Cantor Confusion
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From: Lester Zick on 6 Dec 2006 12:23 On Wed, 06 Dec 2006 10:35:45 -0500, Bob Kolker <nowhere(a)nowhere.com> wrote: >Eckard Blumschein wrote: > >> >> >> They are categorically quite different. According to Peirce, fictitious >> numbers are mere potentialities. According to Brouwer and Pratt they do >> not obey trichotomy. >> Who denies their quite different properties will never understand what >> makes the rationals different from the reals. The thread nightmare real >> numbers has been lasting since September the 9th until now. >> I admit, mathematicians are not trained to grasp the difference between >> quantity and quality. > >And the use of these fictions has lead to solutions of mathematical >problems which have produced useful applications. Without these fictions >there would not be sufficient physics to produce the computers on which >you type your blather. Also sprach der tool and diemaker of truth. ~v~~
From: Eckard Blumschein on 6 Dec 2006 12:27 On 12/6/2006 1:14 AM, Dik T. Winter wrote: > In article <4575B6E1.4010505(a)et.uni-magdeburg.de> Eckard Blumschein <blumschein(a)et.uni-magdeburg.de> writes: > > On 12/5/2006 2:45 PM, Dik T. Winter wrote: > > > > > > I wonder why the mathematicians believe to require one-point > > > > compactification. I consider the rationals as genuine numbers, being as > > > > close as you like to the fictions infinity and real numbers. The exact > > > > numerical representation of pi requires the fiction of actual infinity. > > > > > > I wonder why you are talking about things you know nothing about? Who > > > requires one-point compactification with what goal? > > > > When I presented ideas in connection with > > http://iesk.et.uni-magdeburg.de/~blumsche/M283.html > > I faced scepticism or refusal as well as the hint to compactification. > > Sorry, I can't read Word documents (I have no reader available). Could you read a pdf version? > But again who requires one-point compactification with what goal? > > > I consider my ideas still flawless. I even found plausible answers to > > several questions no mathematician was able to provide a convincing > > answer to. So I doubt about fundamentals which require compactification. > > I understand that no mathematician was able to provide an answer that > could convinve you. It started with the question how to deal with the nil in case of splitting IR into IR+ and IR-. I got as many different and definitve answers as there are possibilities. I expected that there is only one correct answer, and I found a reasoning that compellingly yields just one answer in case of rationals and a different one in case of reals.
From: Eckard Blumschein on 6 Dec 2006 12:30 On 12/6/2006 2:05 AM, Virgil wrote: > In article <1165345832.735910.255620(a)79g2000cws.googlegroups.com>, > "MoeBlee" <jazzmobe(a)hotmail.com> wrote: > >> Bob Kolker wrote: >> > There is no such thing as a countable integer, countable rational or >> > countable real. >> >> In the sense you're trying to get across to the other poster, I >> understand your point. But, just for the record, in a technical sense >> in set theory, as integers, rational numbers, and real numbers are >> themselves sets, it does make sense to say whether one of them is >> countable or not. For example, where integers are defined as >> equivalence classes of natural numbers, each integer is itself a >> denumerable set. I am not necessarily endorsing anything the other >> poster has said; I'm just adding the technical note that in a strict >> set theoretic sense, even numbers are sets and thus it is meaningful to >> talk about the cardinality of a number. >> >> MoeBlee > > In that particular sense, each real number as a Dedekind cut, being a > partition of the rationals into two sets, always has cardinality 2, > while each member of a Dedekind cut is countably infinite. > > But each real number as a set of equivalent Cauchy sequencesis > uncountable. Conclusion: Dedekind cut's are self-delusive.
From: William Hughes on 6 Dec 2006 12:47 mueckenh(a)rz.fh-augsburg.de wrote: Please clarify a point of nomenclature. Do you consider a potentially infinite set contianing only finite elements (e.g. the natural numbers) to be: 1. a set ? 2. a finite set? > William Hughes schrieb: > > > > > > You cannot simply exhange the quantifiers to get > > > > > > > > There exists a line L, such that for every natural number n, > > > > every natural number m<=n, is contained in L. > > > > > > I can simply do that for any finite linear (= totally orderd) set. > > > > You have made this comment several times but you have > > never said *why* this is true. You need to give some > > argument to show why something that is not true in the > > general case is true for a totally ordered set. > > > > Counterexample: > > > > Consider A= [0,1), the set of real > > numbers in greater than or equal to 0 and less than 1. > > This set is totally ordered. This set is > > composed of finite elements. We have the true statement > > > > For every element r in A, there exists an element > > s in A such that r < s. > > > > However, if we simple reverse the quantifiers we get > > the false statement > > This set is not finite. O.K. Replace A. A is now all rational numbers in [0,1]. As there is a bijection between A and the natural numbers, both have the same size. Both are potentially infinite sets. Nothing else in the counterexample changes. So if we have a potentially infinite set we cannot exchange quantifiers. Note that the collection of all lines is a potentially infinite set. So if we have a statement involving the collection of all lines we cannot exchange quantifiers. > > > > There exists an element s in A such that for every > > element r in A, r<s. > > > > So the fact that you have a totally ordered set consisting > > of finite elements, does not mean that you can reverse > > the quantifiers. > > I said: for any finite linear (= totally orderd) set. This means: for > any finte set. > Considering natural numbers, we have the coincidence > between set of (different) finite numbers and finite set. > > Recall my simple example: Every set of even natural numbers like > {2,4,6,...,2n} has a cardinal number which is less than some number in > the set. This theorem does not become invalid if n grows infinitely. No. Something that is true for finite n may or may not be true in the limit. > A > set of even atural numbers cannot have an infinite cardial number. Piffle. The collection of all even natural numbers is a potentially infinite set. Every potentially infinite set has an infinite cardinal number. > > > > > > > So I can do it for every line. > > > > You need to give some other argument to show that L exists. > > > > This you have not done. > > > > > > > > Recall: > > > > The potentially infinite set of natural numbers > > exists. > > That does not involve the existence of every natural number. (Don't ask > me which is missing. Those which can be named can be incorporated.) I have never claimed nor used the existence of every natural number. It is enough to claim the existence of the potentially infinite sets to define bijections between potentially infinite sets. You do not have to claim the existence of every element of a potentially infinite set. > > > > It is possible to have a bijection involving a > > potentially infinite set. > > > > > > > > It is in fact easy to show that L cannot exist. > > > > > > > > If X is such that for every natural number n, n is > > > > an element of X, then X is a potentially infinite set. > > > > > > > > No line L is a potentially infinite set. > > > > > > > > Therefore, there does not exist a line L such that > > > > > > > > for every natural number n, every natural number > > > > m <= n is an element of L. > > > > > > Therefore, there is no set of all finite lines, i.e., there is no set > > > containing all natural numbers as elements. > > > > However the potentially infinite set of all finite lines exists. > > It is possible to have a bijection involving the potentially > > infinite set of all finite lines. > > > > > > > > > > > > > > The fact that there each natural > > > > number is a member *some* line, does not > > > > mean there is one line which contains all > > > > natural numbers. > > > > > > > > The point remains. > > > > > > > > A set with a largest element can have elements > > > > all of which are natural numbers. > > > > > > > > A potentially infinite set without a largest element > > > > can have elements all of which are > > > > natural numbers. > > > > > > Correct, although this set can never be considered complete. > > > > Irrelevent. It does not matter whether we can > > consider a potentially infinite set "complete" or not. All that > > matters is that we can have a bijection involving a potentially > > infinite set. > > Not a complete bijection. A bijection is but a set of ordered pairs. If > the domain is incomplete (and the range too), then the bijection cannot > be complete. Recall this post from Dec 1 We extend this to potentially infinite sets: A function from the potentially infinite set A to the potentially infinite set B is a potentially infinite set of ordered pairs (a,b) such that a is an element of A and b is an element of B. We can now define bijections on potentially infinite sets A bijection on potentially infinite sets is not a set of ordered pairs it is a potentially infintite set of ordered pairs. It is not necessary to show that a bijection is "complete" to show that it exists. > > > > > > > > I claim that every element of the diagonal must be an element of a line. > > > > The statement > > > > Every element of the diagonal must be an element of a line. > > > > is true. But the statement > > > > Every element of the diagonal must be an element of a single line. > > > > is false. > > But it cannot be false for a finite line, because finite sets allow > quantifier reversal. Now there is no infinite line. > However, the statement is not about the elements of a line, but about the elements of the diagonal. The set of elements of the diagonal is potentially infinite. Potentially infinite sets do not allow quantifier reversal. > > Therefore we cannot use the statement > > > > Every element of the diagonal must be an element of a line. > > > > to conclude that there is a bijection between the diagonal and a single > > line. > > > > Do you intend to keep claiming that a bijection can exist > > between the diagonal and a single line?. > > If the diagonal would actually exist, then a line of same length would > actually exist. It is customary to define a term (e.g. "actually exist") before you use it. Below you clarify that for a set or potentially infinite set to actually exist, it must both exist and all of its elements must exist. It is irrelevent whether the diagonal "actually exists" or not. It is enough to know that the diagonal exists to define a bijection involving the diagonal. Now any line L actually exists, so it is possible to ask the question: Does a bijection between the diagonal and L exist? Do you intend to keep claiming that a bijection can exist between the diagonal and a single line?. > > > > > > > > > > If the "infinite set of finite numbers" existed, then we would have a > > > bijection between the diagonal and a line. > > > > > > > We have agreed to disagree on whether the "infinite set of finite > > numbers" > > exists. However, we have agreed that the "potentially infinite set > > of finite numbers" exists. > > But we have not agreed that it exists actually (i.e., is complete). Since I have never used the property that the set of natural numbers is complete, this is irrelevent. [Note by the way you claim that the set of natural numbers exists, but does not actually exist. This doesn't sound any better in German. You need to work on your nomenclature.] > > > > > > > (A diagonal cannot exist without being in a line.) > > > > False. Every element of the diagonal must be in some > > line. However, there is no line that contains all elements of the > > diagonal. So the diagonal is not in a line. > > The elements of the diagonal form the potentially infinite > > set of natural numbers. This potentially infinite set exists, > > so the diagonal exists. So the diagonal can exist without > > being in a line. > > Why don't you directly say that the potentially infinite diagonal > actually exists? Because it is not necessary to say that the potentially infinite diagonal is complete, either to show that it exists, or to define a bijection on it. - William Hughes
From: MoeBlee on 6 Dec 2006 14:17
Han de Bruijn wrote: > When WM uses metaphors, > for the purpose of convincing you of something, these would have been > picked up, and interpreted properly by most normal people. Too bad WM didn't do the same when he came across Fraenkel, Bar-Hillel, and Levy's metaphoric, scare-quote formatted mention of the universe of sets as "growing". MoeBlee |