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MoeBlee
science forum Guru

Joined: 27 Jul 2005
Posts: 493

Posted: Mon Jul 17, 2006 11:12 pm    Post subject: Re: The list of all natural numbers don't exist

Dave Seaman wrote:
 Quote: On Mon, 17 Jul 2006 15:24:56 -0700, Russell Easterly wrote: 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? To define *all* the subsets of N we need only one axiom (the power set axiom) and no predicates.

Maybe I'm wrong, but I took him to mean "all" in the sense of "each one
onto itself". So my answer to him, in that sense, is that there are
subsets of w that are not definable in the theory. Your answer, with
which of course I agree, is that the entire SET of subsets of w is
definable in the theory by virtue of the power set axiom.

 Quote: The purpose of the axiom schema of separation is to allow us to deduce that for each set and for each predicate definable on the members of that set, there is a subset consisting of all the members that satisfy that predicate. Since there are only countably many predicates, there are only countably many different subsets of any given set that are definable via separation.

And that is the sense in which I responded to his previous question.

We should add that 'definable predicate' is made precise by the axiom
schema of separation referring to formulas (and with the crucial
resriction that the the variable "standing for the defined set" dos not
occur free in the "defining formula").

MoeBlee
Virgil
science forum Guru

Joined: 24 Mar 2005
Posts: 5536

Posted: Mon Jul 17, 2006 11:37 pm    Post subject: Re: The list of all natural numbers don't exist

"zuhair" <zaljohar@yahoo.com> wrote:

 Quote: 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.

As Zuhair also would allow one set to have many cardinalities,
one could no longer speak of "the" cardinality of a set.
 Quote: You think that Cantor's cardinality is based all the decision of all injective functions between two sets.

"Decision"?
 Quote: 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. 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.

Then you cannot even have uh-card N = uh-card N, as some, but not all,
injections from N to N are surjections.

 Quote: 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.

And in yours, equality can only exist between finite sets, but infinite
sets cannot even have an uh-card equal to their own uh-card, much less
to any other set.
 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. 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.

There is a theorem, whose name I for the moment do not recall, which
says that if there are injections both ways between any tow sets then
there are also bijections between them, so that if Card(A) <= Card(B)
and Card(B) <= Card(A), in Cantor theory, then Card(A) = Card(B) in
Cantor theory.

 Quote: see the bias.

I see no bias in Cantor theory, but a lot in yours.

 Quote: 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.

No!
 Quote: 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.

What you always say is sometimes wrong. As in this instance.
Virgil
science forum Guru

Joined: 24 Mar 2005
Posts: 5536

Posted: Mon Jul 17, 2006 11:47 pm    Post subject: Re: The list of all natural numbers don't exist

In article <lOGdnTO9wqqmkyHZnZ2dnUVZ_oednZ2d@comcast.com>,
"Russell Easterly" <logiclab@comcast.net> wrote:

 Quote: "MoeBlee" wrote in message news:1153156074.575075.80690@b28g2000cwb.googlegroups.com... Russell Easterly wrote: I never said that. I am most sorry I misquoted you. I will try not to do it again. No. Z set theory avoids Russell's paradox by NOT having an unrestricted comprehension axiom. The purpose of the axiom schema of separation is to prove the existence of sets. Again, we don't AVOID a contradiction by ADDING axioms, let alone an infinite number of them. We add the infinite number of axioms in the axiom schema of separation to OBTAIN the existence of sets that we missed by DROPPING the unrestricted comprehension axiom, but without allowing the formation of sets that are contradictory. Moreover, ZF does not require the axiom schema of separation as an axiom schema, since the schema is derivable in ZF as a theorem schema. Moreover, as to the infinite number of axioms, NBG, which is a conservative extension of ZF, can be axiomatized with a finite number of axioms. And the axiom schema of separation provides only for a recursive infinite set of axioms, so there is still a mechanical method to determine whether a formula is or is not an instance of the schema, and this is just as with the induction axiom in PA. I have a question about the axiom of separation and recursive infinite sets of axioms. If I understand the Axiom of Separation (please correct me if I am wrong), it says we can not define a set, N, unless we have a predicate that defines every subset of the set N.

What it says is, in effect, given any set,
every subclass that is defined by a predicate is itself a set.

http://en.wikipedia.org/wiki/Axiom_of_separation

Fore each predicate P:
Given any set A, there is a set B such that, given any set C, C is a
member of B if and only if C is a member of A and P holds for C.
Note that there is one axiom for every such predicate P; thus, this is
an axiom schema.
To understand this axiom schema, note that the set B must be a subset of
A. Thus, what the axiom schema is really saying is that, given a set A
and a predicate P, we can find a subset B of A whose members are
precisely the members of A that satisfy P. By the axiom of
extensionality this set is unique. We usually denote this set using
set-builder notation as {C \in A: P(C)}. Thus the essence of the axiom
is:
EVERY SUBCLASS OF A SET THAT IS DEFINED BY A PREDICATE IS ITSELF A SET.

 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? Russell - 2 many 2 count
William Hughes
science forum Guru

Joined: 05 May 2005
Posts: 355

Posted: Tue Jul 18, 2006 2:28 am    Post subject: Re: The list of all natural numbers don't exist

albstorz@gmx.de wrote:
 Quote: 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.
So a complete list of finite integers is one that contains every
finite integer. Since there is no last integer (under the usual
ordering)
a complete list of finite integers does not have an end.

-William Hughes

 Quote: Best regards Albrecht S. Storz
Ross A. Finlayson
science forum Guru

Joined: 30 Apr 2005
Posts: 873

 Posted: Tue Jul 18, 2006 3:57 am    Post subject: Re: The list of all natural numbers don't exist Metamath shows the rationals uncountable. That's according to the formulation of Cantor's first, there, CantorMegill style. Metamath verifies that so, and then contradicts itself immediately. They're countable. That's where there are infinitesimal iota-values in the reals. I agree with Russell, that was a good post, from "MoeBlee". I read that Forster book about a "universal set", of which none exist in ZF. That's quite readable. The order type of ordinals would be an ordinal. Thus, you can't quantify over ordinals, there is no set of ordinals in ZF. The cumulative hierarchy is not a set, in ZF. If you can quantify over ordinals, then that would be some unrestriction of comprehension. Look at FOM, recently about paradoxes they are discussing Russell, and similar notions. In ZF you can't prove "ordinals" exists. In ZF there are only sets, "ordinals" is either a set of not, "sets" a set, or not. Sets are defined by their elements. In another thread there is discussion about the "limit of a sum of an infinite series." Except for the trivial cases, eg constants, for no finite value is the partial sum the actual sum. It is said in nonstandard analysis that the sum exists. That's said because it's so, and obviously there's a reason that people would prefer that language. The limit doesn't exist unless there is infinite induction. So, there is the infinite in the reasoning of delta-epsilonics, and without it, only (somewhat) trivial cases are soluble. Look at the infinitesimal analysis (integral calculus), it is exactly the Summation of differentials, with a very specific result that Sum 1 dx from 0 to 1 equals 1. It is exactly that, perfectly that, under no conditions is it not. Calling dx "the" infinitesimal is quite regular. There is no infinitesimal in ZF, and Internal Set Theory is much more Robinsonian that some other systems with notions of infinitesimals, like Leibniz', which are applied in every calculus classroom. (Don't fool yourself that dx isn't an infinitesimal and S isn't a summation sign.) There are the infinite dimensional systems that generally only care about one infinity. When the word "infinitesimal" is used, it's generally known what it means, and the same is so for an "infinity" in many cases. There is no universe in ZF. Infinite sets are infinite. If constructively in naive set theory via the domain principle the universe exists, with much less axiomatization than ZF(C), yet it doesn't, then it does and doesn't, two mutually contradictory things at once. If a theory isn't complete, then there is no theory of the natural integers. A theory is complete. Ross
albstorz@gmx.de
science forum Guru Wannabe

Joined: 11 Sep 2005
Posts: 241

Posted: Tue Jul 18, 2006 8:17 am    Post subject: Re: The list of all natural numbers don't exist

William Hughes schrieb:

 Quote: 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. You apply finite considerations on infinity.

x ) 1 subgroup of "x" / a set of subgroups

x x ) 2 subgroups of "x" / a set of subgroups
xx xx )

x x x )
xx xx xx ) 3 subgroups of "x" / a set of subgroups
xxx xxx xxx )

x x x x
xx xx xx xx
xxx xxx xxx xxx
xxxx xxxx xxxx xxxx

....

Your argument about completeness is not the only one. Why do you think,
only your definition is legal, mine is not?

I argue logical correct, that the structure above is only complete, if
it contains a infinite substructure - which is impossible.

This argumentation is as correct as yours. So, who wants to decide
which argumentation is the one which should be applied and which is the
one which should be dissmissed.

The fault in axiomatic set theory is the arbitrariness of faith, which
must not find a place in logical considerations.

 Quote: So a complete list of finite integers is one that contains every finite integer. Since there is no last integer (under the usual ordering) a complete list of finite integers does not have an end.

You just turn the definition around and around. What makes you think
you say anything about the completeness of the objects you just have
defined like this. To say, every natural number is a natural number or
to say every complete list is a complete list is both needless.

Best regards
Albrecht S. Storz
tez_h@nospam.yahoo.com
science forum beginner

Joined: 27 Jun 2006
Posts: 9

Posted: Tue Jul 18, 2006 10:09 am    Post subject: Re: The list of all natural numbers don't exist

zuhair wrote:
 Quote: 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".

 Quote: 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.

 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?

 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.

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.

 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

 Quote: 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).

 Quote: 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.

 Quote: 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.

 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.

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.

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?

 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.

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.

 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.

 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?

 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.

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?

 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?

 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).

 Quote: 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.

 Quote: see you later

-Tez
Dik T. Winter
science forum Guru

Joined: 25 Mar 2005
Posts: 1359

Posted: Tue Jul 18, 2006 11:22 am    Post subject: Re: The list of all natural numbers don't exist

In article <1153174235.958398.192250@p79g2000cwp.googlegroups.com> "zuhair" <zaljohar@yahoo.com> writes:
....
 Quote: 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. 2) uh-card A < uh-card B if all injective functions from A to B are homomorphically non surjective 3) uh-card A> uh-card B if all injective functions from B to A are homomorphically non surjective.

But this would make it impossible for infinite sets to have their
uh-cardinality compared. That is, given N (the naturals) and Q (the
rational). There exist:
(1) Injective functions from N to Q that are surjective
(2) Injective functions from N to Q that are not surjective
(3) Injective functions from Q to N that are not surkective
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
zuhair
science forum Guru

Joined: 06 Feb 2005
Posts: 533

Posted: Tue Jul 18, 2006 3:23 pm    Post subject: Re: The list of all natural numbers don't exist

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".

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.
 Quote: 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.

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
 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.

they
 Quote: 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?

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

No no you are far away from what I mean, you misunderstood
 Quote: 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. 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. 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? 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. 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? 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? 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
Virgil
science forum Guru

Joined: 24 Mar 2005
Posts: 5536

Posted: Tue Jul 18, 2006 7:09 pm    Post subject: Re: The list of all natural numbers don't exist

albstorz@gmx.de wrote:

 Quote: 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.
David R Tribble
science forum Guru

Joined: 21 Jul 2005
Posts: 1005

Posted: Tue Jul 18, 2006 7:28 pm    Post subject: Re: The list of all natural numbers don't exist

William Hughes schrieb:
 Quote: 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?

William Hughes schrieb:
 Quote: Complete means that nothing is missing. A complete list of finite integers is a list of integers that is not missing anything.

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

So given the set
S = { x, where x is a real },
in what sense is S "incomplete"? Which x's are missing from S that
prevent S from being a "completed" set?

It seems that using the function
f(x) = x, for all real x
we can say that f(x) is in S, for all real x. So it appears that S is
"complete", since it contains all the reals, specifically all the reals
where f(x) is defined. So how is S not complete? Is there some
value of f(x) that is not "yet" a member of S?
Virgil
science forum Guru

Joined: 24 Mar 2005
Posts: 5536

Posted: Tue Jul 18, 2006 7:43 pm    Post subject: Re: The list of all natural numbers don't exist

"zuhair" <zaljohar@yahoo.com> wrote:

 Quote: Tez wrote: 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". 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. 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. 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

For any infinite set, S, there will exist pairs of injections from S to
itself which are neither HS nor not HS

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

If you think that, you are not at all aware of what Cantor actually said

Cantor said that Card(A) <= Card(B) means that there is an injection
from A to B. He then said that
if Card(A) <= Card(B) and Card(b) <= Card(a) then Card(A) = Card(B).
Of course, since a bijection is an injection and has an inverse which
is also an injection, that is enough to show equality of cardinality.
 Quote: 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.

Why should we "get" what is false?

Cantor's definition of equality of cardinality for two sets depends only
on the existence of injections from each to the other.
 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.

One need not look at all functions.
If one can find one injection each way,
f:A ---> B and g:B --> A,
that is quite enough to show Cantor's
Card(A) = Card(B).
albstorz@gmx.de
science forum Guru Wannabe

Joined: 11 Sep 2005
Posts: 241

Posted: Tue Jul 18, 2006 9:39 pm    Post subject: Re: The list of all natural numbers don't exist

Virgil schrieb:

 Quote: 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.
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.

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

Best regards
Albrecht S. Storz
Virgil
science forum Guru

Joined: 24 Mar 2005
Posts: 5536

Posted: Tue Jul 18, 2006 9:54 pm    Post subject: Re: The list of all natural numbers don't exist

albstorz@gmx.de wrote:

 Quote: Virgil schrieb: 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. 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).

On the contrary, mathematics is the one place in which one can do
precisely that, make a definiendum mean exactly what one defines it to
mean.
 Quote: If endlessness logical implies incompleteness

A circle is endless but not in the least incomplete, so that endlessness
does NOT imply incompleteness,
William Hughes
science forum Guru

Joined: 05 May 2005
Posts: 355

Posted: Wed Jul 19, 2006 2:55 am    Post subject: Re: The list of all natural numbers don't exist

albstorz@gmx.de wrote:
 Quote: 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

Which integer is missing from the list {1,2,3,...}

 Quote: since infinity means: incompletable.

No, an infinite set is by definition a set that can be bijected to a
proper subset. There is not mention of incompletable.

 Quote: You apply finite considerations on infinity. x ) 1 subgroup of "x" / a set of subgroups x x ) 2 subgroups of "x" / a set of subgroups xx xx ) x x x ) xx xx xx ) 3 subgroups of "x" / a set of subgroups xxx xxx xxx ) x x x x xx xx xx xx xxx xxx xxx xxx xxxx xxxx xxxx xxxx ... Your argument about completeness is not the only one. Why do you think, only your definition is legal, mine is not? I argue logical correct, that the structure above is only complete, if it contains a infinite substructure - which is impossible.

You claim that the above structue is only complete if it
contains an infinite substructure. Why? What definition of
complete are you using?

 Quote: This argumentation is as correct as yours.

No, I gave a definiton of complete. So far you have failed to.

 Quote: So, who wants to decide which argumentation is the one which should be applied and which is the one which should be dissmissed. The fault in axiomatic set theory is the arbitrariness of faith, which must not find a place in logical considerations. So a complete list of finite integers is one that contains every finite integer. Since there is no last integer (under the usual ordering) a complete list of finite integers does not have an end. You just turn the definition around and around. What makes you think you say anything about the completeness of the objects you just have defined like this.

The fact that I give a definition of complete and then show that an
object satisfies the definition.

 Quote: To say, every natural number is a natural number or to say every complete list is a complete list is both needless.

True, but to say that a list that contains every natural number
is a complete list of natural numbers is not.

-William Hughes

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