Cantor Confusion
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From: mueckenh on 12 Dec 2006 06:38 Dik T. Winter schrieb: > > > > Please give me all the bits of 1/3. Then I will show the bijection. > > > > > > I can't so you can not show a bijection. > > > > Representations (= paths) which do not exist cannot be part of a > > bijection. > > So you can not show a bijection (in your opinion), but nevertheless you > state that you have given a surjection. Do you not think you are > contradicting yourself a bit? I give a surjection on all existing paths. > > No? Even if there are enough parts to gather more than a whole edge to > > be mapped on every path? > > No. Each part of an edge may map to a single path, that does *not* give > a map from each edge to a single path. Since infinity = infinity became an allowed equation, 1/2 + 1/2 = 1 can no longer be true? > I can. And I see that each edge maps, when you do this, to a > plethora of paths, so that is *not* a surjection from edges to paths. > The crucial thing in a surjection (and indeed for a mapping) from A to B > is that each element of A maps to a *single* element of B. So, let me > ask a different question. To which single path does the edge that goes > left from the root map. That one which goes left from the root is not engaged, because we need only half of the set of edges. That one which goes right, could be mapped on that real number (path) which is asked for most frequently, namely 1/3. That edge leaving the node 1 on level 1 left is mapped on the next frequently mentioned number, namely pi. You will not be able to ask for more nmbers than I can name edges. And you will not be able to construct a "diagonal" path. But considering this catalog for another time, I remember that we have twice as much edges as paths. Therefore, we can waste the first countable infinity of edges, and if not a whole infinity, we can at least waste all the edges up to level n (for any finite number n) and notwithstanding this big loss, there remain enough edges to attach one to each path. That is so overwhelming, that I in fact cannot understand how you may dare to defend your deplorably swaying position. My sympathy and condolences. Regards, WM
From: William Hughes on 12 Dec 2006 07:17 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > > > Why should there be a problem as long as *all* lines are finite? > > > > The problem is not with any one line, it is with > > the attempt to combine all lines into one. > > The fact that *all* lines are finite does not > > mean there is a last line. > > But it means that every line is finite. Therefore for every line we can > reverse quantifiers. > I do not understand why you always talk about how many lines there are. Because there are three steps needed to reverse the quantifiers: 1. break up the original statement, giving on statement per line 2. reverse the quantifiers in each statment 3. recombine the statements to make a single statement. Step 3 (which you have not shown to be possible) depends on how many lines there are and whether there is a last line. - William Hughes
From: William Hughes on 12 Dec 2006 07:31 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > William Hughes schrieb: > > > > > > > > > > This is the argument you present. > > > > > > > > > > > > > > 1) The set of lines is also the potentially infinite set of natural > > > > > numbers. > > > > > 2) Every element of the diagonal is in at least one line. > > > > > 3) Every initial segment of the diagonal is (in) a line. > > > > > > > > No, there is one initial segment of the diagonal that is > > > > not in a line. > > > > > > Then name the element of the diagonal please, which supports your > > > claim. > > > > > > There is one initial segment of the diagonal which is not defined > > by a single element. > > > > The potentially infinite sequence > > > > {1,2,3 ...} > > > > is an initial segment of the diagonal, but it is not in a line, nor > > does it have a largest element. > > If it is not in any line then there must be at least one element of it > which is not in any line. No. Consider the trivial A_1 = {1} A_2 = {2} A_3 = {3} B = {1,2,3} There is no element of B which is not in one of the A's, but B is not in any of the A's. We need something more. If we have the situation where A_i contains A_j, for i>j A_1 = {1} A_2 = {1,2} A_3 = {1,2,3} B = {1,2,3} then B is contained in the last A_i. If there is no last A_I, then there is no A_i that contains B Which of the following three do you claim (note that none of them require assuming that the set of natural numbers actually exists). - {1,2,3 ...} is not an initial segment - there is an natural number which is not an elment of {1,2,3 ...} - there is a line for which you cannot find a natural number which is not an element of the line. - William Hughes
From: mueckenh on 12 Dec 2006 08:09 William Hughes schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > William Hughes schrieb: > > > > > > > > Why should there be a problem as long as *all* lines are finite? > > > > > > The problem is not with any one line, it is with > > > the attempt to combine all lines into one. > > > The fact that *all* lines are finite does not > > > mean there is a last line. > > > > But it means that every line is finite. Therefore for every line we can > > reverse quantifiers. > > I do not understand why you always talk about how many lines there are. > > Because there are three steps needed to reverse > the quantifiers: > > 1. break up the original statement, giving > on statement per line > > 2. reverse the quantifiers in each statment > > 3. recombine the statements to make a single > statement. > > Step 3 (which you have not shown to be possible) depends on > how many lines there are and whether there is a last line. There is only *one* important statement: "Every member of the diagonal is a member of at least one line." It is not necessary to break up anything and to recombine anything. The fact that every line is finite guarantees that either the diagonal is finite too or that the diagonal has at least one element which is not in the line. Regards, WM
From: mueckenh on 12 Dec 2006 08:26
William Hughes schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > William Hughes schrieb: > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > William Hughes schrieb: > > > > > > > > > > > > > This is the argument you present. > > > > > > > > > > > > > > > > > 1) The set of lines is also the potentially infinite set of natural > > > > > > numbers. > > > > > > 2) Every element of the diagonal is in at least one line. > > > > > > 3) Every initial segment of the diagonal is (in) a line. > > > > > > > > > > No, there is one initial segment of the diagonal that is > > > > > not in a line. > > > > > > > > Then name the element of the diagonal please, which supports your > > > > claim. > > > > > > > > > There is one initial segment of the diagonal which is not defined > > > by a single element. > > > > > > The potentially infinite sequence > > > > > > {1,2,3 ...} > > > > > > is an initial segment of the diagonal, but it is not in a line, nor > > > does it have a largest element. > > > > If it is not in any line then there must be at least one element of it > > which is not in any line. > > No. Consider the trivial > > A_1 = {1} > A_2 = {2} > A_3 = {3} > > B = {1,2,3} > > There is no element of B which is not in one of the A's, but B is not > in any of the A's. We need something more. And we have it: We are dealing with linear segments. Your set A_3 does not satisfy this important requirement: A_1 is not a subset of A_3. This kind of inappropriate arguing occurs again and again, somtimes with dancers and sometimes with lovers. But it dos not hit the point! Don't you wonder why it is always necessary to fail to argue with linear segments? I can tell you: Because for linear segments this aguing in invalid. Your statement that the union of all natural numbers is more than any natural number is simply wrong. And you could know it by the fact, that you cannot name an element of the union which is in no natural number. According to extensionality then there is no difference. > > If we have the > situation where A_i contains A_j, for i>j > > A_1 = {1} > A_2 = {1,2} > A_3 = {1,2,3} > > B = {1,2,3} > > then B is contained in the last A_i. If there is no last A_I, then > there is > no A_i that contains B That has nothing to do with "last". It has all to do with "every A is finite". > > Which of the following three do you claim > (note that none of them require assuming that > the set of natural numbers actually exists). > > - {1,2,3 ...} is not an initial segment If no actual existence is assumed, what is the meaning of "..."? > > - there is a natural number which is > not an element of {1,2,3 ...} Every natural number is an element of {1,2,3 ...}. > > - there is a line for which you cannot > find a natural number which is not an > element of the line. You can always find a natural number which is not an element of a given line. But by naming this number you name the line to which it belongs. It would beas wrong to state that a line containing every natural number exists, as it would be wrong to state that every natural number exists. But if the diagonal existed for every natural number, then a line containing every natural number had to exist too. Therefore the diagonal (= set of all natural numbers) does not exist. Regards, WM |