Cantor Confusion
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From: William Hughes on 18 Dec 2006 10:16 Han de Bruijn wrote: > William Hughes wrote: > > > Han.deBruijn(a)DTO.TUDelft.NL wrote: > > > >>William Hughes schreef: > >>> > >>>Does the "potentailly infinite set" of natural numbers > >>>exist? A potentially infinite set is a function on sets > >>>that takes on the values true and false. The > >>>potentially infinite set of natural numbers > >>>takes on the value true for a set containing only > >>>natural numbers, and false for any other set. > >>>(i.e. is there a way of recognizing a set of > >>>natural numbers?) > >> > >>I don't understand this question. > > > > Stop me at the first line you don't understand. > > > > [Note if you do not like the term "potentially infinite > > set" substitue "recognizer".] > > > > A "potentially infinite set" is a function on sets. > > It's already here: "function" and "set". > You answered the question "Is there an unbounded set of natural numbers?" No. Now you claim you don't know what I mean by "set". . Clearly you cannot answer simple questions. - William Hughes
From: mueckenh on 18 Dec 2006 10:16 William Hughes schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > William Hughes schrieb: > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > William Hughes schrieb: > > > > > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > William Hughes schrieb: > > > > > > > > > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > > > William Hughes schrieb: > > > > > > > > > > > > > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > > > > > Virgil schrieb: > > > > > > > > > > > > > > > > > > > > > > > (It is contained in the union of all lines, but the > > > > > > > > > > > > > union of all lines is not a line) > > > > > > > > > > > > > > > > > > > > > > > > That is a void assertion unless you can prove it by showing that > > > > > > > > > > > > element by which the union differes from all the lines. > > > > > > > > > > > > > > > > > > > > > > Not quite. In order to achieve that the diagoal is not in any linem all > > > > > > > > > > > that is required is: > > > > > > > > > > > Given any line there is an element of the diagonal not in THAT line. > > > > > > > > > > > It is not requires that: > > > > > > > > > > > There is an element of the diagonal that is not in any line. > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > For linear sets you cannot help yourself by stating that the diagonal > > > > > > > > > > differs form line A by element b and from line B by element a, but a is > > > > > > > > > > in A and b is in B. This outcome is wrong. > > > > > > > > > > > > > > > > > > > > Therefore your reasoning "there is an element of the diagonal not in > > > > > > > > > > THAT line. It is not required that: There is an element of the diagonal > > > > > > > > > > that is not in any line." is inapplicable for linear sets. You see it > > > > > > > > > > best if you try to give an example using a finite element a or b. > > > > > > > > > > > > > > > > > > > > > > > > > > > In every finite example the line that contains > > > > > > > > > the diagonal is the last line. > > > > > > > > > > > > > > > > Every example with natural numbers (finite lines) is a finite example. > > > > > > > > > > > > > > > > > Your claim is that there is a line which contains the diagonal. > > > > > > > > > > > > > > > > Because a diagonal longer than any line is not a diagonal. > > > > > > > > > > > > > > > > > Call it L_D. Question: "Is L_D the last line?" > > > > > > > > > > > > > > > > There is no last line > > > > > > > > > > > > > > Then, there is a line that comes after L_D. > > > > > > > > > > > > > > Therefore :L_D does not contain every element > > > > > > > that can be shown to exist in the diagonal. > > > > > > > > > > > > All elements that can be shown to exist in the diagonal can be shown to > > > > > > exist in one single line. > > > > > > > > > > > > > > > > Call it L_D > > > > > > > > > > L_D contains a largest element. n. > > > > > > > > > > L_D is not the last line, so there is > > > > > a line with element n+1, > > > > > > > > > > Element n+1 can be shown to exist in the diagonal. > > > > > > > > > > > > Element n+1 can be shown to exist in L_D (which is obviously a line > > > > containing n+1). > > > > > > No. L_D is bounded. The largest element of L_D is n. > > > L_D does not contain n+1. > > > > You misinterpret L_D. L_D is that line which contains all numbers > > contained in the diagonal. If your L_D does not contain them, then you > > have the wrong L_D. > > Assume that there exists an L_D which contains all the numbers > contained in the diagonal. L_D is bounded, If an unbounded diagonal exists, then obviously an unbounded line must exist. The conclusion is false, so the antecedent cannot be true. > therefore there > exists a largest element, call it n(L_D). L_D is not the last line > therefore there exists a line containing n(L_D) + 1, therefore > the diagonal contains n(L_D) + 1, L_D does not contain n(L_D) + 1. > Contradiction. Therefore L_D does not exist. I think you can agree to the statement: The diagonal does, by definition, not contain any element which is missing in every line. Now assume that at least two lines were required to contain all the elements of the diagonal. All the lines are finite, so one of them must be smaller than the other. Therefore the assumption that both were required is false. You see we have two proofs with different results. The only possible outcome is to withdraw the assumption of the existence of the diagonal. Regards, WM
From: mueckenh on 18 Dec 2006 10:24 William Hughes schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > Franziska Neugebauer schrieb: > > > > > > > >> > > > All elements that can be shown to exist in the diagonal can be > > > >> > > > shown to exist in one single line. [(P1)] > > > > > > This proposition P1 has _not_ yet been proved (shown). > > > > This proposition is the definition of the diagonal. > > > > > > You misinterpret L_D. L_D is that line which contains all numbers > > > > contained in the diagonal. > > > > > > The Diagonal is unbounded thus any _assumed_ L_D is not bounded, too. > > > > Correct is: > > If the diagonal is unbounded then any _assumed_ L_D is not bounded, > > too. > > > L_D is a line and is therefore bounded. > > Contradiction. > Hence, the _assumed_ L_D does not exist. > Assumed is the diagonal. Premise: "If the diagonal is unbounded." L_D is a line and is therefore bounded. Regards, WM
From: mueckenh on 18 Dec 2006 10:28 Newberry schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > Newberry schrieb: > > > > > Sorry that I joined a bit late. > > Doesn't matter. > > > > >Are you saying that (in an infinite > > > binary tree) the set of paths is uncountable but the set of edges is > > > countable? > > > > The set of edges is obviously countable by, e.g., > > > > 1, 2, > > 3,4,5,6, > > 7,... > > > > As no path can separate from another one without the existence of two > > more edges, the number of edges is an upper bound for the number of > > paths. > > If it is so simple why do you still need the proof by inheritance of > shares? I don't need it. But there are others who do not see the forest because of too many trees blocking the view. Regards, WM
From: Dik T. Winter on 18 Dec 2006 10:25
In article <1166453622.108599.302270(a)j72g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: .... > > And what if m is infinite? How do you divide? How do you recombine? > > Suppose I have an edge that is divided in countably many shares. Now > > suppose I select the even numbered shares and recombine them. Do I > > now have an edge? And what when I do the same with the odd numbered > > shares, do I now also have an edge? If that is not the case, why not? > > And if that is the case, please explain. > > We need not divide anything by infinite numbers, because there are only > finite numbers n in the sum > 1 + 1/2 + 1/4 + ...+ 1/2^n + ... > Every level of the tree has a finite number n. For finite n also 2^n is > finite. I do not understand the connection. Through every edge there go infinitely many paths. So to give each path a share of the edge we have to divide the edge in infinitely many shares. > > I would not like to call such a recombination of shares an edge, because > > it is *not* an edge in your original tree. I would at most call it a > > collection of shares. And it is quite trivial to split an edge in an > > infinite number of shares and to recombine those shares in an infinite > > number of collections of an infinite number of shares. > > There are no infinite numbers in 1 + 1/2 + 1/4 + ... But each edge is divided in infinitely many shares. > > > decimal point here. Let us denote the two equal shares of edge number 1 > > > by 1#1 and 1#2. > > > > What is bewildering is that you are again using a theorem that would be > > valid in the finite case as being also valid in the infinite case without > > proof. > > > > > Consider you have two coins of 50 cent value each. Then you can buy > > > something which costs 1 euro. > > > > Again, the finite case. > > Of course. There is only the finite case for natural numbers. There are > no infinite numbers in N. I do not state so. But through each edge go infinitely many paths. > > > > For this sake we have mathematics, I mean really correct mathematics, > > > not based on dubious infinite sets, which states that 1 + 1/2 + 1/4 + > > > ... = 2. > > > > There are indeed texts that do state that. But what you, apparently, fail > > to see is that that is not an infinite sum, but a limit in disguise. > > I will not discuss whatever you may call it. It is 2. That's enough. > > > > If AC > > is true, > > How could it be false, if being an axiom? You can work with mathematics where AC holds, but also in mathematics where AC does not hold. Just like you can work with geometry where the parallel axiom holds and with geometry where it does not hold. > > there exists an Hamel basis for the reals as a vector space over > > Q. And using that basis it is reasonably simple to well order the reals. > > Why does such an ordering definitely *not* exist? No book in any > library of the world (incuding all extraterrestrical cultures) contains > it, neither a formula, nor a recursion and, obviously, not a list. Obviously you need AC to make it existing. So only in that version of mathematics were AC holds does it exist. > > In geometry the > > parallel axiom tells us there is only single line. Abandoning that > > axiom gives us different kinds of geometry. It is similar with AC. > > If we accept it we get a different kind of set theory (where every set > > can be well-ordered) from a version where we do not accept AC. > > > > Obviously with AC it has been shown that there is a Hamel basis and that > > immediately leads to a well-ordering. > > Could you post it, please? Given the Hamel basis, do some lexicographic ordering of the reals as a vector space over the rationals. > > But when AC is false there is not > > necessarily a Hamel basis, so there is not necessarily a well-ordering. > > The only difference with the parallel axiom is that that axiom is a much > > more direct statement. And when Lobachevsky developed his geometry he > > was generally thought to be mad. > > That does *not* prove that, after Cohen, the advocates of AC are *not* > mad. Did I say so? I only state there are two versions of set theory, one version where AC holds and one version where it does not hold. > > > Revisit the 3rd class of the elementary school. There the teacher will > > > teach you how money can be divided into smaller pieces and different > > > pieces can be collected to yield a desired value, for instance euro > > > 28.50 to order my book. If you found a company and every member gives 5 > > > euro, then 6 members will suffice to buy the book including postage. > > > > He never did tell me how I can divide money among an infinite number of > > persons. > > Perhaps you missed that lesson? Ah, may be. Could you educate me how I can divide money among an infinite number of persons? > > But now you are apparently stating that some of the shares can > > be zero. In your tree, what shares are zero and what shares are non-zero? > > It is possible to divide an object into equal shares or into non-equal > shares. Compare a birthday cake and some guests. In principle it is > possible to give nothing to some guest. Then his share does not exist > or, by modern terms, is empty. But all the shares of an edge in my tree > are divided in equal parts, because of justice and because of ease of > calculation. And there are infinitely many shares of each edge. So how do you divide? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |