Cantor Confusion
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From: mueckenh on 4 Feb 2007 13:09 On 4 Feb., 17:33, Carsten Schultz <cars...(a)codimi.de> wrote: > >> > On Feb 2, 5:12 pm, Carsten Schultz <cars...(a)codimi.de> wrote: > >> ... > >> > > From that it follows that there are exactly four sets with four > >> > > elements, since these are the elements of 4. > > > Is this intended to be a serious statement? > > Of course I am serious in pointing out that this is a consequence of > what you have written, and had you not snipped your statement from which > this is derived, evryone could see this. I did not snip anything here. (I did not answer you.) But what ever you may have thought, it must be mad. > > >> > > It also follows that there > >> > > is only one set with four elements, namely four. > > > With respect to the property "set having elements" one can in fact > > take this position. There is only one set with four elements. > > So for example {1,2,3,4} = {7,42,666,1984}? With respect to the property "set having n elements" (= cardinal number), yes, of course. Cardinal number is 4 in both cases. > > > Therefore "a set of four elements" is the same as "every set of all > > sets of four elements". > > Good to know. > This is new for you? > > >>> > You should > >> > > write a book about this. > > > That book is available, > > Who would have guessed this! And in it you prove that 1=4? Usually I do no longer read your writings. But I have looked at your "conclusion": quote: > 4 is the set of all sets with 4 elements. 4 is also every set with 4 > elements. >From that it follows that there are exactly four sets with four elements, since these are the elements of 4. quote end. I do not see a valid conclusion. Regard, WM
From: Virgil on 4 Feb 2007 13:25 In article <1170581930.094813.165200(a)s48g2000cws.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 2 Feb., 12:10, Franziska Neugebauer <Franziska- > Neugeba...(a)neugeb.dnsalias.net> wrote: > > mueck...(a)rz.fh-augsburg.de wrote: > > > On 2 Feb., 02:42, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > [...] > > >> However, when we construct the path: > > >> p = {n_1} U {n_1, n_2} U {n_1, n_2, n_3}, ... > > >> we get as path: > > >> {n_1, n_2, n_3, n_4, ...} > > >> the path length in this case is *not* a natural number. It is the > > >> cardinality of N. > > > > > The pathlength is not a natural number but it is a number (by > > > definition), namely omega. You see that there is no infinite set N > > > without an infinte number in it. > > > > As omega is not member of N I don't see that. > > The total pathlength is the union of all single pathlengths. If all those separate pathlengths are finite ordinal numbers and there are more than any finite number of them, then the union is the first limit ordinal. > The total > pathlength cannot be omega without omega being a pathlengt (in the > union). False! It is quite possible for a union not to be any one of the sets being unioned. But omega cannot be in the union of finite pathlengths. It can be, and is, everywhere except possibly in WM's weird world. > Hence > there is no actually infinite pathlength. It can be, and is, everywhere except possibly in WM's weird world. > "Infinite pathlength" means > only that every finite pathlength is surpassed by another finite > pathlength. Which requires a path with no end.
From: mueckenh on 4 Feb 2007 13:39 On 4 Feb., 17:33, Fuckwit <nomail(a)invalid> wrote: > On 4 Feb 2007 08:14:30 -0800, mueck...(a)rz.fh-augsburg.de wrote: > > > > > What is 4 in mathematics? > > In axiomatic set theory it's (usually) the set > > {0, 1, 2, 3}. That is circular. What is 3, what is 2, wat is 1, what is 0, what is the empty set? > > But 4 contains IIII --- not 0,1,2,3. > > No. As you can see, 4 has exactly the elements 0,1,2,3. IIII is only > an element of 4 if IIII is either 0,1,2 or 3. That is set theory, not mathematics. > > Henry VIII had 7 predecessors --- only including him they were 8 > > Henries. > > Was Henry VIII a natural number? No, but VIII is. > > > > >> ... for computable numbers some representation does exist. > > > But this representation does not necessarily enable us to determine > > the trichotomy relation with numbers which are really numbers. > > "All numbers are real, but some numbers are more real than > others." (Animal Farm) As far as I remember, there is not reality in question but equality. That is in fact an important point for numbers. > > > I want to define *the natural numbers* [the following way] No, you misinterpret me. The definition is done with more or less success by Peano or Dedekind. The success is measured by comparing how closely the numbers, intuitively known as natural numbers, sketched here: > I > II > III > ... are reprodced. The result is tha Peano is not so good. > Rules for the construction of these sequences of symbols: > > Rule 1: We may construct I. > Rule 2: If n is constructed, we may construct nI. > > Yes. This definition is better than that by Peano. > >> The non-existence of a pink elephant in the realm of the axiom that > >> states that there is a pink elephant is not mathematics but philosophy. > >> Axioms state what things exist (or do not exist) in their realm. > > > No. > > Yes. No. > > > > >>> Whether it exists remains to be investigated. > > >> Your existence is not a mathematical existence. > > > This form of existence is the only possible existence. > > No. No to no. Regards, WM
From: Franziska Neugebauer on 4 Feb 2007 13:49 mueckenh(a)rz.fh-augsburg.de wrote: > On 4 Feb., 14:34, Franziska Neugebauer <Franziska- > Neugeba...(a)neugeb.dnsalias.net> wrote: >> mueck...(a)rz.fh-augsburg.de wrote: >> > On 2 Feb., 12:10, Franziska Neugebauer <Franziska- >> > Neugeba...(a)neugeb.dnsalias.net> wrote: >> >> mueck...(a)rz.fh-augsburg.de wrote: >> >> > On 2 Feb., 02:42, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: >> >> [...] >> >> >> However, when we construct the path: >> >> >> p = {n_1} U {n_1, n_2} U {n_1, n_2, n_3}, ... >> >> >> we get as path: >> >> >> {n_1, n_2, n_3, n_4, ...} >> >> >> the path length in this case is *not* a natural number. It is >> >> >> the cardinality of N. >> >> >> > The pathlength is not a natural number but it is a number (by >> >> > definition), namely omega. You see that there is no infinite set >> >> > N without an infinte number in it. >> >> >> As omega is not member of N I don't see that. >> >> > The total pathlength is the union of all single pathlengths. >> >> The path-length of the infinite path is the supremum of the set of >> paths-lengths of all finite paths. The latter is the supremum of the >> set of natural numbers. This supremum exists. > > No. What does it consist of? It "consists" of (i.e. contains as members) every and all natural numbers, each of which is finite. > A supremum exists in many places but not here. Not where? > The simplest reason is that omega - n = omega for all n in N. Where did you get that from? Reference? EB? F. N. -- xyz
From: Virgil on 4 Feb 2007 13:52
In article <1170582481.992415.182840(a)s48g2000cws.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 2 Feb., 14:29, "William Hughes" <wpihug...(a)hotmail.com> wrote: > > On Feb 2, 2:14 am, mueck...(a)rz.fh-augsburg.de wrote: > > > > > > > certainly depends on the cardinal number of the set. Sets of type > > > > I have a cardinal which is a (finite) natural number. Sets of type > > > > II do not. > > > > > My questions are: > > > > > What numbers are subject to complete induction? > > > What is the set the elements of which are subject to complete > > > induction? > > My questions are: > > What numbers are subject to complete induction? > What is the set the elements of which are subject to complete > induction? As WM has not explained how "complete induction" differs from standard "induction", we have no idea of what he means by it. For standard induction, based on the set of natural numbers, N: If S is a subset of N such that the first member of N is a member of S and whenever a member of N is a member of S so its successor, then S = N. > > > > > > For example, every one of the elements > > > > of A is finite. This can be shown by induction. > > > > However, this does not show that the set > > > > A is not (potentially) infinite. > > > > > It does. > > > > No. Let P be the set of prime numbers. > > Every prime number is finite. P is potentially > > infinite. > > Yes. But no prime number is the largest. Irrelevant, as usual. > > > So showing that every element of a set is > > finite does not show that the set is not potentially infinite. > > This is already the case for sets of two elements: Showing that every > element is a single elements does not show that the set is a single > element. > > Every potentially infinite set is finite. False by any standard definition of finiteness. What definition of finiteness does WM claim? I. Every finite set can be ordered so as to have both a first ands a last member. A set which cannot be so ordered is not finite. The set of naturals is, by this rule, not finite. II. A set is finite if every injection from it to itself is a surjection. A set having an injection to itself which is not a surjection is not finite. The set of naturals is, by this rule, not finite. > This can be proved by induction. > Therefore induction is sufficient for all statements about potentially > infinite sets. > > On the other hand potential infinity can be proved by induction: > If there is a set with n numbers, then there is a set with n+1 > numbers. > Similarly one can prove by induction: Every set of even positive > numbers contains numbers which are larger than the cardinal number of > the set. One can prove that for finite sets, but not for sets which are not finite. So that WM is assuming as an axiom that all sets are finite. WM cannot prove any of his silly claims without that assumption. > > > > > > Of course every set of natural numbers is finite, as proved > > > by induction over all initial segments. (Potential infinity.) Not without first having assumed it. So all such "proofs" are circular. > > > > If you define a potentially infinite set to be finite, and > > say that there are only sets with a fixed maximum and > > potentially infinite sets, then all set are finite. > > So what? Calling a potentially infinite set finite does > > not change its properties. > > Calling it infinite does not change its properteis either. But proving that a set must have properties that finite sets cannot have makes it not finite. What is WM's definitions of a set being finite? It seems to mean only that it is a set, at least to WM, and nothing else, since WM assumes every set to be finite. > Every set which is given by its finite numbers is finite. Which still does not define finiteness of sets, nor finiteness of numbers. > If this does > not exclude that a larger set of this kind can be given, then the set > of such sets (usually abbreviated as "the set of such elements") is > called potentially infinite. This can be proved by induction. In fact, > induction was mainly devised for potentially infinite sets, because > for finite sets we do not need it. Induction was not devised for sets at all, but for proofs. > > > > > > > This > > > simple truth has only been veiled by the arbitrary assumption that > > > there is an actual infinity. And corrupted by the arbitrary assumption that all sets must be finite, although WM never makes clear what he means by a set being finite. > > > > No. Sets with a fixed maximum and potentially infinite sets > > have different properties, whether or not you assume an > > "actual infinity". > > Potentially infinite sets are subject to induction. There is no such thing as a merely potentially infinite set. A set is either finite (according to some standard definition of finiteness like having no injection to a proper subset) or it is not finite. Sets which are not finite are called infinite, and there is no mere potentiality about it, every set is either a finite set or is a set which is not finite, there is no third alternative, regardless of what WM dreams. |