Cantor Confusion
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From: mueckenh on 4 Feb 2007 04:48 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? > > > > 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. > 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. 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. > > > Of course every set of natural numbers is finite, as proved > > by induction over all initial segments. (Potential infinity.) > > 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. Every set which is given by its finite numbers is finite. 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. > > > This > > simple truth has only been veiled by the arbitrary assumption that > > there is an actual infinity. > > 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. > > > As this assertion cannot be proved, some > > have argue that induction was not valid for the whole set. > > No. There are two different questions. > Let A be a set of natural numbers. > > i: Can induction show something > about the elements of A Can induction show something about *every* element of A. > > ii: Can induction show something about A. If A is infinite, induction can show that A is infinite. > > Note, for some sets A, the answer to i can be Yes > and the answer to ii can be No. That is true, but it does not depend on the infinity of a set. It is already the case for sets of two numbers. Induction is devised for infinite sets and is required for infinite sets only. But it cannot be used for extra-matzhematical properties like actual infinity. Regards, WM
From: Franziska Neugebauer on 4 Feb 2007 08:34 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. 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. > The total pathlength cannot be omega without omega being a pathlengt Omega is the path-length of the infinite path. > (in the union). Non sequitur. This is merely your claim X is not finite -> there must be an x in X which is infinite > But omega cannot be in the union of finite pathlengths. Straw man. Nobody claimed that omega is (a member) in N. > Hence there is no actually infinite pathlength. Non sequitur. > "Infinite pathlength" means only that every finite pathlength is > surpassed by another finite pathlength. This is the wrong meaning in the framwork of contemporary set theory. Perhaps WH may demonstrate that you are also in error with respect to a "potentially infinity" framework. F. N. -- xyz
From: Franziska Neugebauer on 4 Feb 2007 09:00 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: [...] >> No. There are two different questions. >> Let A be a set of natural numbers. >> >> i: Can induction show something >> about the elements of A > > Can induction show something about *every* element of A. It was devised for this purpose. >> >> ii: Can induction show something about A. > > If A is infinite, induction can show that A is infinite. 1. Where did you read that? Reference? 2. It reminds me of the M�ckenheim-Axiom X is not finite -> there must be an x in X which is infinite >> Note, for some sets A, the answer to i can be Yes >> and the answer to ii can be No. > > That is true, but it does not depend on the infinity of a set. > It is already the case for sets of two numbers. So there is no reason why this should be different for infinite sets. > Induction is devised for infinite sets Induction is also applicable to elements of finite sets. > and is required for infinite sets only. Required to achieve what? > But it cannot be used for extra-matzhematical properties > like actual infinity. In contemporary set theory there is no notion of "actual" (vs. "potential") infinity at all. F. N. -- xyz
From: William Hughes on 4 Feb 2007 09:31 On Feb 4, 4:48 am, mueck...(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? > > > > > > > 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. In other words my statement showing that every element of a set is finite does not show that the set is not potentially infinite. is correct and your It does is wrong. >But no prime number is the largest. equivalent to saying the set of prime numbers is potentially infinite. > > > 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. > 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. > No. You make your usual mistake. Let E be the potentailly infinite set of even numbers. Two statements i: For every element e of E, the set F(e) = {2,4,6,...,e} contains cardinal numbers which are larger than the cardinal number of F(e) ii E contains numbers which are larger than the cardinal number of E Statement i is true and can be shown by induction. Statement ii is false. The fact that statment i is true does not mean that statement ii is true. - William Hughes
From: mueckenh on 4 Feb 2007 11:07
On 4 Feb., 03:36, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1170470698.824513.309...(a)m58g2000cwm.googlegroups.com> cbr...(a)cbrownsystems.com writes: > > > 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? > > > 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. Therefore "a set of four elements" is the same as "every set of all sets of four elements". > > > You should > > > write a book about this. That book is available, for instance from Amazon or from the German National Library or directly from the publisher http://www.shaker.de/Online-Gesamtkatalog/details.asp?ID=1471993&CC=21646&ISBN=3-8322-5587-7 > > > > Dik is even reviewing it, I think. > > I am. I am now halfway chapter 8 and only found one serious error (in > chapter 7). And one place where I have serious doubts (in chapter 8, but > I have to look thoroughly at that). But until this point it is an > excellent review about the history about the thinking about the infinite. I am afraid, in chapter 7 (which gives an uncritical introduction to set theory) there is a multitude of serious errors. Regards, WM |