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The list of all natural numbers don't exist
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William Hughes
science forum Guru


Joined: 05 May 2005
Posts: 355

PostPosted: Wed Jul 19, 2006 3:03 am    Post subject: Re: The list of all natural numbers don't exist Reply with quote

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

In article <1153210672.175232.170890@35g2000cwc.googlegroups.com>,
albstorz@gmx.de wrote:

William Hughes schrieb:

albstorz@gmx.de wrote:
William Hughes wrote:

Why do you think that something without an end must
be incomplete or unfinished? What is wrong with saying that
a list of finite integers is complete if it contains every finite
integer?



Think!


Complete means that nothing is missing. A complete list of
finite integers is a list of integers that is not missing anything.

There is no infinite list which isn't missing nothing since infinity
means: incompletable.

Endless does not mean incompletable when using an axiom set like ZF or
NBG which specifically requires endless but completed sets.

Infinity means endlessness means incompletable processes or
incompletable considerations or incompletable extensions.

No it doesn't. What makes you think it does?


Quote:
You can't put a word out of his context and claim: now it's math and it
means what I want (and use just the aspects of its meaning, you like to
use).

If endlessness logical implies incompleteness it does imply it in the
usual understanding, in ZF and in NGB and in any other considerations.

And if my grandmother had wheels she'd be a bus. Endlessness does
not imply incompleteness.

Quote:

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


Why? Do you think this nonsense is the usual definition of
infinity?

-William Hughes
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albstorz@gmx.de
science forum Guru Wannabe


Joined: 11 Sep 2005
Posts: 241

PostPosted: Wed Jul 19, 2006 3:24 pm    Post subject: Re: The list of all natural numbers don't exist Reply with quote

William Hughes schrieb:

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

In article <1153210672.175232.170890@35g2000cwc.googlegroups.com>,
albstorz@gmx.de wrote:

William Hughes schrieb:

albstorz@gmx.de wrote:
William Hughes wrote:

Why do you think that something without an end must
be incomplete or unfinished? What is wrong with saying that
a list of finite integers is complete if it contains every finite
integer?



Think!


Complete means that nothing is missing. A complete list of
finite integers is a list of integers that is not missing anything.

There is no infinite list which isn't missing nothing since infinity
means: incompletable.

Endless does not mean incompletable when using an axiom set like ZF or
NBG which specifically requires endless but completed sets.

Infinity means endlessness means incompletable processes or
incompletable considerations or incompletable extensions.

No it doesn't. What makes you think it does?


You can't put a word out of his context and claim: now it's math and it
means what I want (and use just the aspects of its meaning, you like to
use).

If endlessness logical implies incompleteness it does imply it in the
usual understanding, in ZF and in NGB and in any other considerations.

And if my grandmother had wheels she'd be a bus. Endlessness does
not imply incompleteness.


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


Why? Do you think this nonsense is the usual definition of
infinity?


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

After this first step we may see if the propertie "infinite" make sense
in connectivity with sets (as wheels make sense in the connectivity
with grandmothers - or not).


Best regards
Albrecht S. Storz
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zuhair
science forum Guru


Joined: 06 Feb 2005
Posts: 533

PostPosted: Wed Jul 19, 2006 4:43 pm    Post subject: Re: The list of all natural numbers don't exist Reply with quote

Tez wrote:
Quote:
zuhair wrote:
Tez wrote:

This is a strange view. As I mention, cardinality requires that we
consider *all* functions between sets A and B, whereas yours depends
entirely on which functions we might be able to consider at the time.
This makes cardinality objective, and Z-card totally subjective.

This is a false idea the one you had their.

You think that Cantor's cardinality is based all the decision of all
injective functions between two sets.

I will prove to you that you are wrong.

Let us define two functions to be homomorphic if they are injective and
have the same surjective state.

ie if g:A->B and f:A->B are injective then if both are surjective or
both are not surjective then they are called homomorphic functions.

Ok. Eg, if g is injective and not surjective, and f is injective and
not surjective, they are "homeomorphic". If f and g are both
bijective, they are "homeomorphic".

Now I will define a cardinality called unanamouselly homomorphic
cardinality "uh-card"

1) uh-card A = uh-card B if all injective functions between A and B
are homomorphically surjective.

For example, if all functions between A and B are, say, injective and
and not surjective, let me say (I pressume this is what you mean) they
are all pairwise "homeomorphic". Then we can say uh-card A = uh-card
B. Note, this would not agree with cardinality.

2) uh-card A < uh-card B if all injective functions from A to B are
homomorphically non surjective

What? You now seem to be using "homeomorphic" in a different way. I
think you've now made the meaning of "homeomorphic" redundant. I
thought "homeomorphically surjective" would mean all funcs are
surjective, or all funcs are not surjective. What does
"homeomorphically non surjective" mean?

3) uh-card A> uh-card B if all injective functions from B to A are
homomorphically non surjective.

This would be a non biased defintion of cardinality based on the
unanamouse decision of all injective functions from one set to the
other.

You see that the definition of inequality here is exactly similar to
Cantors inequality.

But the definition of equality is not similar to Cantor's. In cantors
theorm it is biased to bijection. and doesn't take all the injective
functions into consideration as you thought.

So the definition is similar, and not similar. Which is it? Is it
both? I'm starting to think it's not set theory, or z-card that's
inconsistent, it's *you*!

In your definitions, you explicitly say "... if all injective
functions...", which should indicate to most people aged 8 to 110 that
we are considering "all injective functions". Do you not read what you
write? Perhaps you confused yourself when you said the definitions
were similar, and then started the very next paragraph with the words
"not similar".

And anyway, with cardinalities, we either prove the existence of a
function with certain properties, eg. we consider the set of all
functions between A and B and use that knowledge to work out which
equalities/inequalities are true between the 2 sets, or we consider the
set of all functions between A and B and prove that a
injective/surjective/bijective function does not exist in that set.

Again, the existence of different kinds of functions between sets A and
B never gives rise to contradictory cardinality comparisons, and to
prove or disprove the existence of some kind of function, we consider
the set of all functions betwen sets A and B.

So if I know for example that for sets A and B ther exist f:A->B that
is injective and not surjective then according to cantor this only
means that card A <=card B. so knoweldge of the surjective state of a
function is not enought to determine a strict inequality untill we see
what other funcitons say.

Exactly. We only know more *once we consider all functions* between A
and B. Is that not what I claimed before?? Cardinality requires that
we consider all functions. That we don't know enough to get strict
cardinality information if we *don't* consider all functions doesn't
contradict anything.

But notice that if I know that for sets A and B there exist f:A->B that
is injective and surjective, then here knoweldge of the surjective
function is enought to say that card A=card B. see the bias.

Huh? Bias? I don't understand. The existence of a bijection implies
(rather trivially) the existence of a surjection and an injection. The
existence of an injection from A to B let's us say |A| => |B|, and a
surjection, |B| => |A|. Wouldn't it be rather inconsistent, therefore,
if I couldn't say |A| = |B|? Are you claiming that it *should* be this
way?! I thought you were trying to eliminate inconsistencies!

Anyway. How did you find this bijection? You either constructed it
(plucked it out of the set of all functions between A and B), or you
proved it's existence indirectly (probably by considering the nature of
all functions between sets A and B).

The
knoweldge of a negative surjective state of an individual injective
function from A to B is not enought to make a strict determination of
there inequality, while knowledge of a positive surjective state of an
individual injective function from A to B IS enought to make a strict
determination of equality. see the bias man.

So your only objection to cardinality is that you are sometimes unable
to work out the existence of injections, surjections, or bijections
between sets? That seems rather, uh, lazy. And to then claim this is
some kind of failing of cardinality that your knowledge is incomplete
seems rather disingenuous.

I always say that Cantor's defintion for cardinal equality is not
congrous with its definition of cardinal inequality, the last depends
on unanamou homomphisim while the first not.

No. What you don't seem to understand is that bijection is a stronger
property than injection or surjection. It's stronger in the sense that
bijection implies injection, and bijection implies surjection. To know
that a bijection exists between sets is to inherently know more than if
you only know of an injection (if you didn't know whether it was
surjective or not). Also, you've probably done a lot more work and
proof if you've found a non-obvious bijection between 2
distantly-related sets, than had you only proved the existence of an
injective function, say.


So let me understand this. You find cardinality unintuitive. Given
you sets A and B above, and your function f: A -> B where f(x) = 2 *
floor(x/2), you say that we can see Z-card(A) > Z-card(B).

What if I consider the function g: A -> B where g(x) = 2x ? Then using
your definitions, Z-card(A) = Z-card(B). Not only is this unintuitive,
it seems rather inconsistent. Am I not allowed to consider this
function g? Maybe you mean to say Z-card_f(A) > Z-card_f(B) and
Z-card_g(A) = Z-card_g(B), where I've put the "generational function"
as a subscript to Z-card. But now all you've done is made a rather
case-by-case, subjective version of cardinality, and one that looks a
little inconsistent (especially if you stick to the strict inequality,
and expect people not to be confused).

*Perhaps* you think that the set B_f "generated" by your function f is
different from the set B_g "generated" by my function g. I am using
membership and the axiom of extensionality to determine set equality
(whereby B_f = B_g). How do you determine set equality?

bye the same why you do, by the axiom of extentionality. I am not
saying that set B_f is different from set B_g, I am only saying that
z-card B-f can be different from z-card B-g if f is different from g.
Though B is the same set in both cases . got it.


While if the defining function of B from A is for example f:A->B , f(x)
= 2x

Then z-card A = z-card B , since f^(-1):B->A is one valued.

Oh right. You realise everything I've just written. You don't find
that counterintuitive?

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.

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.

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.

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.


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.

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 ).

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.

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 ).

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.

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.

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

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

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

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.

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


Quote:
Note further, if we're keeping count, that cardinality is unique, and
is an equivalence relation. Very useful. Does your idea of size have
this feature? Is it as useful as cardinality?


But how do you know whether all converjective functions give the same
z-card? Can you give me an (non-trivial!) example proof where all
functions between 2 sets that show the same Z-card relation between
them? This seems a rather important point and distinction, and an
example may highlight everyone's misunderstandings.

I will tell you lateron. for the time being just assume it is right.

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.
Quote:


Ok, so I can't determine Z-card on arbitrary sets. Fine. But then you
must see that Z-card is far less useful than cardinality, since we lose
the fact that it is an equivalence relation, thus losing charateristics
that make it model "size" well. For example, consider the set N of
natural numbers, and P of primes. Using the usual definitions, |N| =
|P|. There is no ambiguity. How would you determine the Z-card
relation between N and P? Show us, if it isn't at least intuitive,
that Z-card is _useful_.

here you r right , that is why I am thinking of standarizing this
z-cardinality by assumptions as below:

Assumption 1) The set of all natural numbers N={0,1,2,3,4,.} if
mentioned without refereing to how it is generated then it is to be
assumed that it is generated directly from Peano's five axioms.

I'm not sure what I'm supposed to take from this. The only intersting
fact might be that you realise the successor function succ: N -> N-{0}
is bijective.

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.
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.
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.
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).

got it.
Quote:

Also the z-card of the even numbers inside N is less than N itself.
etc.

I'm still not clear why this is so. What is the implicit function
here? It seems to me that the implicit function f: N -> E between the
naturals, N, and the evens E would be f(n) = 2n. Since f, the implicit
generational function, is a bijection, I can only come to the
conclusion that z-card(N) = z-card(E).

So z-card are not personal and they are generalized by the assumptions
above.

Well, anything with the word "implicit" without a way of determining
how to make it explicit falls under "personal and subjective" in my
book.

see you later

-Tez
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Virgil
science forum Guru


Joined: 24 Mar 2005
Posts: 5536

PostPosted: Wed Jul 19, 2006 5:47 pm    Post subject: Re: The list of all natural numbers don't exist Reply with quote

In article <1153322686.631623.151980@m73g2000cwd.googlegroups.com>,
albstorz@gmx.de wrote:

Quote:
William Hughes schrieb:

albstorz@gmx.de wrote:
Virgil schrieb:

In article <1153210672.175232.170890@35g2000cwc.googlegroups.com>,
albstorz@gmx.de wrote:

William Hughes schrieb:

albstorz@gmx.de wrote:
William Hughes wrote:

Why do you think that something without an end must
be incomplete or unfinished? What is wrong with saying that
a list of finite integers is complete if it contains every
finite
integer?



Think!


Complete means that nothing is missing. A complete list of
finite integers is a list of integers that is not missing anything.

There is no infinite list which isn't missing nothing since infinity
means: incompletable.

Endless does not mean incompletable when using an axiom set like ZF or
NBG which specifically requires endless but completed sets.

Infinity means endlessness means incompletable processes or
incompletable considerations or incompletable extensions.

No it doesn't. What makes you think it does?


You can't put a word out of his context and claim: now it's math and it
means what I want (and use just the aspects of its meaning, you like to
use).

If endlessness logical implies incompleteness it does imply it in the
usual understanding, in ZF and in NGB and in any other considerations.

And if my grandmother had wheels she'd be a bus. Endlessness does
not imply incompleteness.


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


Why? Do you think this nonsense is the usual definition of
infinity?


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?

Why should you now be right when you are so used to being wrong.

Your definition is, as best circular, and therefore meaningless.

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

A set is infinite if it can be injected to a proper subset of itself.
Quote:

After this first step we may see if the propertie "infinite" make sense
in connectivity with sets

The mapping n -> n+1 injects N to a proper subset of N.

So OUR definition makes sense.
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Virgil
science forum Guru


Joined: 24 Mar 2005
Posts: 5536

PostPosted: Wed Jul 19, 2006 6:19 pm    Post subject: Re: The list of all natural numbers don't exist Reply with quote

In article <1153327390.221880.7720@m79g2000cwm.googlegroups.com>,
"zuhair" <zaljohar@yahoo.com> 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.

Since the axioms of every set theory in common usage say that a set is
entirely determined by what are and are not its members and is not in
any way dependent on how those members are "generated", Cantor is doing
precisely what all those set theories require, relying on what are
members of a set not why they are members.
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.

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

Actually not on all of them but only whether the set of them is empty or
not empty, which only requires one of them.
Quote:

How can you prove this statement ?

Why should WE have to prove YOUR statement?
Quote:

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
Actually we do not say this, you are the only one saying it.



Quote:
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.

It is the being empty or not being empty of your F:A->B or F :B->A that
is all that is needed for investigating Cantor cardinality.
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.

how?

If all functions in F are surjective then this is called unanamous
surjective homomorphism

This only can occur with finite sets. For any infinite set mapped to
itself, there are, by definition, injections that are not surjections.

Since for finite sets, Cantor cardinality is totally satisfactory and
for infinite sets, we already see that Zuhair's cardinality sucks, the
sensible mathematician will stick to Cantor's.

At least until something better than Zuhair's notions comes along.

Quote:
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).

And suffers from that non-equivalence.
Quote:

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 ).


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.

What Cantor himself says it that Card A = Card B if and only if there
is at least one injection from each to the other.
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 ).

Wrong. Cantor's definition depends on the existence of an injection from
A to B which need not be a bijection and another injection from B to A
which also is an injection.

While there is a theorem which says that when two such injections both
exist there will be at least one bijection, there is nothing in Cantor's
definition that requires one.
Quote:

Why I am saying all of that?

Because you do not know what Cantor's definitions say?
Quote:

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.

Cantor's Card(A) <= Card(B) is based on the existence of a single
injection from A to B.
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.

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

Actually the definition only requires injections both ways , but of
course a bijection is an injection both ways, so is sufficient, but not,
at least according to the statement of the definition, neccessary.
Quote:

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

Only by whether the set of injections is empty or not,
Quote:

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

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.

Since you are wrong about what Cantor's equality of cardinalities
requires, your whole castle in air collapses.
Quote:

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

Until you get right what Cantor actually says, you erring judgement of
what he says is sufficient to explain any bias.
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David R Tribble
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Joined: 21 Jul 2005
Posts: 1005

PostPosted: Wed Jul 19, 2006 10:09 pm    Post subject: Re: The list of all natural numbers don't exist Reply with quote

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


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


Albrecht Storz wrote:
Quote:
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.

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?


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

How about:
Set S is infinite if it is not finite.
Set S is finite if there exists a bijection between it and some
natural n.

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

There are other, more formal definitions, of course.
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Peter Niessen
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Joined: 05 Jun 2006
Posts: 19

PostPosted: Wed Jul 19, 2006 10:43 pm    Post subject: Re: The list of all natural numbers don't exist Reply with quote

Am Mon, 17 Jul 2006 15:24:56 -0700 schrieb Russell Easterly:

Quote:
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?

\forall m P(m) is a Set.
Short enought?
--
Mit freundlichen Grüssen
Peter Nießen
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Christian Stapfer
science forum beginner


Joined: 05 Jun 2006
Posts: 15

PostPosted: Thu Jul 20, 2006 3:50 am    Post subject: Re: The list of all natural numbers don't exist Reply with quote

Peter Niessen wrote:
Quote:
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"...

Quote:

\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.

Regards,
Christian
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Russell Easterly
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Joined: 27 Jun 2005
Posts: 199

PostPosted: Thu Jul 20, 2006 5:21 am    Post subject: Re: The list of all natural numbers don't exist Reply with quote

"Christian Stapfer" <nil@dev.nul> wrote in message
news:99646$44befd7b$54497364$21350@news.hispeed.ch...
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. 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.


Russell
- 2 many 2 count
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Virgil
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Joined: 24 Mar 2005
Posts: 5536

PostPosted: Thu Jul 20, 2006 7:10 am    Post subject: Re: The list of all natural numbers don't exist Reply with quote

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

Quote:
"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.
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albstorz@gmx.de
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Joined: 11 Sep 2005
Posts: 241

PostPosted: Thu Jul 20, 2006 7:40 am    Post subject: Re: The list of all natural numbers don't exist Reply with quote

Virgil schrieb:

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.

Who needs the ones which are inaccesible?

AS
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tez_h@nospam.yahoo.com
science forum beginner


Joined: 27 Jun 2006
Posts: 9

PostPosted: Thu Jul 20, 2006 11:35 am    Post subject: Re: The list of all natural numbers don't exist Reply with quote

zuhair wrote:
[snip]
Quote:

If g is homomorphic to f and f is surjective Then g is surjective
and
If g is homomorphic to f and f is not surjective Then g is not
surjective

if g is heteromorphic to f and f is not surjective Then g is surjective
if g is hetermorphic to f and f is surjective Then g is not surjective.

Right. Two injective functions, f and g, are homomorphic if either
they are both surjective, or both not surjective.

[snip]
Quote:

homomorphically surjective mean that that both functions are injective
and surjective. got it.

while homomorphically non surjective mean that both functions are
injective and not surjective . got it

I note here that "homomorphic" and "homomorphically surjective" are
subtly different. Two functions are homomorphically surjective if they
are both injective and surjective? So homomorphically surjective is
shorthand for bijective? That seems slightly redundant terminology,
but fine.

Quote:

For example, if all functions between A and B are, say, injective and
and not surjective, let me say (I pressume this is what you mean)

No wrong that is not what I mean.

Uh, I haven't got to the bit where I attempt to explain what I think
you mean...

Quote:
they
are all pairwise "homeomorphic". Then we can say uh-card A = uh-card
B. Note, this would not agree with cardinality.

*This* is where I indicate what I think you mean.

Anyway, given that "homomorphically surjective" means "bijective", I
agree rule (1) of uh-card is like cardinality.

Quote:

2) uh-card A < uh-card B if all injective functions from A to B are
homomorphically non surjective

What? You now seem to be using "homeomorphic" in a different way. I
think you've now made the meaning of "homeomorphic" redundant. I
thought "homeomorphically surjective" would mean all funcs are
surjective, or all funcs are not surjective. What does
"homeomorphically non surjective" mean?

it means that both injective functions f and g are injective and not
surjective.

Cool, I understand now that "homomorphic" and "homomorphically
non-surjective" use 'homomorphic' in different ways, and should not be
confused.

Quote:
3) uh-card A> uh-card B if all injective functions from B to A are
homomorphically non surjective.

This would be a non biased defintion of cardinality based on the
unanamouse decision of all injective functions from one set to the
other.

You see that the definition of inequality here is exactly similar to
Cantors inequality.

But the definition of equality is not similar to Cantor's. In cantors
theorm it is biased to bijection. and doesn't take all the injective
functions into consideration as you thought.

I have an additional comment to make here. You know you can determine
cardinal equality without bijections at all, right? You need only
consider injections. See
http://planetmath.org/encyclopedia/SchroederBernsteinTheorem.html , and
its very very elegant proof
http://planetmath.org/encyclopedia/SchroederBernsteinTheorem.html .

Perhaps then you will realise that your objection is baseless, and why
I mentioned this theorem posts ago.

Quote:

So the definition is similar, and not similar. Which is it? Is it
both? I'm starting to think it's not set theory, or z-card that's
inconsistent, it's *you*!

In your definitions, you explicitly say "... if all injective
functions...", which should indicate to most people aged 8 to 110 that
we are considering "all injective functions".

So do I.

Do you not read what you
write? Perhaps you confused yourself when you said the definitions
were similar, and then started the very next paragraph with the words
"not similar".

No they are similar regarding inequality, but not equality. see it
yourself. think man.

Yes, I missed the "in-" part where you wrote "inequality". My
apologies. Doesn't make you right though...

Quote:

And anyway, with cardinalities, we either prove the existence of a
function with certain properties, eg. we consider the set of all
functions between A and B and use that knowledge to work out which
equalities/inequalities are true between the 2 sets, or we consider the
set of all functions between A and B and prove that a
injective/surjective/bijective function does not exist in that set.

Again, the existence of different kinds of functions between sets A and
B never gives rise to contradictory cardinality comparisons, and to
prove or disprove the existence of some kind of function, we consider
the set of all functions betwen sets A and B.

And that what exactly not done by Cantor, Cantor only biased to the
bijective funciton.

I agree with you that Cantor's definition of inequality depends on all
injective functions between two sets. But Cantor's definition of
equality depends only on the bijective function so it is determined by
a single function , got it.

Again, this is not the case due to the Cantor-Schroeder-Bernstein
theorem.

Quote:

So if I know for example that for sets A and B ther exist f:A->B that
is injective and not surjective then according to cantor this only
means that card A <=card B. so knoweldge of the surjective state of a
function is not enought to determine a strict inequality untill we see
what other funcitons say.

Exactly. We only know more *once we consider all functions* between A
and B. Is that not what I claimed before?? Cardinality requires that
we consider all functions. That we don't know enough to get strict
cardinality information if we *don't* consider all functions doesn't
contradict anything.

But you didn't consider all function at arriving at the equality
definition, that's what is contradictive.

I didn't when *defining* equality? What does that mean? If this is
your objection, why don't you object to the "bias" in defining "<=" by
injections, or the "bias" in defining ">=" by surjection?? Objecting
to definitions is quite strange, especially a definition of a totally
uncommon word, "cardinality". If we were defining "round" to mean "has
4 corners", I might agree with an objection that the word "round" is
not being used intuitively. But we are not defining a common word
here.


[snip]
Quote:

So your only objection to cardinality is that you are sometimes unable
to work out the existence of injections, surjections, or bijections
between sets? That seems rather, uh, lazy. And to then claim this is
some kind of failing of cardinality that your knowledge is incomplete
seems rather disingenuous.

No no you are far away from what I mean, you misunderstood

Well, I thought I understood. Now I see that you're talking about a
"bias" toward bijection. At this point, I am completely baffled. Can
you explain in what way bijections are "favoured", or perhaps
"inappropriate", given the definition of cardinality? Perhaps you
think "is even" is biased towards integers or natural numbers. Or
perhaps you think "is rotationally symmetrical" discrimanates against
translations. Or that maybe "down" is biased towards gravity.

[snip]
Quote:


So let me understand this. You find cardinality unintuitive. Given
you sets A and B above, and your function f: A -> B where f(x) = 2 *
floor(x/2), you say that we can see Z-card(A) > Z-card(B).

What if I consider the function g: A -> B where g(x) = 2x ? Then using
your definitions, Z-card(A) = Z-card(B). Not only is this unintuitive,
it seems rather inconsistent. Am I not allowed to consider this
function g? Maybe you mean to say Z-card_f(A) > Z-card_f(B) and
Z-card_g(A) = Z-card_g(B), where I've put the "generational function"
as a subscript to Z-card. But now all you've done is made a rather
case-by-case, subjective version of cardinality, and one that looks a
little inconsistent (especially if you stick to the strict inequality,
and expect people not to be confused).

*Perhaps* you think that the set B_f "generated" by your function f is
different from the set B_g "generated" by my function g. I am using
membership and the axiom of extensionality to determine set equality
(whereby B_f = B_g). How do you determine set equality?

bye the same why you do, by the axiom of extentionality. I am not
saying that set B_f is different from set B_g, I am only saying that
z-card B-f can be different from z-card B-g if f is different from g.
Though B is the same set in both cases . got it.


While if the defining function of B from A is for example f:A->B , f(x)
= 2x

Then z-card A = z-card B , since f^(-1):B->A is one valued.

Oh right. You realise everything I've just written. You don't find
that counterintuitive?

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.

I'd still like to see how cardinality makes things inconsistent.
Remember, showing inconsistency is as simple as proving P and proving
~P for some statement P in set theory.

[snip]
Quote:

But how do you know whether all converjective functions give the same
z-card? Can you give me an (non-trivial!) example proof where all
functions between 2 sets that show the same Z-card relation between
them? This seems a rather important point and distinction, and an
example may highlight everyone's misunderstandings.

I will tell you lateron. for the time being just assume it is right.

Again, this seems rather crucial to your argument. I'd like to see
some inconsistent examples.

[snip]
Quote:
here you r right , that is why I am thinking of standarizing this
z-cardinality by assumptions as below:

Assumption 1) The set of all natural numbers N={0,1,2,3,4,.} if
mentioned without refereing to how it is generated then it is to be
assumed that it is generated directly from Peano's five axioms.

I'm not sure what I'm supposed to take from this. The only intersting
fact might be that you realise the successor function succ: N -> N-{0}
is bijective.

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?

You haven't explained how you can assume the existence of a bijection
between arbitrary sets, how you determine a generational function, the
existence of a converjective function between two sets, etc, etc. If
these don't exist, are the z-cards for those sets undefined?

[lots of confused stuff about explicating "implicit" generational
functions between sets]

-Tez
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Patricia Shanahan
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Posts: 214

PostPosted: Thu Jul 20, 2006 12:18 pm    Post subject: Re: The list of all natural numbers don't exist Reply with quote

albstorz@gmx.de wrote:
Quote:
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?

Patricia
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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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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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