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The list of all natural numbers don't exist
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albstorz@gmx.de
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Joined: 11 Sep 2005
Posts: 241

PostPosted: Fri Jul 21, 2006 1:45 pm    Post subject: Re: The list of all natural numbers don't exist Reply with quote

David R Tribble schrieb:

Quote:
Albrecht Storz wrote:
If you don't talk about infinity = endlessness = impossibility of
completion, use another word or symbol.


William Hughes schrieb:
Why? Do you think this nonsense is the usual definition of infinity?


Albrecht Storz wrote:
Nonsense? In what kind of world do you live in?

Let's see:

You claim: "An infinite set is by definition a set that can be bijected
to a proper subset."
Yes, that's the usual definition.

But I think, infinite sets are at first a special kind of sets which
have the propertie to be infinite. I'm right?

Depends on what the "property of being infinite" means.
Do you mean that infinite sets have the property of being infinite?
Seems like a pretty vacuous definition to me.

??? Definition? Why?

Quote:

Can we also say that finite sets have the property of being finite,
and that round sets have the property of being round, and that red
sets have the property of being red?


So please tell me your definition of the propertie "infinite".

How about:
Set S is infinite if it is not finite.

Not good.

Quote:
Set S is finite if there exists a bijection between it and some
natural n.

For me, this is a very good definition for sets. If there is no
bijection, it's not a set.

Quote:

Or, more simply:
A set is infinite if there does not exist a bijection between it and
any natural.

Sounds like: if a unicorn is invisible, you can't see it.

Quote:

There are other, more formal definitions, of course.


I hope so.

Best regards
Albrecht S. Storz
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Lady Musketeer
science forum beginner


Joined: 10 Jul 2006
Posts: 3

PostPosted: Fri Jul 21, 2006 1:44 pm    Post subject: Re: The list of all natural numbers don't exist Reply with quote

hello,
I dont know but from your logic if the Subject is true
then
logically then the list of all
emails on this thread also does not exist .

``````LM

stephen@nomail.com wrote:
Quote:
Dave Seaman <dseaman@no.such.host> wrote:
On Tue, 27 Jun 2006 09:40:48 EDT, phyti wrote:
Those who say the list does not exist are correct.
By definition (ZF, Peano, etc.), the natural (basic)
integers are constucted using an infinite process.

The axioms don't say anything about "construction" or "processes". An
infinite set is postulated to exist. That is all.

The idea of "construction" and "processes" seems to be
at the heart of many people's misconceptions about set
theory. Or perhaps more accurately, it is at the heart
of their view of mathematics and makes it difficult,
if not impossible to understand the standard position.

Based on the arguments used by people like Russell, Tony
and others, their idea of sets is based on computer
implementations. Sets are containers to which objects
can be added, or from which objects can be removed.
Sets are created by set operations. Given the sets A and B,
A union B "creates" a new set, or perhaps it modifies
A or B.

I am wondering if they think the same away about numbers?
My guess would be no, but who knows? For example, given
the following code
x=7;
x=x-2;
I do not think they would think we changed 7. x had
the value of 7, and after the operation x has the value
of 5, but 5 and 7 did not change, and they did not
need to be "created". But given the code
x={ 2, 3, 5, 7, 11 }; // set assignment
x=x - { 2 }; // set difference
they think the set x has been modified. Instead
of seeing x as a variable whose value has changed
from one set to another set, they see x as the set,
and operations on x change the set. In the standard
view the set { 2, 3, 5, 7, 11 } still "exists" after the operation,
it is just the case that x no longer has that value.

Stephen
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albstorz@gmx.de
science forum Guru Wannabe


Joined: 11 Sep 2005
Posts: 241

PostPosted: Fri Jul 21, 2006 12:52 pm    Post subject: Re: The list of all natural numbers don't exist Reply with quote

Patricia Shanahan schrieb:

Quote:
albstorz@gmx.de wrote:
Virgil schrieb:

In article <_eydndK1Kp9-jyLZnZ2dnUVZ_uudnZ2d@comcast.com>,
"Russell Easterly" <logiclab@comcast.net> wrote:

"Christian Stapfer" <nil@dev.nul> wrote in message
news:99646$44befd7b$54497364$21350@news.hispeed.ch...
Peter Niessen wrote:
Am Mon, 17 Jul 2006 15:24:56 -0700 schrieb Russell Easterly:

If N is the set of all natural numbers, the number of subsets of N
is the powerset of N. Does this mean we need an uncountable
number of axioms to define all the subsets of N?
Um, it depends on what you mean by "define"...

\forall m P(m) is a Set.
Short enought?
That axiom does not give you *all* the subsets
of NI. It just gives you the "existence" of the
set of all subsets of N. Which is rather an
empty shell of a mere tag, in and of itself.
If your set theory is consistent at all, it
admits of having a merely countable interpretation
(Skolem's paradox). So there is something illusory
about your suggestion that the powerset axiom does
the job of *really* capturing an "uncountable"
P(NI) in its entirety.

Nearly all the elements of P(N) can't be defined.
Nearly all real numbers are inaccessible, too.

But not the ones we need.

Who needs the ones which are inaccesible?

AS


Anyone who wants a system simple enough to allow general statements?



Ockham's razor blade seems to be out of the time.


Best regards
Albrecht S. Storz
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Aatu Koskensilta
science forum Guru Wannabe


Joined: 17 May 2005
Posts: 277

PostPosted: Fri Jul 21, 2006 11:57 am    Post subject: Re: The list of all natural numbers don't exist Reply with quote

David R Tribble wrote:
Quote:
Russell Easterly wrote:
Nearly all the elements of P(N) can't be defined.

We can _define_ all the elements of P(N). (E.g., using a simple
bijection between the binary real fractions in [0,1) and the subsets
of N.)

What notion of defining you have in mind here?

--
Aatu Koskensilta (aatu.koskensilta@xortec.fi)

"Wovon man nicht sprechen kann, daruber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
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Dave Seaman
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Joined: 24 Mar 2005
Posts: 527

PostPosted: Fri Jul 21, 2006 11:29 am    Post subject: Re: The list of all natural numbers don't exist Reply with quote

On Thu, 20 Jul 2006 19:59:27 -0700, Russell Easterly wrote:

Quote:
David R Tribble wrote:
We can _define_ all the elements of P(N). (E.g., using a simple
bijection between the binary real fractions in [0,1) and the subsets
of N.)

Russell Easterly wrote:

Binary real fractions?
ZF has binary real fractions?
Which axiom says there are real fractions, binary or otherwise.

All of the axioms together. The real numbers are a defined concept, and
they are defined in terms of set theory.

The reals are similar to the ordinals in that respect. Neither is
actually mentioned in the axioms. Both are defined concepts.

Quote:
David R Tribble wrote:
We can only _compute_ a countable number of those elements
(or reals), though.

Russell Easterly wrote:

Actually, you can only compute a finite number of them,
as I proved with my Zeno machine proof.
(More precisely, no single computation can compute
more than a finite number of naturals or more than a
finite number of digits in a real number.)

Nevertheless, a countable infinity of real numbers are computable. In
fact, there is a Turing machine that computes the identity function. For
each input n, the machine halts and returns the result n. This machine
enumerates the natural numbers, which are an infinite subset of the
reals.

Quote:
I really question definitions of computability that
say we can compute any irrational number.

Not every irrational number. Just a denumerable subset.

Quote:
Turing defined computable numbers in his original paper
as binary strings formed by TM's that never halt.
He realized it was impossible to compute a real number
with a finite number of operations.

Unfortunately for Turing, a sequential machine can't
ever compute an infinite string, even if the TM never halts.

But for any n, the machine can compute the n'th digit in finite time.
That's all the definition requires.

Quote:
It seems to me the Axiom of Separation doesn't
prevent the existence of any set, it only guarantees
that any predicate of set theory defines a
subset of a set.

Yes.

Quote:
How does the Axiom of Separation prevent the
existence of contradictory sets like Russell's
paradox or the set of all sets?

It doesn't. How could it? It has been explained to you repeatedly that
you cannot avoid inconsistencies by adding axioms.

Quote:
It is obvious ZF without the Axiom of Regularity
is inconsistent, even if set theorists claim the statement
"there exists a set that is a member of itself" is
undecidable in ZF-R. I can still form Russell's
paradox in ZF-R not to mention the Ordinal
of all Ordinals.

Ok, let's see your proof that the Russell "set" is actually a set in
ZF-R. If you can prove it in ZF-R, then you can also prove it in ZF.
Just use the same proof. It's not required that you use every axiom in
order for a proof to be valid, you know.

If you can prove a contradiction in ZF-R, you can prove the same
contradiction in ZF. Do you see why you can't avoid inconsistency by
adding axioms?

Quote:
Can I prove anything in ZF?
It seems ZF maintains consistency by making
every question undecidable.

Lots of questions are decidable in ZF. More, in fact, than in ZF-R or in
ZF-I.

Quote:
I find it hard to believe any axiomatic system
with an Axiom of Union and a definition
of "successor" can fail to prove every natural number
is finite. Yet, ZF-I says "every natural is finite" is undecidable.

Wrong. By definition, a natural number is a finite ordinal. Hence,
natural numbers are finite.

I think you mean, ZF-I can't prove that every *ordinal* is finite. That
is correct.

--
Dave Seaman
U.S. Court of Appeals to review three issues
concerning case of Mumia Abu-Jamal.
<http://www.mumia2000.org/>
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Russell Easterly
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Posts: 199

PostPosted: Fri Jul 21, 2006 2:59 am    Post subject: Re: The list of all natural numbers don't exist Reply with quote

"David R Tribble" <david@tribble.com> wrote in message
news:1153437864.293836.256990@b28g2000cwb.googlegroups.com...
Quote:
Peter Niessen wrote:
\forall m P(m) is a Set.
Short enought?


Christian Stapfer wrote:
That axiom does not give you *all* the subsets
of NI. It just gives you the "existence" of the
set of all subsets of N. Which is rather an
empty shell of a mere tag, in and of itself.
If your set theory is consistent at all, it
admits of having a merely countable interpretation
(Skolem's paradox). So there is something illusory
about your suggestion that the powerset axiom does
the job of *really* capturing an "uncountable"
P(NI) in its entirety.


Russell Easterly wrote:
Nearly all the elements of P(N) can't be defined.


David R Tribble wrote:
We can _define_ all the elements of P(N). (E.g., using a simple
bijection between the binary real fractions in [0,1) and the subsets
of N.)

Russell Easterly wrote:

Binary real fractions?
ZF has binary real fractions?
Which axiom says there are real fractions, binary or otherwise.

Quote:
David R Tribble wrote:
We can only _compute_ a countable number of those elements
(or reals), though.

Russell Easterly wrote:

Actually, you can only compute a finite number of them,
as I proved with my Zeno machine proof.
(More precisely, no single computation can compute
more than a finite number of naturals or more than a
finite number of digits in a real number.)

I really question definitions of computability that
say we can compute any irrational number.
Turing defined computable numbers in his original paper
as binary strings formed by TM's that never halt.
He realized it was impossible to compute a real number
with a finite number of operations.

Unfortunately for Turing, a sequential machine can't
ever compute an infinite string, even if the TM never halts.


It seems to me the Axiom of Separation doesn't
prevent the existence of any set, it only guarantees
that any predicate of set theory defines a
subset of a set.

How does the Axiom of Separation prevent the
existence of contradictory sets like Russell's
paradox or the set of all sets?

It is obvious ZF without the Axiom of Regularity
is inconsistent, even if set theorists claim the statement
"there exists a set that is a member of itself" is
undecidable in ZF-R. I can still form Russell's
paradox in ZF-R not to mention the Ordinal
of all Ordinals.

Can I prove anything in ZF?
It seems ZF maintains consistency by making
every question undecidable.
I find it hard to believe any axiomatic system
with an Axiom of Union and a definition
of "successor" can fail to prove every natural number
is finite. Yet, ZF-I says "every natural is finite" is undecidable.


Russell
- 2 many 2 count
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David R Tribble
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Joined: 21 Jul 2005
Posts: 1005

PostPosted: Thu Jul 20, 2006 11:24 pm    Post subject: Re: The list of all natural numbers don't exist Reply with quote

Peter Niessen wrote:
Quote:
\forall m P(m) is a Set.
Short enought?


Christian Stapfer wrote:
Quote:
That axiom does not give you *all* the subsets
of NI. It just gives you the "existence" of the
set of all subsets of N. Which is rather an
empty shell of a mere tag, in and of itself.
If your set theory is consistent at all, it
admits of having a merely countable interpretation
(Skolem's paradox). So there is something illusory
about your suggestion that the powerset axiom does
the job of *really* capturing an "uncountable"
P(NI) in its entirety.


Russell Easterly wrote:
Quote:
Nearly all the elements of P(N) can't be defined.

We can _define_ all the elements of P(N). (E.g., using a simple
bijection between the binary real fractions in [0,1) and the subsets
of N.)

We can only _compute_ a countable number of those elements
(or reals), though.
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David R Tribble
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Joined: 21 Jul 2005
Posts: 1005

PostPosted: Thu Jul 20, 2006 11:22 pm    Post subject: Re: The list of all natural numbers don't exist Reply with quote

Zuhair wrote:
Quote:
My objection to Cantor's cardinality is that it doesn't take into
consideration how a set is generated, that is my main objection, you
thought I objected to Cantor's cardinality because it is
counter-intuitive, and this is not may main objection, in reality
z-cardinality is as counter-intuitive as Cantor's cardinality, so I
will address it again , my main objection is that Cantor's cardinality
doesn't take how a set is generated into consideration.

And how is a set "generated"?

I think you mean to say that your Z-cardinality is some kind of
comparison between the enumeration of the members of two sets.
You say the members of set B are "generated" from the members
of set A, and that this "generating" relation defines the relative
Z-cardinalities of both sets.

In actually, this is nothing more than comparing how the members of
two sets are counted as they are paired (mapped) with each other,
giving you some sort of relative "measure" of the sets with respect
to that particular mapping. It's not cardinality, nor set "size", but
something related to denumerating ("generating") the elements.
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Markus2
science forum beginner


Joined: 20 Jul 2006
Posts: 1

PostPosted: Thu Jul 20, 2006 8:27 pm    Post subject: Re: The list of all natural numbers don't exist Reply with quote

"Albrecht"
Quote:
Assumption: The list of all natural numbers exists.

lol wasn das fürn Mörderthread hier =)
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Peter Niessen
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Joined: 05 Jun 2006
Posts: 19

PostPosted: Thu Jul 20, 2006 8:19 pm    Post subject: Re: The list of all natural numbers don't exist Reply with quote

Am Thu, 20 Jul 2006 05:50:17 +0200 schrieb Christian Stapfer:

Quote:
Peter Niessen wrote:
Am Mon, 17 Jul 2006 15:24:56 -0700 schrieb Russell Easterly:

If N is the set of all natural numbers, the number of subsets of N
is the powerset of N. Does this mean we need an uncountable
number of axioms to define all the subsets of N?

Um, it depends on what you mean by "define"...


\forall m P(m) is a Set.
Short enought?

That axiom does not give you *all* the subsets
of NI.

No Problem. I have only said:
If m a Set than is P(m) also a Set. Thinking about konsquenzes is an other
Story.
--
Mit freundlichen Grüssen
Peter Nießen
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zuhair
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Joined: 06 Feb 2005
Posts: 533

PostPosted: Thu Jul 20, 2006 7:46 pm    Post subject: Re: The list of all natural numbers don't exist Reply with quote

Tez wrote:
Quote:
zuhair wrote:

[snip]
I agree with you that the same set having many z-cardinalities
according to there generation is a counter-intutive idea, but it is not
incosistent and it doesn't violate any of the set theory axioms.

This paragraph should be saved for posterity. So now it's fine to come
up with definitions and formalisms that are counter-intuitive. I
agree. Note, though, that this is the only objection you have against
cardinality (that it's counter-intuitive, since you haven't found any
inconsistencies). This admittance makes you seem rather crankish.

Thanks alot.

Please try to snip content that isn't referred to or isn't relevant to
your replies.

My objection to Cantor's cardinality is that it doesn't take into
consideration how a set is generated, that is my main objection, you
thought I objected to Cantor's cardinality because it is
counter-intuitive, and this is not may main objection, in reality
z-cardinality is as counter-intuitive as Cantor's cardinality, so I
will address it again , my main objection is that Cantor's cardinality
doesn't take how a set is generated into consideration.

The only definition of "generational function" I can find is from
earlier, where you say:

Define: g:A ->B is called the "injective generational function"
of B from A, if g is injective and if B is defined in terms of g from
A. ie every member in B is a one valued function of a member
of A as determined by g.


But let me clarify that some of my objections are not to Cantor's
cardinality perse, but to your understanding of the essence of Cantor's
cardinality, you attribute properties to Cantor's cardinality that
cantor himself didn't state.

I'm sure there are many theorems and lemmas that Cantor did not prove,
state, or was even aware of that can be deduced from the things he
*did* state. Luckily for mathematics, we are allowed to use such
lemmas and theorems.

You stated : Cantor's cardinality depends on all injective functions
from one set to another.

How can you prove this statement ?

For instance let me denote {f:f is injective} :A ->B to mean the set of
all injective functions from A to B.

In order to my symbology easier lets say that F={f:f is injective }

so when I write F:A->B , it means the set of all injective functions
from A to B.

Now you say that determining Cantor's cardinality between these two
sets depends of F:A->B or F :B->A , now formally speaking there
should be some property P that these F should determine , and this
property should then determines the cantor cardinality of these two
sets.

Yes, ok. Using capital letters to denote the set of all injective
functions to and from 2 sets, consider F: A -> B, and G: B -> A.
If F is non-empty, we can say |A| <= |B|
If G is non-empty, we can say |B| <= |A|
If both F and G are non-empty, by the Cantor-Schroeder-Bernstein
theorem, |A| = |B|.

Satisfied?

In my model ( which is only an explanatory model ) I proposed that the
surjective state of every injective function from A to B or the
surjective state of every injective funciton from B to A , is that
property P that will determine the relative cardinalities between sets
A and B.

Hmm. I'm not sure this is phrased accurately (and *please* do not use
"cardinality" when you mean z-card). It seems to me, given previous
exposition on z-card, that the surjective state of *an* injective
function from A to B (or vice-versa) is the property that will
determine *a* z-card relation between sets A and B. Is that not more
precise?

how?

If all functions in F are surjective then this is called unanamous
surjective homomorphism
this means that F determines a positive state of surjection and this
positive state of surjection whould determine that Card A = Card B. (
this is not eqivalent to cantor's definition though).

while if F determines a negative state of surjection from A to B ( ie
every injective funciton in F is not surjective )( this is called
unanamous non surjective homomphism) then this negative state of
surjection determines that Card A < Card B ( this definition is
eqivalent to cantors cardinal inequality ).

So (simplifying the terms, if I may), If all injections are
surjections, we may say z-card A = z-card B (and *please* use "z-card"
when you mean z-card). And if all injections are non-surjective, we
may say z-card A < z-card B. What if some injections are surjections,
but some are non-surjective. Does this leave z-card undefined in those
circumstances?

I mean if a model like the one above is followed then I can understand
clearly that Cardinality is determined from the set of all injective
functions from A to B or the set of all injective functions from B to
A.

But let me tell you how Cantor's equality is determined using this
terminology.

Card A = Card B , if there is unanamous surjective homomphism XOR
there is heteromorphism in set F.

You're saying that |A| = |B| if every function injective from A to B is
bijective, XOR there exists 2 functions that are injective and one is
surjective and the other isn't. Is that correct?

Are you saying this is equivalent to the usualy definition of cardinal
equality (which relies only on the existence of one bijection)? Can
you show me how this is equivalent? Can you use your framing of
cardinality to show me whether |P| = |E|, where P is the set of all
prime numbers, and E is the set of all even naturals?

so what I wanted to say that Cantor's Equality of the infinite sets
especially doesn't in reality depend as you think on All injective
functions between set A and B. in Reality it is determined on the
EXISTANCE of ONE injective function that is bijective at the same time
( ie an injective function that is surjective at the same time ).

Well, that is the definition, but as I demonstrated above, it is
equivalent to considering all injective functions from A to B, and all
injective functions from B to A (again, Schroeder-Bernstein theorem).
That seems equitable.

Why I am saying all of that?

Because it was you how stated that z-cardinality is a weak form of
cardinality because it is determined by a single function that is the
generational funtion. and you think any definition of cardinality based
on a single function is not to be trused.

I didn't say it was a weak form of anything. I said that compared to
cardinality, it isn't an equivalence relation, it is non-unique, it can
be indeterminate depending on which sets you're considering. If you
want my to compare it to cardinailty *as a model of size*, at *best* I
would say it is lacking and incomplete. At worst, it is inconherent.

while you think that Cantor's cardinality is more rigorouselly defined
because it takes all the injective functions between A and B into
consideration and therefor you think it is more solid than
z-cardinality.

No. Z-card is less solid because half your terms are undefined, and
you haven't considered possibilities where some injective functions are
surjective, and some are not. Well, in some cases you have, and said
that this means that a set can have different sizes. This doesn't
model "size" well, since usually, we consider static objects to
maintain their "size".

I wanted to tell you that even Cantor's cardinality is determined by a
single function when it comes to determining equality.

It *can* be, because sometimes we are lucky enough to find such a
function. But it isn't necessary, and therefore doesn't *rely* on it.
And since I can do it without considering bijections at all, I have a
hard time seeing how it is *determined* by bijections. It is *defined*
by bijection, but the determination can and is done otherwise.

But I agree with you that Cantor's inequality is determined by all
injective functions between A and B.

And (using the Cantor-Schroeder-Bernstein theorem once again), I can
also determine equality this way.

what You can say is that Cantors inequality is stronger than z-cardinal
inequality because z-cardinality inequality is determined by single
function.

Well, that wasn't my objection. My objection was that consideration of
different functions between sets give differing z-cards. And since
sets are static objects, determined by their memership, I can't see how
z-card models size well.

but you cannot say that Cantor's equality is stronger than z-cardinal
equality since both are determined by a single function, in Cantor's it
is the bijective function while in z-cardinality it is the generational
function weather bijective or not.

Again, it is defined that way, and it isn't a single bijection, it can
be any bijection. But the determination can be totally devoid of
bijections, if you want.

I want to propose that equality in z-cardinality is less biased than
equality in Cantor's cardinality. That was my whole purpose.

Again, you will have to qualify this "bias". I don't see any "bias",
and even if there were some kind of "favouratism" going on, I don't see
how that has any bearing on anything. We are concerned with
consistency, usefulness, totality, simplicity, intuitiveness. I don't
see how "bias" impacts or is affected by any of these things.


Huh? You can postulate an example where every function between sets A,
B show the same z-card. I not disputing right or wrong here. I want
an example of, once given these 2 sets A and B, *how* you determine
that the z-card relation is the same for all functions between the
sets.

I will tell you the answer I was trying to get out. It's that you've
considered all the functions between the sets you're comapring.
Surely. Just like you would if you were thinking about cardinality.

Only when the generational function is not defined.

Right. So in cases where you don't define the generational function,
the only way to determine z-card is by considering all functions to and
from the sets you're comparing.


Assumption 2) for sets A and B if we don't know how one is generated
from the other then we assume that the bijective function between them
is the generational function, since bijectiion is the simplist
converjective function.

But what if a bijection doesn't exist between A and B? How do you know
one exists? Would it involve considering all functions between A and
B, perchance?

If bijection doesn't exist there is no need for assumption 2.

Huh? So if a bijection doesn't exist (and how do you determine that?)
between the set of regular polyhedra and primes, we don't use
assumption 2. We use assumption 1, the one about the peano postulates?
How does that apply to regular polyhedra and primes? How do you
determine their z-card reationship?


One last word, I want to confirm that z-cardinality is not personal,
the generational function is either mentioned in the question
explicitely or implicitelly.

example what is the z-card of a set of parents (in pairs_ mother and
father)and a set of there children, knowing that each parent should
have only one children.

The generational function here is similar to the floor fucntion, so
z-card children < z-card parents.

Implicit? Was that supposed to be a demonstration of how you determine
the "implicit" generational function?? It seems rather, uh, explicit
really. In fact, you mention it when you say knowing each parent-pair
has only one child.

Ok so I made a mistake , it is that kind of implicit that I was talking
about.

Oh. The explicit kind of implicit. Silly me. What if an explicit
generational function isn't given, and we are considering sets not
defined by the peano postulates?


I'd like to see a less obvious example. What is the implicit
generational function between the set of all primes, P, and the set of
all regular polyhedra, R?

Forget implicit.

Then show me how to explicitly construct a/the generational function.


While if I say for example that each parents whould have three children
then z-card children > z-card parents.

What if there are an infinite number of parents? What if I can come up
with a mapping between parent-pairs and children that's bijective?
Would I be allowed to say z-card children = z-card parents? If I find
an explicit function, does the implicit one trump it?

If the mapping is generational as explicitly mentioned in the question
given then yes z-card children =z-card parents

example if each pair of parents had two children , then this is a
bijective generational mapping , of coarse accordingly z-card children
= z-card parents.

what trumph is the generational fucntion explicity mentioned in the
question.

For example lets take the first question where each pair of parents had
only one child, so the generational function here is 2- to - 1 fucntion
from parents to children, here in that situation even if you find a
bijective function between parents and children , it has no
significance , because what determines the z-cardinality of children is
the generational fucntion which is not bijective here so z-card
children < z-card parents.

Only in conditions when say we don't explicitly now the relation
between parents and children , say for example we are confronted with a
set of parents and a set of there children but it is not mentioned in
the question how many children are there per pair of parents, then here
it is the bijective function that will be regareded as generational (
Assumption 2).

And how do you know that there even *exists* a bijection between the
two sets? What if there isn't a bijection?? How do apply assumption 2?

got it.

Apparently not.

[snip]

Please note also, that you still haven't generally defined generational
functions for arbitrary sets, so my objections still stand, whereas I
think i've clearly demonstrated that you can easily remove all bias
toward bijections when determining cardinality. Note further, I still
don't understand how one generational function is preferred over
another when we are comparing, say, the z-card of the naturals against
the z-card of the even naturals. Are you claiming at this point that a
set is only well-defined if we supply a function that generates it
(this needs defining anyway)?

-Tez

Ok , let us stop talking about Cantor's cardinality since this was not
the original subject.

Definitions:

Definitions:

For sets A and B.

1) f:A->B is called "one valued" if for every x in A there is one and
only one y=f(x) , in B.
so if n e A and m e A, then if n=m then f(n) must equal f(m). But if
n<>m then is doesn't imply that f(n) should be inequal to f(m).
Similarly if f(n) <> f(m) then n <> m is a must.
While if f(n) = f(m) then m=n is not a must.

2) f:A->B is called "Bijective" if it is one valued and its inverse
f^(-1):B->A is one valued.

3) f:A->B is called "converjective " if it is one valued and
surjective.

4) f:A->B is called "Generational" if it is converjective and
mentioned explicitely in the definition of B from A.

So if B is explicitely defined as B = { y: y=f(x) , f:A->B }
and if f:A->B is converjective then it is generational of B from A.

and the set B is expressed as B_f , it means set B as generated by
function f.

5) z-card A_h = z-card B_f , if there exist f:A-B that is generational
of B
from A_h. and that has a one valued inverse f^(-1):B->A .

6) z-card A_h> z-card B_f, if there exist f:A->B that is generational
of B
from A _h and that has an inverse f^(-1):B->A that is not one valued.

7) z-card A_f < z-card B_g, if there exist f:B->A that is generational
of A
from B_g and that has an inverse f^(-1): A->B that is not one valued.

8)Axiom 1) for sets A and B, among all the converjective functions
between them, one should be generational.

9) Assumption 1) If the set of all natural numbers N is not defined
explicitely by a generational function f, then it is to be considered
as the set which is primiarly derived from the Peano's five premisses
and it is to be symbolized as N' which means Primary natural numbers
set.

10) Assumption 2) for sets A and B , if the generating fucntions of A
and B, or at least the generating function of one of them from the
other, Is not defined, and there are converjective functions from one
of them to the other that differe in the existance of a one valued
inverse function ( for example g:A->B were g^(-1):B->A is not one
valued, and f:A->B were f^(-1): B->A is one valued ). Then It is the
bijective function that will be considered as the generational
function.

11) If A_f is not identical to B_f , then either A is a proper subset
of B or B is a proper subset of A. Provided of course that f = f. Then
sets A and B are called different equigenerational sets.

Example, the even numbers in the primary natural numbers set N',
denoted as E'( the primary even numbers set ).

12) if the generating funciton of set A is f(x)=x , f:A_g->A, then A_f
= A_g

and A_f is called the copy of A_f , this means that g=f.

Any suggestions.

Zuhair
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Patricia Shanahan
science forum Guru Wannabe


Joined: 13 May 2005
Posts: 214

PostPosted: Thu Jul 20, 2006 7:13 pm    Post subject: Re: The list of all natural numbers don't exist Reply with quote

Christian Stapfer wrote:
Quote:
"Patricia Shanahan" <pats@acm.org> wrote in message
news:OwKvg.2282$157.260@newsread3.news.pas.earthlink.net...
albstorz@gmx.de wrote:
Virgil schrieb:

In article <_eydndK1Kp9-jyLZnZ2dnUVZ_uudnZ2d@comcast.com>,
"Russell Easterly" <logiclab@comcast.net> wrote:

"Christian Stapfer" <nil@dev.nul> wrote in message
news:99646$44befd7b$54497364$21350@news.hispeed.ch...
Peter Niessen wrote:
Am Mon, 17 Jul 2006 15:24:56 -0700 schrieb Russell Easterly:

If N is the set of all natural numbers, the number of subsets of N
is the powerset of N. Does this mean we need an uncountable
number of axioms to define all the subsets of N?
Um, it depends on what you mean by "define"...

\forall m P(m) is a Set.
Short enought?
That axiom does not give you *all* the subsets
of NI. It just gives you the "existence" of the
set of all subsets of N. Which is rather an
empty shell of a mere tag, in and of itself.
If your set theory is consistent at all, it
admits of having a merely countable interpretation
(Skolem's paradox). So there is something illusory
about your suggestion that the powerset axiom does
the job of *really* capturing an "uncountable"
P(NI) in its entirety.

Nearly all the elements of P(N) can't be defined.
Nearly all real numbers are inaccessible, too.

But not the ones we need.
Who needs the ones which are inaccesible?

AS

Anyone who wants a system simple enough to allow
general statements?

Maybe, who knows. But where is your *proof* of that
assertion? (Besides: you only need to be able to
talk *as*if* you had those inaccessible ones, you
to not actually *need*, i.e. "access" them...)

If I'm allowed to talk as if they existed, what consequences, if any,
does their non-existence have?

Patricia
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Christian Stapfer
science forum beginner


Joined: 05 Jun 2006
Posts: 15

PostPosted: Thu Jul 20, 2006 6:55 pm    Post subject: Re: The list of all natural numbers don't exist Reply with quote

"Patricia Shanahan" <pats@acm.org> wrote in message
news:OwKvg.2282$157.260@newsread3.news.pas.earthlink.net...
Quote:
albstorz@gmx.de wrote:
Virgil schrieb:

In article <_eydndK1Kp9-jyLZnZ2dnUVZ_uudnZ2d@comcast.com>,
"Russell Easterly" <logiclab@comcast.net> wrote:

"Christian Stapfer" <nil@dev.nul> wrote in message
news:99646$44befd7b$54497364$21350@news.hispeed.ch...
Peter Niessen wrote:
Am Mon, 17 Jul 2006 15:24:56 -0700 schrieb Russell Easterly:

If N is the set of all natural numbers, the number of subsets of N
is the powerset of N. Does this mean we need an uncountable
number of axioms to define all the subsets of N?
Um, it depends on what you mean by "define"...

\forall m P(m) is a Set.
Short enought?
That axiom does not give you *all* the subsets
of NI. It just gives you the "existence" of the
set of all subsets of N. Which is rather an
empty shell of a mere tag, in and of itself.
If your set theory is consistent at all, it
admits of having a merely countable interpretation
(Skolem's paradox). So there is something illusory
about your suggestion that the powerset axiom does
the job of *really* capturing an "uncountable"
P(NI) in its entirety.

Nearly all the elements of P(N) can't be defined.
Nearly all real numbers are inaccessible, too.

But not the ones we need.

Who needs the ones which are inaccesible?

AS


Anyone who wants a system simple enough to allow
general statements?

Maybe, who knows. But where is your *proof* of that
assertion? (Besides: you only need to be able to
talk *as*if* you had those inaccessible ones, you
to not actually *need*, i.e. "access" them...)

Regards,
Christian
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Christian Stapfer
science forum beginner


Joined: 05 Jun 2006
Posts: 15

PostPosted: Thu Jul 20, 2006 6:52 pm    Post subject: Re: The list of all natural numbers don't exist Reply with quote

"Virgil" <virgil@comcast.net> wrote in message
news:virgil-475117.01100320072006@news.usenetmonster.com...
Quote:
In article <_eydndK1Kp9-jyLZnZ2dnUVZ_uudnZ2d@comcast.com>,
"Russell Easterly" <logiclab@comcast.net> wrote:

"Christian Stapfer" <nil@dev.nul> wrote in message
news:99646$44befd7b$54497364$21350@news.hispeed.ch...
Peter Niessen wrote:
Am Mon, 17 Jul 2006 15:24:56 -0700 schrieb Russell Easterly:

If N is the set of all natural numbers, the number of subsets of N
is the powerset of N. Does this mean we need an uncountable
number of axioms to define all the subsets of N?

Um, it depends on what you mean by "define"...


\forall m P(m) is a Set.
Short enought?

That axiom does not give you *all* the subsets
of NI. It just gives you the "existence" of the
set of all subsets of N. Which is rather an
empty shell of a mere tag, in and of itself.
If your set theory is consistent at all, it
admits of having a merely countable interpretation
(Skolem's paradox). So there is something illusory
about your suggestion that the powerset axiom does
the job of *really* capturing an "uncountable"
P(NI) in its entirety.


Nearly all the elements of P(N) can't be defined.

Nearly all real numbers are inaccessible, too.

But not the ones we need.

So, may I infer from this, that we do not *need*
"uncountably many" of anything? ;-)

Regards,
Christian
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tez_h@nospam.yahoo.com
science forum beginner


Joined: 27 Jun 2006
Posts: 9

PostPosted: Thu Jul 20, 2006 2:27 pm    Post subject: Re: The list of all natural numbers don't exist Reply with quote

zuhair wrote:

[snip]
Quote:
I agree with you that the same set having many z-cardinalities
according to there generation is a counter-intutive idea, but it is not
incosistent and it doesn't violate any of the set theory axioms.

This paragraph should be saved for posterity. So now it's fine to come
up with definitions and formalisms that are counter-intuitive. I
agree. Note, though, that this is the only objection you have against
cardinality (that it's counter-intuitive, since you haven't found any
inconsistencies). This admittance makes you seem rather crankish.

Thanks alot.

Please try to snip content that isn't referred to or isn't relevant to
your replies.

Quote:
My objection to Cantor's cardinality is that it doesn't take into
consideration how a set is generated, that is my main objection, you
thought I objected to Cantor's cardinality because it is
counter-intuitive, and this is not may main objection, in reality
z-cardinality is as counter-intuitive as Cantor's cardinality, so I
will address it again , my main objection is that Cantor's cardinality
doesn't take how a set is generated into consideration.

The only definition of "generational function" I can find is from
earlier, where you say:

Quote:
Define: g:A ->B is called the "injective generational function"
of B from A, if g is injective and if B is defined in terms of g from
A. ie every member in B is a one valued function of a member
of A as determined by g.


Quote:
But let me clarify that some of my objections are not to Cantor's
cardinality perse, but to your understanding of the essence of Cantor's
cardinality, you attribute properties to Cantor's cardinality that
cantor himself didn't state.

I'm sure there are many theorems and lemmas that Cantor did not prove,
state, or was even aware of that can be deduced from the things he
*did* state. Luckily for mathematics, we are allowed to use such
lemmas and theorems.

Quote:
You stated : Cantor's cardinality depends on all injective functions
from one set to another.

How can you prove this statement ?

For instance let me denote {f:f is injective} :A ->B to mean the set of
all injective functions from A to B.

In order to my symbology easier lets say that F={f:f is injective }

so when I write F:A->B , it means the set of all injective functions
from A to B.

Now you say that determining Cantor's cardinality between these two
sets depends of F:A->B or F :B->A , now formally speaking there
should be some property P that these F should determine , and this
property should then determines the cantor cardinality of these two
sets.

Yes, ok. Using capital letters to denote the set of all injective
functions to and from 2 sets, consider F: A -> B, and G: B -> A.
If F is non-empty, we can say |A| <= |B|
If G is non-empty, we can say |B| <= |A|
If both F and G are non-empty, by the Cantor-Schroeder-Bernstein
theorem, |A| = |B|.

Satisfied?

Quote:
In my model ( which is only an explanatory model ) I proposed that the
surjective state of every injective function from A to B or the
surjective state of every injective funciton from B to A , is that
property P that will determine the relative cardinalities between sets
A and B.

Hmm. I'm not sure this is phrased accurately (and *please* do not use
"cardinality" when you mean z-card). It seems to me, given previous
exposition on z-card, that the surjective state of *an* injective
function from A to B (or vice-versa) is the property that will
determine *a* z-card relation between sets A and B. Is that not more
precise?

Quote:
how?

If all functions in F are surjective then this is called unanamous
surjective homomorphism
this means that F determines a positive state of surjection and this
positive state of surjection whould determine that Card A = Card B. (
this is not eqivalent to cantor's definition though).

while if F determines a negative state of surjection from A to B ( ie
every injective funciton in F is not surjective )( this is called
unanamous non surjective homomphism) then this negative state of
surjection determines that Card A < Card B ( this definition is
eqivalent to cantors cardinal inequality ).

So (simplifying the terms, if I may), If all injections are
surjections, we may say z-card A = z-card B (and *please* use "z-card"
when you mean z-card). And if all injections are non-surjective, we
may say z-card A < z-card B. What if some injections are surjections,
but some are non-surjective. Does this leave z-card undefined in those
circumstances?

Quote:
I mean if a model like the one above is followed then I can understand
clearly that Cardinality is determined from the set of all injective
functions from A to B or the set of all injective functions from B to
A.

But let me tell you how Cantor's equality is determined using this
terminology.

Card A = Card B , if there is unanamous surjective homomphism XOR
there is heteromorphism in set F.

You're saying that |A| = |B| if every function injective from A to B is
bijective, XOR there exists 2 functions that are injective and one is
surjective and the other isn't. Is that correct?

Are you saying this is equivalent to the usualy definition of cardinal
equality (which relies only on the existence of one bijection)? Can
you show me how this is equivalent? Can you use your framing of
cardinality to show me whether |P| = |E|, where P is the set of all
prime numbers, and E is the set of all even naturals?

Quote:
so what I wanted to say that Cantor's Equality of the infinite sets
especially doesn't in reality depend as you think on All injective
functions between set A and B. in Reality it is determined on the
EXISTANCE of ONE injective function that is bijective at the same time
( ie an injective function that is surjective at the same time ).

Well, that is the definition, but as I demonstrated above, it is
equivalent to considering all injective functions from A to B, and all
injective functions from B to A (again, Schroeder-Bernstein theorem).
That seems equitable.

Quote:
Why I am saying all of that?

Because it was you how stated that z-cardinality is a weak form of
cardinality because it is determined by a single function that is the
generational funtion. and you think any definition of cardinality based
on a single function is not to be trused.

I didn't say it was a weak form of anything. I said that compared to
cardinality, it isn't an equivalence relation, it is non-unique, it can
be indeterminate depending on which sets you're considering. If you
want my to compare it to cardinailty *as a model of size*, at *best* I
would say it is lacking and incomplete. At worst, it is inconherent.

Quote:
while you think that Cantor's cardinality is more rigorouselly defined
because it takes all the injective functions between A and B into
consideration and therefor you think it is more solid than
z-cardinality.

No. Z-card is less solid because half your terms are undefined, and
you haven't considered possibilities where some injective functions are
surjective, and some are not. Well, in some cases you have, and said
that this means that a set can have different sizes. This doesn't
model "size" well, since usually, we consider static objects to
maintain their "size".

Quote:
I wanted to tell you that even Cantor's cardinality is determined by a
single function when it comes to determining equality.

It *can* be, because sometimes we are lucky enough to find such a
function. But it isn't necessary, and therefore doesn't *rely* on it.
And since I can do it without considering bijections at all, I have a
hard time seeing how it is *determined* by bijections. It is *defined*
by bijection, but the determination can and is done otherwise.

Quote:
But I agree with you that Cantor's inequality is determined by all
injective functions between A and B.

And (using the Cantor-Schroeder-Bernstein theorem once again), I can
also determine equality this way.

Quote:
what You can say is that Cantors inequality is stronger than z-cardinal
inequality because z-cardinality inequality is determined by single
function.

Well, that wasn't my objection. My objection was that consideration of
different functions between sets give differing z-cards. And since
sets are static objects, determined by their memership, I can't see how
z-card models size well.

Quote:
but you cannot say that Cantor's equality is stronger than z-cardinal
equality since both are determined by a single function, in Cantor's it
is the bijective function while in z-cardinality it is the generational
function weather bijective or not.

Again, it is defined that way, and it isn't a single bijection, it can
be any bijection. But the determination can be totally devoid of
bijections, if you want.

Quote:
I want to propose that equality in z-cardinality is less biased than
equality in Cantor's cardinality. That was my whole purpose.

Again, you will have to qualify this "bias". I don't see any "bias",
and even if there were some kind of "favouratism" going on, I don't see
how that has any bearing on anything. We are concerned with
consistency, usefulness, totality, simplicity, intuitiveness. I don't
see how "bias" impacts or is affected by any of these things.


Quote:
Huh? You can postulate an example where every function between sets A,
B show the same z-card. I not disputing right or wrong here. I want
an example of, once given these 2 sets A and B, *how* you determine
that the z-card relation is the same for all functions between the
sets.

I will tell you the answer I was trying to get out. It's that you've
considered all the functions between the sets you're comapring.
Surely. Just like you would if you were thinking about cardinality.

Only when the generational function is not defined.

Right. So in cases where you don't define the generational function,
the only way to determine z-card is by considering all functions to and
from the sets you're comparing.


Quote:
Assumption 2) for sets A and B if we don't know how one is generated
from the other then we assume that the bijective function between them
is the generational function, since bijectiion is the simplist
converjective function.

But what if a bijection doesn't exist between A and B? How do you know
one exists? Would it involve considering all functions between A and
B, perchance?

If bijection doesn't exist there is no need for assumption 2.

Huh? So if a bijection doesn't exist (and how do you determine that?)
between the set of regular polyhedra and primes, we don't use
assumption 2. We use assumption 1, the one about the peano postulates?
How does that apply to regular polyhedra and primes? How do you
determine their z-card reationship?


Quote:
One last word, I want to confirm that z-cardinality is not personal,
the generational function is either mentioned in the question
explicitely or implicitelly.

example what is the z-card of a set of parents (in pairs_ mother and
father)and a set of there children, knowing that each parent should
have only one children.

The generational function here is similar to the floor fucntion, so
z-card children < z-card parents.

Implicit? Was that supposed to be a demonstration of how you determine
the "implicit" generational function?? It seems rather, uh, explicit
really. In fact, you mention it when you say knowing each parent-pair
has only one child.

Ok so I made a mistake , it is that kind of implicit that I was talking
about.

Oh. The explicit kind of implicit. Silly me. What if an explicit
generational function isn't given, and we are considering sets not
defined by the peano postulates?


Quote:
I'd like to see a less obvious example. What is the implicit
generational function between the set of all primes, P, and the set of
all regular polyhedra, R?

Forget implicit.

Then show me how to explicitly construct a/the generational function.


Quote:
While if I say for example that each parents whould have three children
then z-card children > z-card parents.

What if there are an infinite number of parents? What if I can come up
with a mapping between parent-pairs and children that's bijective?
Would I be allowed to say z-card children = z-card parents? If I find
an explicit function, does the implicit one trump it?

If the mapping is generational as explicitly mentioned in the question
given then yes z-card children =z-card parents

example if each pair of parents had two children , then this is a
bijective generational mapping , of coarse accordingly z-card children
= z-card parents.

what trumph is the generational fucntion explicity mentioned in the
question.

For example lets take the first question where each pair of parents had
only one child, so the generational function here is 2- to - 1 fucntion
from parents to children, here in that situation even if you find a
bijective function between parents and children , it has no
significance , because what determines the z-cardinality of children is
the generational fucntion which is not bijective here so z-card
children < z-card parents.

Only in conditions when say we don't explicitly now the relation
between parents and children , say for example we are confronted with a
set of parents and a set of there children but it is not mentioned in
the question how many children are there per pair of parents, then here
it is the bijective function that will be regareded as generational (
Assumption 2).

And how do you know that there even *exists* a bijection between the
two sets? What if there isn't a bijection?? How do apply assumption 2?

Quote:
got it.

Apparently not.

[snip]

Please note also, that you still haven't generally defined generational
functions for arbitrary sets, so my objections still stand, whereas I
think i've clearly demonstrated that you can easily remove all bias
toward bijections when determining cardinality. Note further, I still
don't understand how one generational function is preferred over
another when we are comparing, say, the z-card of the naturals against
the z-card of the even naturals. Are you claiming at this point that a
set is only well-defined if we supply a function that generates it
(this needs defining anyway)?

-Tez
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