An uncountable countable set
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From: Daryl McCullough on 1 Jul 2006 11:16 mueckenh(a)rz.fh-augsburg.de says... >If this non-existence can be proved. But up to date it has not been >proved. I think that you are using the word "proved" to mean something different from anyone else here. A claim is proved if it is shown to follow from the mathematical and logical principles that you are using. That is the case for Cantor's theorem. It follows from the generally accepted principles of set theory, so it is proved. Now, you can complain that those aren't very *good* principles, or that they are contradictory. If you want to show that they are contradictory, then derive a proof of a contradiction using those principles. You haven't done that. What you've done is to show that set theory contradicts some of *your* principles. So either we must reject set theory, or we reject your principles. Each of us must decide which set of principles to reject. >> And in the set of all mappings from N to P(N), there are injections but >> no surjections. > >The latter is an empty assertion. It is a provable fact. Yes, you don't accept the proof, because you don't accept the basic principles of mathematics, but the proof certainly follows from those principles. They aren't *your* principles, but why should anyone besides you care about that? >> There is no surjection from any set S to its power set P(S). >> Proof: >> For every set S and every finction f: S --> P(S), >> K(f) = {x in S: not x in F(S)} is a member of P(S). > >That leads to the same paradox as in my example with |N and P(|N). It doesn't lead to any paradox at all. It leads to the conclusion that for any function f from S to P(S), f is not a surjection from S to P(S). Do you at least agree for *finite* sets, that there is no surjection from a finite set S to P(S)? In particular, if S = { 0 }, then P(S) = { {}, {0} }. Do you agree that there is no surjection from S to P(S) in this case? Is there a finite S such that there is a surjection from S to P(S)? If so, what is it? >Look up *my* definition. I did define the set K(f): >Map |R (including |N) on P(|N) with the only condition that a natural >number has to be mapped on that set K e P(|N) which contains all >natural numbers which are not mapped on sets containing them. You are just reiterating the same claim over and over, but in a convoluted way. Yes, there is no pair of objects <f,k> such that all three of the following hold: 1. f is a function 2. k is an element of N 3. f(k) = { x in N | x is not an element of f(x) } The fact that f is assumed to be defined on all of R is irrelevant. If K(f) is defined as { x in N | x is not an element of f(x) }, then the only thing that is important is the way f behaves on the naturals. For *any* mapping f whatsoever, no matter what its domain, we can conclude: K(f) is not in the image of N under the mapping f. So f does not map N onto P(N). We have 1. For any function f, K(f) exists and is well-defined. 2. For any f, K(f) is an element of P(N). 3. For any f, there is no k in N such that f(k) = K(f). 4. (from 3) For any f, f does not map any element of N to K(f). 5. (from 4 and 2) For any f, there is an element K of P(N) such that no element of N is mapped to K. 6. (from 5) For any f, f is not a surjection from N to P(N) >Look up *my* definition of K(f): >Map |R (including |N) on P(|N) with the only condition that a natural >number has to be mapped on that set K e P(|N) which contains all >natural numbers which are not mapped on sets containing them. > >This K(f) is not defined for any surjective mapping of one or some or >all natural numbers on sets containing sets of natural numbers. K(f) is defined for any f whatsoever, as long as f is defined. There is a surjection from R to P(N), but there is no surjection from N to P(N). So if f is a surjection from R to P(N), what we can conclude is If f(k) = K(f), then k is not a natural number. -- Daryl McCullough Ithaca, NY
From: Virgil on 1 Jul 2006 16:09 In article <1151758813.109076.7170(a)b68g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > > I wrote: P(k) = lim{k -> oo} (k,k+1)(k,k+2)...(k,k+n) where I explicitly > > wrote that I used some intuitive form of limit here. This is the only > > way I can make sense of "infinitely many transpositions". > > A transposition is an element of the set of transpositions, i.e. a set > of pairs of numbers. The set af rationals is also a set of pairs of > numbers. Do you raise any questions concerning this infinite set? Do > you need some limit in order to well-order the set of rationals? If one stars with the normally ( and denslely) ordered rationals, no sequence of transpositions will make it any less densely ordered. If one stars with the rationals well ordered, no sequence of transpositions will make it any less well ordered. Thus you alleged conversion form one form to the other does not and cannot occur. > The set of rationals is taken to be well-ordered. The following > transpositions operate on that set simultaneously (given are the > indices): > (1,2), (3,4), (5,6), ... > where Elements q 2n-1 and q 2n are interchanged, if they deviate from > order by size. > In the next step the pairs > (2,3), (4,5), (6,7), ... > are ordered by size, in the next step the pairs > (1,2), (3,4), (5,6), ... > are ordered by size, and so on. There are exactly as many steps as are > required to define the diagonal of a Cantor list. And there is the same > definition of "infinitely" as in Cantor's diagonal proof. Then there must be a ->first<- transposition after which there is no longer any first element in the current ordering of the rationals, if the final result is to be the usual ordering of the rationals. But that is clearly impossible. > > So you define a series of mappings. But the series is infinite. How do > > you define such an infinite series? > > Ask Cantor. The difference being that Cantor's sequence of non-diagoal digits need not be created seqeuentially, but can be done independently and simultaneously, whereas yours cannot. > > > Again you are starting with a conclusion about the finite and extending it > > to the infinite. You have to prove that you can do that. > > Ask Cantor, how he can be sure that his diagonal is different from > every number of the list in case of infinitely many numbers. One does not have to ask Cantor to explain what is transparently clear. All Cantor needs do is specify a rule which can be applied simultaneously to ever number in a given list to produce something that cannot be a member of the list. > I am not an advocate of infinity. You are not an advocate of sanity either. > > Regards, WM
From: Virgil on 1 Jul 2006 16:16 In article <1151759090.073603.122700(a)m79g2000cwm.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > > Eh? K(f) is defined regardles the kind of mapping of f. How can you > > come to the conclusion that it is not defined for some particular f? > > What part of the definition fails for some particular f? > > > Remember what I said: > Map |R (including |N) on P(|N) with the only condition that a natural > number has to be mapped on that set K e P(|N) which contains all > natural numbers which are not mapped on sets containing them. > You should be able to find out that in case of a surjective mapping, K > is undefined. If one sets conditions on the mappings allowed, perhaps, but what if there are no such restrictions, other than having domain R and codomain P(N). In that case nothing prohibits surjection, and, in fact, surjections have been constructed. I have, myself, even constructed a bijection. > > > What part of the definition fails? > > Remember what I said: > Map |R (including |N) on P(|N) with the only condition that a natural > number has to be mapped on that set K e P(|N) which contains all > natural numbers which are not mapped on sets containing them. Why not map it without that restriction? If one specifically takes a set of mapping which are known not to be surjections and complains that none of then are surjections, one is a fool. So what does that say about "mueckenh"?
From: Virgil on 1 Jul 2006 16:25 In article <1151759464.595165.88640(a)75g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > > > 5. Therefore, f is not a surjection. > > > > > > The existence of one or more non-surjective mappings doesn't tell us > > > anyting about cardinalities. > > > > But the non-existence of any surjective mappiings, combined with the > > existence of at least one injective map, does! > > If this non-existence can be proved. But up to date it has not been > proved. The non-existence, at least in ZFC or NBG, of any surjection from N to P(N) or from N to R has been adequately enough proved to satisfy anyone and everyone who accepts ZFC or NBG as a workable system of axioms. THose who do not accept this, put themselves beyond the pale. > > > > And in the set of all mappings from N to P(N), there are injections but > > no surjections. > > The latter is an empty assertion. It is a provable, and throroughly proved, assertion in both ZFC and NBG. > > > > > > > There is no surjection from any set S to its power set P(S). > > Proof: > > For every set S and every finction f: S --> P(S), > > K(f) = {x in S: not x in F(S)} is a member of P(S). > > That leads to the same paradox as in my example with |N and P(|N). It > does not prove anything. Except the stupidity of those denying the validity of the proof.
From: Virgil on 1 Jul 2006 16:47
In article <1151759780.766369.176910(a)m79g2000cwm.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > Cantor's diagonal proof does not prove that all numbers of the list are > > > included. > > > > All number of the list are certainly included in the list. What else are > > they supposed to be included in??? > > Cantor's diagonal proof does not prove that all numbers of the list are > included in that sense that all are covered by the diagonal. If you claim that any member of any list is not differentiated from the number created for that list by any of uncountably many variations of Cantor's constructions, then you must be declaring there there exists a first number in some list which is not differentiated according to some version of Cantor's rules. So Give us an example of this alleged situation. If you can. If I declare that there re fairies at the bottom of my garden, such a declaration will be rejected without solid evidence. "Mueckenh" declares the equivalent of fairies at the bottom of Cantor's garden, but cannot produce any evidence of their existence. We will not take his word for it, and we will not accept the lame argument he has presented as anything but lame arguments. >Only to > give you an example: If the list was longer than wide As the list is supposed to be endlessly long and the decimal (or other base) representation of each member is supposed to be endlessly "wide", why should either endlessness be greater than the other? > > > Cantor's diagonal proof does not prove that all numbers of the list are > > > included. > > > > Perhaps not to you, but it does to most. > > That is not an argument for truth, rather the opposite. There are those to whom the truth is anathema. "Mueckenh" seems more and more to be one of them. If so, then his rejection of a statement would be rather more evidence of its truth than its falsity. > > > > The society of mathematicians is limited of those who agree to accept > > whatever is proved according to certain generally accepted standards of > > logical proof and not to insist on things whose proofs cannot meet that > > standard. > > The diagonal proof is not valid for an infinite list. There is no > definition at all by Cantor. He only stated continuation. On the contrary, Cantor, and others, have specifically defined valid rules for constructing a real number not in any given list of real numbers. > > > You reject things that have been proved within that standard and insist > > on things whose proof does not meet that standard. > > Same did Cantor 130 years ago. Only in your opinion, not in fact. And your opinions have proved flawed. > > > > So that what you claim is not acceptable as mathematics. > > > > Either equal rights for all or for none. > > > > Standards of proof are not democratic. > > Correct. And your arguing few lines above about the society of > mathematicans is nonsense. It is mathematicians collectively who decide what mathematics is, the rest of the world does not have a vote. And you are one of those with no vote. > > > > Then the set of rationals never becomes naturally ordered, which would > > > > require exactly what you say never occurs. > > > > > > Hence there must be one assumption which is wrong, isn't it? > > > > The assumption that any sequence of transpostions can convert a well > > ordered set into a dense set, or vice versa. > > That is the result. Which is the wrong assumption? > > > > > Hence there must be one assumption which is wrong, isn't it? > > > The only assumption I made is the existence of infinite sets. > > > > You also assumed that you could convert between dense ordering and well > > ordering by a sequence to transpostions. > > That is the result. Which is the wrong assumption? > > > > > > > > > Do you have an idea why? (What is wrong with my assumptions?) > > > > You assumed, in contradiction to what Cantor said, > > that you could > > convert between dense ordering and well ordering by a sequence to > > transpostions. > > That is the result. Which is the wrong assumption? The assumption that that result can be valid is wrong in the face of the proofs that it is not valid. |