Cantor Confusion
in [Math]
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From: mueckenh on 18 Feb 2007 04:15 On 17 Feb., 03:05, Michael Press <rub...(a)pacbell.net> wrote: > In article > > <1171468278.284645.273...(a)v33g2000cwv.googlegroups.com>, mueck...(a)rz.fh-augsburg.de wrote: > > On 13 Feb., 21:17, Virgil <vir...(a)comcast.net> wrote: > > > In article <1171364856.226197.135...(a)l53g2000cwa.googlegroups.com>, > > > mueck...(a)rz.fh-augsburg.de wrote: > > [...] > > > > > What about all existing sets with 3 objects, i.e., the fundamenal set > > > > of 3? > > > > What fundamental set of 3 does WM refer to? > > > Olease read before writing. The set of all existing sets with 3 > > objects. > > How many sets with three elements are there? No idea. Less than 10^100. But that is unimportant. Important is that there is at least one set with 3 elements. Regards, WM
From: mueckenh on 18 Feb 2007 04:29 On 18 Feb., 02:51, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1171702351.590890.177...(a)h3g2000cwc.googlegroups.com> mueck....(a)rz.fh-augsburg.de writes: > > > On 16 Feb., 15:58, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > ... > > > > This is not a representation of 3 other than in a perverted system, > > > > which calls 0 the first number, 1 the second and so on. Of course > > > > {{{{}}}}, or better and easier {{{{, denotes the fourth number which > > > > is 4 and not 3. > > > > > > Can you tell me a form of set theory where 0 is *not* the first ordinal > > > or cardinal number? If so, how many elements does the empty set have > > > in such a system? > > > > 0 may be the first (or better the zeroest) ordinal or cardinal number > > (if you wish to have the empty set in the theory). Nevertheless it is > > not the first natural number and not a natural number at all. > > But you said "perverted system which calls 0 the first number". The only > place where it is called the first *natural* number is in Bourbaki and its In all set theory 0 is called the first ordinal number, but in fact it is the zeroth one. Why do you start counting ordinals with 0 but start counting ordinally with 1? > followers, but that is only a simple renaming of the term "natural number". > But following your reasoning, {{{}}} is the third number, which is 3. > BTW, I can quote you as saying: > > > > > {{{}}} > > > > It was page 93 of my book. > > > I have seen it earlier than that. > > By the way, above is only number 2 given. > So earlier you said it is 2. What is it? 2 is the number called in set theory which I referred to here and in chapter 7 of my book (I hope without too many errors). 3 is the number called in any reasonalbe system. > > > > So how many natural numbers precede the first natural numbers? We are > > > counting natural numbers, so it should be a natural number? > > > > Natural numbers are counting the elements of natural sets, i.e., of > > sets which exist in reality (in nature, as Cantor woud have said). > > If my house contains no dogs, in what way does the set of dogs in my house > not exist in reality? There is no dog and no set of dogs, because it would be the same set of cats. Your dog would be your cat. That is impossible. They would not live together, let alone form the same (identical) set. > > > > |{1,2,3,...}| = ... i.e. potentially infinite, not fixed, capable of > > > > growing without bound, denoted by oo but not by a fixed number omega. > > > > > > And |{}| = ? > > > > A set which "also streng genommen als solche gar nicht vorhanden > > ist" (Cantor) > > A set which "ist also vermöge der Definition von S. 4 gar keine > > Menge" (Fraenkel) > > An unnatural set cannot have a natural number of elements. > > If you think {} to be an unnatural set, so be it (that is not mathematics, > because there is no mathematical definition of natural set). There not even a mathematical definition of an unnatural set. > But if > somebody asks how many coins I have in my purse, he is asking for the > cardinality of the number of coins in my purse. And I can correctly > answer 0 at some times. In that case the set is the set of coins in > my purse, and that can be empty (and is quite often in reality). > According to current definitions and axioms in set theory, {} *is* a > set. That is known to me but will not prevent me from criticising it. Regards, WM
From: William Hughes on 18 Feb 2007 09:53 On Feb 18, 4:01 am, mueck...(a)rz.fh-augsburg.de wrote: > On 16 Feb., 15:19, "William Hughes" <wpihug...(a)hotmail.com> wrote: > > > On Feb 16, 6:58 am, mueck...(a)rz.fh-augsburg.de wrote: > > > > On 16 Feb., 00:32, "William Hughes" <wpihug...(a)hotmail.com> wrote: > > > > > True and absolutely irrelevent. Which of the following do > > > > you disagree with? > > > > > All initial segments of E have a fixed maximum > > > > yes. > > > > > E does not have a fixed maximum > > > > yes. > > > > > If every initial segment of E has a fixed maximum > > > > then E has a fixed maximum > > > > no. > > > Your claim is > > > M: If all initial segments of E have [a] property and if > > M: no element of E is outside of every initial segment, then E has > > that > > M: property. > > > So if > > i: All initial segments of E have a fixed maximum > > and > > ii: no element of E is outside of every initial segment > > then > > > iii: E has a fixed maximum > > > You have claimed that i is true, and iii is false. > > Given the ii is clearly true, we have i [true] plus ii [true] implies > > [Claim above] iii. How do you explain this contradiction? > > You are right. The claim in its generality is clearly wrong, So stop using it. Stop claiming This holds for every initial finite segment therefore it holds for the set. If you want to prove something holds for the set you cannot use induction. Induction can only show that something holds for every initial segment. As you have now noted, this says nothing one way or the other about the set. In particular. The statement: It is true for every finite ... says nothing about something that is not finite. A statement of this form cannot be used to argue that assuming the existence of something that is not finite leads to a contradiction. >as one > can see already by the fact that every finite set is even or odd and > bounded and the infinite set is neither even nor odd and is unbounded > from above. And even in Wolkenmeuckenheim, a potentially infinite set is neither even nor odd. > The claim holds only for certain properties (like the > existence of elements larger than the cardinal number in any set of > even numbers which has a cardinal number). Nonsense. The set E may or may not have a property that every initial segment has, but even if it does the "proof" This holds for every initial finite segment therefore it holds for the set. is wrong (the conclusion may be true, but the conclusion does not follow from the premise). If you want to prove that E has some property you need a different argument. Note. The statement A: Any set of finite even numbers that has a cardinality contains an element larger that its cardinality says nothing about sets that do not have a cardinality. (Yes, I know that such sets do not "actually exist" in Wolkenmuekenheim, the question is whether assuming that such sets exist leads to a contradiction. If we assume that such a set exists, then either it does not have a cardinality or we have extended the concept of cardinality, in either case A does not apply. A only applies to the non-extended concept of cardinality) - William Hughes
From: Andy Smith on 18 Feb 2007 11:04 mueckenh(a)rz.fh-augsburg.de writes >On 16 Feb., 17:30, Andy Smith <A...(a)phoenixsystems.co.uk> wrote: > >> As I understand it : >> >> P(1) is the set of all numbers of 1 bit length - 2 elements (0.0 & 0.1) >> P(2) is the set of all numbers of 2 bits length - 4 elements (0.00, >> 0.01, 0.10, 0.11) >> P(n) is the set of all numbers of n bits length - with 2^n elements. >> >> The set P(n) includes all P(m) for m<n, if a right fill of 0 for the >> bits is inferred (i.e. 0.0 is the same as 0.00 is the same as 0.00..) > >The set P(n) contains only the n-bits numbers. In the tree with n >levels as I defined it, there are no shorter paths. (We could change >that definition without impairing the proof, but it would confuse the >readers.) > >> For any bit-length (path-length) n, P(n) is countable, so for any finite >> n, P(n) is countable. > >Yes. >> >> But, real numbers require an infinite number of bits to represent them >> (or, if you prefer, we can define the set of reals as the set of numbers >> with an infinite random binary expansion). But if the number of bits is >> not finite, neither is P(oo), and the reals are uncountable. > >The question is simply as follows: >The union of all finite segments of natural numbers {1}, {1,2}, >{1,2,3},... is the infinite set N {1,2,3,...}. >Why is the union of all finite paths {0.}, {0.0}, {0.00}, ... not the >infinite path {0.000...} = p(oo)? > >If it were the union, then the union of all finite paths would contain >all subsets required to "be", when united, all infinite paths like >p(oo) above. Then the reals were uncountable and countable. > With respect, this is circular - the union of all finite paths in the binary tree is the power set of N , which is what you wish to assert is countable? What I was trying to say, not very clearly, is that P(n) looks like nothing more than counting all the real numbers defined by the first n bits? So it looks as if , when you "let n->oo", that it seems P(oo) is then still countable? But it isn't, because you haven't included nearly all the real numbers which have an infinite bit length; because n is always finite, there are always an infinite number of bits left to define, and the limit does not exist. I probably misunderstand your argument. regards -- Andy Smith
From: mueckenh on 18 Feb 2007 11:33
On 18 Feb., 17:04, Andy Smith <A...(a)phoenixsystems.co.uk> wrote: > >The question is simply as follows: > >The union of all finite segments of natural numbers {1}, {1,2}, > >{1,2,3},... is the infinite set N = {1,2,3,...}. > >Why is the union of all finite paths {0.}, {0.0}, {0.00}, ... not the > >infinite path {0.000...} = p(oo)? > > >If it were the union, then the union of all finite paths would contain > >all subsets required to "be", when united, all infinite paths like > >p(oo) above. Then the reals were uncountable and countable. > > With respect, this is circular - the union of all finite paths in the > binary tree is the power set of N , which is what you wish to assert is > countable? The set of all finite subses of N is easily proved countable. It is only the infinite subsets of N which make P(N) uncountable. Compare the set of algebraic numbers resulting from polynoms with degree n for *all* n in N. Only the series which really have infinitely many terms, yield transcendental numbers. > > What I was trying to say, not very clearly, is that P(n) looks like > nothing more than counting all the real numbers defined by the first n > bits? So it looks as if , when you "let n->oo", that it seems P(oo) is > then still countable? That is the problem of set theory. If you let n --> oo then n nevertheless remains finite. Set theory needs an nfinite set which s larger than every set you get from letting n --> oo. The set of all fractions of the form 1/n, for instance, can be divided in subsets, as we do when proving the divergence of the harmonic series. The n-th set has 2^n elements. If aleph_0 was reached by the number of subsets, then 2^aleph_0 was reached by the fractions 1/n and, therefore, there were 2^aleph_0 natural numbers. > But it isn't, because you haven't included nearly > all the real numbers which have an infinite bit length; because n is > always finite, there are always an infinite number of bits left to > define, and the limit does not exist. The limit {1,2,3,...} = N does exist according to set theory. Why should the limit {1}, {1,2}, {1,2,3}, ... = N not exist? There is no arguing in set theory that the infinite binary tree would not contain all the real numbers of the unit interval. Regards, WM |