Cantor Confusion
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From: David Marcus on 9 Nov 2006 19:11 Lester Zick wrote: > On Thu, 9 Nov 2006 00:09:14 -0500, David Marcus > <DavidMarcus(a)alumdotmit.edu> wrote: > > >mueckenh(a)rz.fh-augsburg.de wrote: > >> William Hughes schrieb: > > > >> > It is not true that a set cannot exist unless it has a maximum element. > >> > >> It is true, at least if all of its elements do exist. > > > >The set of natural numbers does not have a maximum element, so what you > >just wrote is clearly false. Want to try again? > > > >> But in order to force you to understand that simple truth, > > > >Calling something a "simple truth" does not make it more believable, > >especially when it is so easy to give an example to show that it is > >false. > > Look who's talking. Is believability your criterion of truth in > mathematics? Nope. In mathematics, we use proof. > Is "no contrary examples to show that it is false" your > criterion of truth in mathematics? Nope. But, a contrary example is a proof that the statement is false. > I suspect mathematics requires some > criterion of truth more substantial than your naive credulity. Indeed. That is why we prove things using axioms. -- David Marcus
From: David Marcus on 9 Nov 2006 19:21 mueckenh(a)rz.fh-augsburg.de wrote: > > William Hughes schrieb: > > > > It is necessary to distinguish between the lengths of the > > lines and columns and the indexes of the lines and columns. > > > > Let the set of line indexes be LI. > > Then LI is just |N. For every natural number n we have a line > > n. There is no last line so there is no line infinity. > > > > Let the set of column indexes be CI. > > Then CI is just |N. For every natural number > > n there is a column n. > > > > Let the set of column lengths be CL > > Then CL has only one element, aleph_0. > > The supremum of CL (also the maximum) is > > aleph_0. > > There are two sets the bijection of which has not yet been considered > but sould be: > A = The set of lengths of initial segments of the (first) column. > B = The set of lines. > Here we can set up a bijection, expressing the lines by the natural > indexes of the elements: > > 1 1 > 2 1,2 > 3 1,2,3 > ... ... > n 1,2,3,...,n > ... ... > omega 1,2,3,... > > which shows that omega or aleph_0 does not exist. Do you agree with the following statements? For each natural number, we have a line. Each line is assigned a natural number. The length of the first column is omega. The number of lines is omega. -- David Marcus
From: David Marcus on 9 Nov 2006 19:27 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > mueckenh(a)rz.fh-augsburg.de wrote: > > > William Hughes schrieb: > > > > > > > The supremum of W is omega. The width of the > > > > matrix is the supremum of the set of widths of the lines, i.e. > > > > omega. Thus the width of the matrix is equal to the height of the > > > > matrix. > > > > > > But this supremum is not taken. > > > > > > The diagonal connects width and length by a bijection. The element d_nn > > > of the diagonal maps the n-th line on the first n elements of the first > > > column. As long as n is a natural number, this is no problem. Only for > > > aleph_0 the diagonal has to map a not existing maximum on an existing > > > maximum. This is a hard task. > > > > No. You are confusing the length of the columns with the set of line > > indexes contained in each column. These are not the same. > > Explain how in a square matrix (squarity proven by the existence of a > diagonal) the columns can be longer than the lines for *every* line > and every column. What is your definition of a "square matrix"? > > The set of line indexes contained in the first column is |N. > > The length of the first column is the size of |N, aleph_0. > > > > The set of column indexes is |N. > > > > The bijection defined by the diagonal connects the line indexes > > with the column indexes. Finding a bijection between |N and |N > > is not hard. > > Of course not, because you can do that for finite indexes only. But > finding that there is a column with aleph_0 indexes while there is no > line with aleph_0 indexes shows that the idea of the existence of > aleph_0 is false. So, you are saying that a column can't be longer than every line. Does your argument also apply to the following (finite) list? 1 1 1 Here we have an array that has three lines and one column. The length of the column is three, which is greater than the length of every line (the latter being one). > > In other words, the fact that there are aleph_0 lines does not mean > > that there is a line with index aleph_0. Such a line would have > > to be the last line, however, there is no last line. > > If there were aleph_0 lines with aleph_0 a (non-natural) number, then > we would need a last line. Consider the following list. 1 1 1 .... There is a line for each natural number. How many lines do you say there are? According to you, is there a last line? -- David Marcus
From: Dik T. Winter on 9 Nov 2006 19:58 In article <1163068502.005252.83170(a)k70g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > In article <1162820871.886446.129490(a)k70g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: .... > > > are not omega? I was under the impression that there are aleph_0 lines > > > with less than aleph_0 columns. > > > > There are alpeh_0 lines and aleph_0 columns. But neither is a maximum. > > Both are supprema. There is no aleph_0-th line as there is no > > aleph_0-th column. So, where is there a maximum involved? > > There is the following difference: > The first and (any other) column has actually aleph_0 1's. > No line has aleph_0 1's. > So the first column is the maximum which is not reached by the lines. When you consider the columns, yes, each column has the same length, and so the height of the triangle is the maximum of these lengths, which actually is each and every length. But each length is a supremum, namely the supremum of the length of the columns in the finite triangles. The lengths of the lines are all finite, so the width of the triangle is the supremum of these lengths. But you may also compare it with stacking blocks on each other, each one 1 wide and 1/2^n high. When "completed" the stack will be 1 wide and 1 high. Just like in the reals the lengths of the intervals [0, 1] and [0, 1) are equal. > > > The number 0.111... or the diagonal has alep_0 1's as has the n-th > > > column but there i sno line having aleph_0 1's. > > > > Yes. What does that prove? *Not* that the infinite set of natural numbers > > contains a non-natural number. Because that is obviously nonsense. > > And this obvious nonsense results from the assumption that the set N > does exist. Pray give a proof starting with basics, that the set N contains a non-natural number. It is extremely simple to prove, based on the axiom of infinity, that the set N does not contain such a number. So if you can prove it (with the axiom of infinity in your set of assumptions), go ahead, and you have found a contradiction. But pray, a real proof, not some handwaving. > > > > But let's take the actually infinite set of natural numbers. > > > > Suppose it contains your non-natural number. Now remove that > > > > non-natural number from that set. Isn't the set still actually > > > > infinite? > > > > > > That shows only the self-contradiction of the notion actual infinity. > > > > No self-contradiction at all. What *is* the self contradiction? > > It is the set which is equinumerous with the 1's in the column, but not > with the 1's in any line. Why is that a contradiction? > > > You see not difference between the infinite man 1's of 0.111... and the > > > finite many 1's in every line? > > > > I see a difference, but I have no problem with that difference. > > The problem is that the bijection n <--> n fails. There are n numbers > in the set {1,2,3,...,n} (Pray, in the future do not use tabs, my tab-settings are different from yours, so I had to reformat the table.) > > 1 1 > 2 1,2 > 3 1,2,3 > ... ... > n 1,2,3,...,n > ... ... > omega 1,2,3,... > omega + 1 1,2,3,...,omega > ... ... > omega + n 1,2,3,...,omega + n + 1 This one is wrong, it should be: omega + n 1,2,3,...,omega,omega+1,...,omega+n-1 But I fail to see why it is not a bijection. This is a bijection between the union of the set of natural numbers and the set of ordinal numbers >= omega to the sets of ordinal numbers starting at 1 and terminating at either the natural number or at the ordinal number minus one. Except for the limit ordinals, where the mapping is to a non-terminating set. But omega is not a natural. When you want a logical mapping do use ordinals only, so let's do it with ordinals (0 is a proper ordinal, it is the ordinal of the empty set) 0 -> {} 1 -> {0} 2 -> {0,1} 3 -> {0,1,2} ... w -> {0,1,2,...} See? Each ordinal maps to the set of ordinals smaller than itself, and it is a bijection. > > > > No, not mixing at all. Supremum only. > > > > > > Then 0.111... is not different from the finite sequences of 1's? > > > > How do you conclude that? Given finite initial segments of the triangle > > we get a length, a width and a length of the diagonal that are all equal. > > When we look at the complete triangle (which exists by the axiom of > > infinity) all three become the supremum of the finite quantities, and > > so still are equal. No maximum involved. > > > The diagonal and the left side of triangle are complete with infinitely > many ones, but no line has infinitely many 1's. And, what exactly is the problem? > > Yes. What is the problem with that? The length is realised as omega as > > the supremum of all finite lengths, so is the width realised as omega as > > the supremum of all finite widths. Where is the difference? > > The length is not only a supremum but exists as a maximum (if all > natural numbers exist). No. When we start with finite triangles, also each column has a finite length. So when we have "completed" the triangle, the height of the triangle it the supremum of the set of the number of lines, and is also the supremum of the lengths of the finite columns. The width is the supremum of the lengths of the lines. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Virgil on 9 Nov 2006 20:17
In article <MPG.1fbd858a35c3b49198988b(a)news.rcn.com>, David Marcus <DavidMarcus(a)alumdotmit.edu> wrote: > Virgil wrote: > > In article <1163067640.060706.37510(a)k70g2000cwa.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Virgil schrieb: > > > > > > > > > > The length of the diagonal is less than omega. > > > > > > > > > > > > The number of 1's in the diagonal is aleph_0. So, what you wrote is > > > > > > false. In fact, I have no clue what you could possibly be thinking > > > > > > that > > > > > > would lead you to make such an obviously incorrect statement. The > > > > > > length > > > > > > of the diagonal is clearly the same as the length of the first > > > > > > column, > > > > > > and you just wrote above that the length of each column is omega. > > > > > > > > > > The length of the diagonal is clearly not more than the length of any > > > > > line. > > > > > > > > If every line is finite but at least of length equal to is line number, > > > > then even more clearly, the diagonal is longer that every line. > > > > > > That is outright nonsense. > > > > How is it nonsense when that diagonal contains a position for every n to > > say that it is longer that any line of length less than n? > > > > It is outrageous nonsense to deny it. > > > > > Its acceptance by set theorists makes set > > > theory suspicious as a theory without any value. > > > > Its denial by EB makes EB a laughing stock. > > I think you mean "WM", not "EB". My Bad. They both are, but in this case it is specifically WM, nor EB. > > > > > So that WM could hardly be more wrong if he tried. |