Cantor Confusion
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From: Dik T. Winter on 17 Dec 2006 20:51 In article <1166361677.475397.105840(a)l12g2000cwl.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > > The notion "rational relation" is a further development of the notion > > > "relation". While a relation connects elements of a domain to elements > > > of a range, the rational relation connects shares of the elements of > > > the domain to (shares of the) elements of the range. A simple example: > > > > Again, what are "shares"? > > An edge is considered to be a set of shares. You can divide it into m > equal shares. If you divide n edges into m equal shares each, then you > have m * n shares and you can combine m abitrary shares to have one > full edge. As an example use euro and cent. 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. 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. > > > A) Consider the relation: > > > 1 ---> a > > > 2 ---> b > > > B) Now a rational relation could be: > > > 1.1 ---> a > > > 1.2 ---> b > > > > > > 2.1 ---> a > > > 2.2 ---> b > > > > > > If 1.1 and 1.2 are the only shares 1, i.e., if 1.1 u 1.2 = 1 and if 2.1 > > > and 2.2 are the only shares of 2, i.e., if 2.1 u 2.2 = 2, then one full > > > element of the domain is available for every element of the range. > > > > Makes barely sense. What is 1.1 united 1.2? I have no definition, so I > > have no idea what that means. > > I wrote "1.1 u 1.2 = 1". But perhaps it is bewildering to use the > 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. > And even going into your reasoning, I see > > still nothing that allows the "then one full element of the domain is > > available for every element of the range". I would like to ask for a > > proof. You are still reasoning in the finite case. Whether it is > > applicable in the infinite case remains a question. > > 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. > > > This proof is similar to the proof of a well-ordering of the real > > > numbers. It cannot be done (and it can even be proved that it cannot be > > > done) but it has been proved to exist. > > > > You are entirely wrong here. > > (1) Axiom of choice valid: it can be done (and I think it has been done). > > LOL. Would you be so kind to tell me in which form this well-ordering > has been done? Is a catalog available at some bibliothy? Or is there a > formula stating which real comes first and which comes after that > number aleph * omega^omega^omega? aleph * omega^omega^omega is neither a real, nor a number. Unless you define something first. But you should learn some mathematics. If AC is true, 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. > > (2) Axiom of choice not valid: you can prove that it can not be done. > > But you confluate the two together and apparently do not see the > > difference. > > In *mathematics*, either it can be done or it cannot be done. And if it > has ben done, one need not forget it after abolishing AC. In *mathematics*, either there is a single line parallel to another line through a point, or there is none or many. And if it has been shown, one need not forget it after abolishing the parallel axiom. You fail (or are not willing) to see the similarity. 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. 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. > > > The rational relation devised in the binary tree proves that a > > > surjection of edges onto paths exists. In so far my assertion "there is > > > a surjective mapping" was not quite wrong. > > > > Not so. Because you are using a lot of terms that lack definition. > > 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. 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? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Newberry on 18 Dec 2006 00:33 mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > > > The notion "rational relation" is a further development of the notion > > > "relation". While a relation connects elements of a domain to elements > > > of a range, the rational relation connects shares of the elements of > > > the domain to (shares of the) elements of the range. A simple example: > > > > Again, what are "shares"? > > An edge is considered to be a set of shares. You can divide it into m > equal shares. If you divide n edges into m equal shares each, then you > have m * n shares and you can combine m abitrary shares to have one > full edge. As an example use euro and cent. > > > > > A) Consider the relation: > > > > > > 1 ---> a > > > 2 ---> b > > > > > > B) Now a rational relation could be: > > > > > > 1.1 ---> a > > > 1.2 ---> b > > > > > > 2.1 ---> a > > > 2.2 ---> b > > > > > > If 1.1 and 1.2 are the only shares 1, i.e., if 1.1 u 1.2 = 1 and if 2.1 > > > and 2.2 are the only shares of 2, i.e., if 2.1 u 2.2 = 2, then one full > > > element of the domain is available for every element of the range. > > > > Makes barely sense. What is 1.1 united 1.2? I have no definition, so I > > have no idea what that means. > > I wrote "1.1 u 1.2 = 1". But perhaps it is bewildering to use the > decimal point here. Let us denote the two equal shares of edge number 1 > by 1#1 and 1#2. > > Consider you have two coins of 50 cent value each. Then you can buy > something which costs 1 euro. > > And even going into your reasoning, I see > > still nothing that allows the "then one full element of the domain is > > available for every element of the range". I would like to ask for a > > proof. You are still reasoning in the finite case. Whether it is > > applicable in the infinite case remains a question. > > 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. > > > > > This proof is similar to the proof of a well-ordering of the real > > > numbers. It cannot be done (and it can even be proved that it cannot be > > > done) but it has been proved to exist. > > > > You are entirely wrong here. > > (1) Axiom of choice valid: it can be done (and I think it has been done). > > LOL. Would you be so kind to tell me in which form this well-ordering > has been done? Is a catalog available at some bibliothy? Or is there a > formula stating which real comes first and which comes after that > number aleph * omega^omega^omega? > > > (2) Axiom of choice not valid: you can prove that it can not be done. > > But you confluate the two together and apparently do not see the > > difference. > > In *mathematics*, either it can be done or it cannot be done. And if it > has ben done, one need not forget it after abolishing AC. > > > > > The rational relation devised in the binary tree proves that a > > > surjection of edges onto paths exists. In so far my assertion "there is > > > a surjective mapping" was not quite wrong. > > > > Not so. Because you are using a lot of terms that lack definition. > > 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. > > Regards, WM Sorry that I joined a bit late. Are you saying that (in an infinite binary tree) the set of paths is uncountable but the set of edges is countable?
From: mueckenh on 18 Dec 2006 01:38 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. Regards, WM
From: mueckenh on 18 Dec 2006 01:53 cbrown(a)cbrownsystems.com schrieb: > Your proof is supposed to be a /proof/ that the theorem in ZFC that the > reals are uncountable /does/ constitute a contradiction in ZFC; becuase > your proof constructs a bijection between N <-> R. It is pointless to > start by asserting "obviously, the reals are countable" until you have > actually /proven/ that the reals are countable. I did not start so. I tried only to explain to you why 2^omega is a countable rdinal in ZFC. > > You have claimed that you can use the binary tree and your "rational > relation" to prove this. This proof should be independent of any > /other/ proof that the reals are countable. And so it is. > > > Your edges are not > > > being broken into 2 "shares" each; they are being divided into an > > > infinite number of shares, and that is very different from 2 shares! > > > > 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. > > > > "Really correct mathematics" doesn't simply /state/ that "1 + 1/2 + 1/4 > + ... = 2". Mathematics /defines/ what the series of symbols "1 + 1/2 + > 1/4 + ... = 2" means, and then /proves/ that it is true from these > definitions. > That is the nonsense en vogue today. Really correct mathematics computes that the sum is 2. Already Euclid and Archimedes did not know that. But however, even the mathematics taught today should come to this conclusion. > Similarly, I ask that you don't simply /state/ "there is a rational > relation which is a surjection of edges onto paths". I ask that you > /define/ what that statement means; and then /prove/ that that > statement is true from those definitions. I define and proved. That I did not use the language of set theory is because this theory has no value. > > So the above "drawing" is /not/ sufficient to say "(C) proves a > surjection"; we must /also/ know the /sizes/ of the shares, as well as > "what they are mapped to", before we can say "we have proven a > surjection". O course. In my appliation of his technique the shares are equally divided. Shares of different sizes are a complication which I did not introduce in order to keep things as simple as possible (and because they are not required for my purposes). > > > > Does this (C) prove the surjection {1,2} -> {a,b} "is possible, even if > > > it cannot be found"? If instead 1#3 -> a, then does it prove it? Could > > > you explain why? > > > > If 1#3 is a share which contains half of 1, and if all the shares of 2 > > are of equal size, then we have a surjection. > > So 1#3 can "contain 'half of' 1". Are shares sets then? Can 1#3 also > contain two thirds of 1 at the same time, but not one quarter? > > > > > > Wouldn't it be simpler to just give a clear /definition/ of what you > > > mean by your "rational relation", instead of endless examples? > > > > > I gave a clear definition at the biginning of this discussion. I did > > not give it in the language of ZFC because it seems not possible to do > > so. > > > > Feel free to point me to such a "clear definition", in ZFC or any other > mathematical system. > > For example, when first you stated: > > > > > 1#1 ---> a > > > > 1#2 ---> a > > you did not also state "and 1#1 has size 1/2, as does 1#2"; and that > lead to some confusion. > > I'm not asking you to use exclusively the language of ZFC; but it is > reasonable to ask you to include /all/ the properties of "shares" in > your definition at /one time and one place/, and not scatter random > aspects of the definition until your are forced to reveal them through > repeated questioning. > > Clearly, we cannot understand a "rational relation" until we can first > understand what "shares" are. Here's what I can deduce from your > examples and occasional fragments of statements so far: > > Given an element x of a set X, we call a set x' "a subdivision of x > into m shares" if: > > (1) m is a natural number. > (2) if y is an element of x', it is called "a share of x". > (3) there are m distinct shares of x in x'; i.e., card(x') = m. > (4) each y in x' is associated with a real number shareSize(y) >= 0. > (5) sum (y in x') shareSize(y) = 1. > > Is that your definition, when m is a natural number? m need not be a natural number. We can also caculate limits. > Are there any > other properties of "shares" that we need to know about, or does this > cover everything you want to say about "shares" (we will address > "rational relations" when our knowledge of "shares" is complete)? > > If so, then what is your definition when m is /not/ finite, e.g., when > there is one distinct share of x for each natural number, or when there > are "as many shares as there are real numbers"? You can read i from the sum of the series. Do you doubt this sum? > > How should we determine "sum (y in x') shareSize(y) = 1" in these > cases? I know how to determine it in the case when m is finite, and I > can /guess/ what it means when m =|N|. > > But what does it mean when m = |R|? How do you define it, without first > /assuming/ that |N| = |R|? We do not need this assumption. We need not talk about R at all as long as arguing in the tree. We argue only 1 + 1/2 + 1/4 + ... = 2. Nothing else! It is you who tries to doubt this simple and clear formula on the grounds of unjustified assumptiobn about real numbers. Regards, WM
From: Franziska Neugebauer on 18 Dec 2006 02:06
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. [(P1)] This proposition P1 has _not_ yet been proved (shown). >> > > > >> > > >> > > 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. The Diagonal is unbounded thus any _assumed_ L_D is not bounded, too. Hence L_D cannot be a line of the list (meaning: cannot be _in_ the list) for any line in the list is bounded (proof by induction). Hence contradiction to P1. P1 must be dropped. > If your L_D does not contain them, then you have the wrong L_D. Upto now we have no line L_D at all since P1 has not been proved. F. N. -- xyz |