, you are right we need choice to prove this
equivalence.
However Moe, just contemplate these definitions.
1) x is infinite <-> Ey( y/\x=0 & y equinumerous to x & ~y=x &
yux equinumerous to x ).
2) x is infinite <-> Ey( ~y=x & y equinumerous to x & yux
equinumerous to x ).
3) x is infinite <-> Ebc(~ b=0 & ~ c=0 & b/\c=0 & b equinumerous with
x equninumerous with c & x = buc).
4) x is Dedekind infinite.
Two questions:
Without any form of choice:
Is 1 is equivalent to 3 or 4 ?
Is 2 is equivalent to 3 or 4 ?
Zuhair
MoeBlee
You don't need to plaster a bunch of captiousness and "???" just to
ask me whether a line in my proof also works in NFU. You know, I
figure if NFU is really what you wanted to know about, then you would
have SAID that. So I don't appreciate your disingenuousness now as you
claim that your pique was just a way of eliciting me to say something
NFU that you didn't even mention.
When I state a theorem or proof, it is in Z-R and if I do use
regularity, replacement, or any version of choice, then I'll mention
that. And, again, as I ALREADY told you, I don't know enough about NF
or NFU to be able to say whether my proofs also work in those or
other theories.
MoeBlee
MoeBlee
> 1) x is infinite <-> Ey( y/\x=0 & y equinumerous to x & ~y=x &
> yux equinumerous to x ).
>
> 2) x is infinite <-> Ey( ~y=x & y equinumerous to x & yux
> equinumerous to x ).
>
> 3) x is infinite <-> Ebc(~ b=0 & ~ c=0 & b/\c=0 & b equinumerous with
> x equninumerous with c & x = buc).
>
> 4) x is Dedekind infinite.
>
> Two questions:
>
> Without any form of choice:
>
> Is 1 is equivalent to 3 or 4 ?
> Is 2 is equivalent to 3 or 4 ?
What happens when you try to prove or disprove those equivalences?
MoeBlee
Herman Rubin
zuhair <zalj...@yahoo.com> wrote:
>On May 8, 12:59 pm, hru...@odds.stat.purdue.edu (Herman Rubin) wrote:
>> In article <1178375530.054046.82...@n76g2000hsh.googlegroups.com>,
>> zuhair <zaljo...@yahoo.com> wrote:
>> >On May 4, 9:11 am, "Dave L. Renfro" <renfr...@cmich.edu> wrote:
>> >> Dave L. Renfro wrote:
>> >> 4. No proper subset of F has the same
>> >> cardinality as F.
>> >This is Dedekind's finite.
>> >> 5. F is not the union of two disjoint
>> >> nonempty sets, each of which has the
>> >> same cardinality as F.
>> >5 is actually my definition that I have presented in this thread and
>> >that many people in this group say that they prove that it is
>> >equivalent to 4 ( i.e. Dedekind's ).
>> This is not so. If there is a Dedekind finite
>> set which is not finite in the strongest form,
>> the union of a transfinite (at least aleph_0)
>> set and that does not satisfy 5.
>No please this is an important point. One should be clear about it.
>The question is: do the following definition of 'x is ordinal' is
>equivalent to 5, or equivalent to 4?
>x is finite <-> Ayz((y subset_of x & y/\z=0 & ~z=0)->
>yUz supernumerous to y)).
This has nothing to do with the definition of ordinal.
It is equivalent to Dedekind finite.
>Both John Maurera and Moe Blee claim that it is equivalent to 4
>( i.e.Dedekind's).
>And they have presented proofs of their claim, even you yourself had
>said that it is equivalent to 4.
>While to me it looks equivalent to 5. Which is stated to be distinct
>from 4.
>And by then it should be distinct from 4.
To show how strong 5 is, let A be Dedekind finite and B
be an infinite well-ordered set disjoint from A. Then
since B is infinite (all definitions agree for well-ordered
sets), we would like A U B to be infinite. So by 5, consider
breaking up A U B into two sets, C U D and E U F, where
A = C U E and B = D U F. But if A is Dedekind finite,
we cannot have C and E having the same cardinality as
A, and they would have to differ from A by at most a finite
(well-ordered) set. So we have the contradiction.
Statement 5 is equivalent to all infinite cardinals being
equal to their doubles; Shelah showed that this is weaker
than the Axiom of Choice, but it is quite strong.
zuhair
>
> Statement 5 is equivalent to all infinite cardinals being
> equal to their doubles; Shelah showed that this is weaker
> than the Axiom of Choice, but it is quite strong.
So what you are actually saying is that 5 is equivalent to the
following statement:
x is infinite <-> Ey( y.x=0 & y equinumerous to x & ~y=x &
yux equinumerous to x ).
In words: a set x is said to be infinite if and only if there exist a
set y that is disjoint from x and equinumerous to x and different from
x, such that
y u x is equinumerous to x.
This is simply stating that every non empty set that equal its double
is infinite, and every set that is either empty or non empty set but
smaller than its double is finite.
If this is weaker than axiom of choice, then what it is called,
Axiom of cardinal choice. or 'CC'.
What would be the effect of replacing choice with this axiom in ZFC
such as to have ZFCC.
Several questions presents themselfs here:
1) in ZFCC is Dedekindian finite equivalent to the subset maximal
definition of finite?
2) in ZFCC is the subset maximal definition of finite equivalent to
CC?
Actually even a better question that sums them all is: would all five
definitions mentioned by Dave L.Renfro ( see above ), be equivalent
in ZFCC.
The last question , why CC is weaker than choice?
Zuhair
Dave L. Renfro
a reference to a 1924 paper by Tarski in which
several variations of "finite" are defined.
The first post below gives the referrence and
the second post below gives (in English) 5 different
variations that Tarski discusses.
http://mathforum.org/kb/message.jspa?messageID=5680068
http://groups.google.com/group/sci.math/msg/2c17388f7e33a3fe
http://mathforum.org/kb/message.jspa?messageID=5691673
http://groups.google.com/group/sci.math/msg/35703c6cdeb6aa77
In the second post I wrote:
> Tarski gives these in his 1924 paper, according to
> Moore, and Tarski proves 1 implies 2 implies 3
> implies 4 implies 5, where neither the axiom of
> choice nor the axiom of infinity is assumed.
>
> Tarski wasn't able to show that the notions are
> distinct (in the absence of the axiom of choice
> and the axiom of infinity), but Moore says that
> mathematicians later managed to show that all
> 5 of these are distinct notions of "finite".
The other day I happened to come across a paper,
in English and on the internet, where I think some
(all?) of these notions are proved to be distinct
(in some, presumably large, fragment of ZF).
A. Levy, "The independence of various definitions
of finiteness", Fundamenta Mathematicae 46 (1958), 1-13.
http://matwbn.icm.edu.pl/ksiazki/fm/fm46/fm4611.pdf
Dave L. Renfro
MoeBlee
> The other day I happened to come across a paper,
> in English and on the internet, where I think some
> (all?) of these notions are proved to be distinct
> (in some, presumably large, fragment of ZF).
>
> A. Levy, "The independence of various definitions
> of finiteness", Fundamenta Mathematicae 46 (1958), 1-13.http://matwbn.icm.edu.pl/ksiazki/fm/fm46/fm4611.pdf
Great! Thanks, Dave. I can definitely make use of that! The Tarski
paper is unusable for me since, unfortunately, I don't know French;
I've always wished to see it translateed or some other article in
English like it. A lot of the stuff is in Suppes's textbook, but the
paper you've linked to has even more.
MoeBlee