Cantor Confusion
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From: mueckenh on 6 Jan 2007 04:47 William Hughes schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > William Hughes schrieb: > > > > > I do not assume that 0.111... exists. I only assume that > > > given that 0.111...1 exists you can show that 0.111...11 exists. > > > > > That position is correct. It is called potential infinity. > > > > > > > > That is the same with the lines. Why should the diagonal exist actually > > > > but the system of lines should not exist actually? (1/9 = 0.111...) > > > > > > Indeed. It is the same with lines. It is always possible to find > > > another line. However, iii says nothing about whether the > > > diagonal "actally exists" (or equivalently whether the > > > system of lines "acutally exists"). > > > > But you always assert that the the complete diagonal exists, i.e., a > > diagonal which could not be extended. Then you must also accept a line > > system which cannot be extendend. > > No. At no time do I assume that the complete diagonal exists. > > > > > > > > > > > iv: Not all the elements of the diagonal exist > > > > > > > > Not all natural numbers in unary representation 0.1, 0.11, 0.111, ... > > > > exist. If all elements of the diagonal exist (which are the last digits > > > > of the unary numbers) then these unary numbers must exist too, as far > > > > as I understand existence. > > > > > > Any discussion of iv is completely irrelevant. iv is neither needed > > > nor used. > > > > Then drop it. > > I have never used it ("neither needed nor used"). > > > > > You used to use ~iv: all the elements of the diagonal exist > > > > Only to point out that I neither need nor use the > assumption ~iv. > > > > The contradiction is: > > > > > > A: L_D must have the property that given any set of elements > > > that exist in L_D one can always find another element of > > > L_D > > > [This follows immediately from i ii and iii] > > > > It is not different for the lines. > > > > > > B: L_D has a largest element. [this follows from the fact the > > > L_D > > > is a line] > > > > This follows from the assumption (iv) of complete existence of L_D, > > which implies complete existence of the system of lines. If you drop > > it, then the contradiction vanishes. > > No, it follows from the fact that L_D is a line. A line has a largest > element. The diagonal has not a largest element, unless you refer to potential infinity, where it always has a largest element but is not a fixed, complete infinity. > L_D is not the system of lines. > No assumption about the complete existence of the > system of lines is needed or used. What do you mean to have proved? The diagonal L_D is a line? The diagonal has a largest element? Potential infinity is actual infinity? Regards, WM
From: mueckenh on 6 Jan 2007 04:54 Virgil schrieb: > > > But the distinction is apparently difficult (although about first year > > > at University for mathematics). The set of terminating binary expansions > > > is countable, the set of non-terminating binary expansions is no > > > countable. (You may replace terminating binary expansions with binary > > > expansions terminating with either a continuous stream of 0'z or of 1's.) > > > > But these streams are *not* present in all paths of the union of all > > rational trees! > > No one, except possibly WM, is claiming that an infinite binary tree is > the same as the union of all binary rational, and therefore finite, > trees. > What else could it be? If it is something else: which nodes are distinguishing it from the union of all trees? > Those who have thinking brains are > capable of distinguishing between the union of infinitely many finite > trees and a single infinite tree. But they strictly refuse to tell what the difference is. > > > > > > > > You are wrong. sqrt(2) has a pretty good representation: "sqrt(2)". > > > > > > > > That is a name. Name ist Schall und Rauch. > > If we cannot refer to a number by its name, we cannot refer to numbers > at all. Here are some numbers: I -- ooo 0000000000 ***************** =========================================== > Do unnamable numbers have any existence at all? What numbers are ever > referenced in serious mathematics except by being named? According to set theory, most numbers have no names, no addresses and, therefore, most probably, no existence. ========================================== >> So where are the nodes and edges which make the infinite paths longer >> than any finite path? > If you don't know, no one can tell you. That's my impression too. ========================================== > I do not know what sort of trees WM is dreaming about, but in my > infinite binary trees, at each node there are infinitely many paths > branching left and infinitely many others branching right. Is in the union of all finite trees one node through which *not* infinitely many paths are branching left and infinitely many others are branching right? ========================================== > And the only object that can be a member of that union of trees is an > object which is a member of one of those trees. And every path in a > finite tree is a finite path. What about the union of all finite path? =========================================== > That infinite union of finite trees contains infinitely many nodes and > infinitely many edges in infinitely many finite paths, but no infinite > paths. No infinite path? Then every path is finite. Yes? The tree has oly a finite number of levels. The number of finite path in a tree with n levels is 2^n. Therefore: Finite number of paths <==> Finite number of levels (and finite number of nodes) Infinite number of levels <==> Infinite number of paths. ==> Infinite paths. Regards, WM Regards, WM
From: Andy Smith on 5 Jan 2007 20:17 >Read the chapter on mathematical definitions in Patrick Suppes's >'Introduction To Logic'. This is the best general textbook introduction >to the subject I've seen, and just technical enough without being too >technical. It will cast a great deal of light for you on how >definitions work in mathematical theories. Thanks, I will do that.
From: Andy Smith on 5 Jan 2007 20:23 >> I forgot to say, regarding the existance of an >> actually infinite integer, if you consider the >> sequence >> 1 + 1/2 + 1/4 + 1/8 ... > and label each element of the series 1,2,3, ... etc > then when you look at the interval [0,2] representing > the sum of the series I can point to the point at 2.00... > and say with confidence that the label of the series > member at that point is an actually infinite number? The n-th term of the series is a_n = 1/2^n and the n-th partial sum is S_n = sum_{k=1}^n a_k = 2 - 1/2^n for each natural number n. The sum of the series is S = lim_{n->oo} S_n = 2. Although the symbol "oo" appears in that last statement, it goes away completely if we replace the statement by its definition, which is: >For each epsilon > 0, there exists N > 0 such that | S_n - 2 | < epsilon for every n > N. >There are no infinite integers here. The numbers N and n are finite, and >epsilon is a real number, not an integer. Thanks. I thought that you would say that, and understand the analysis. It says that the series sum can become as close to 2.00.. as you like, but denies the infinity as a finished thing. If you have an idealised bouncing ball that loses 1/2 its height on each bounce and takes 1/2 the time, how many bounces has it done in 2.00.. seconds? Hasn't it achieved an actual infinity of bounces as a finished thing?
From: Dave Seaman on 6 Jan 2007 09:08
On Sat, 06 Jan 2007 06:23:03 EST, Andy Smith wrote: >>> I forgot to say, regarding the existance of an >>> actually infinite integer, if you consider the >>> sequence >>> 1 + 1/2 + 1/4 + 1/8 ... >> and label each element of the series 1,2,3, ... etc >> then when you look at the interval [0,2] representing >> the sum of the series I can point to the point at 2.00... >> and say with confidence that the label of the series >> member at that point is an actually infinite number? > The n-th term of the series is a_n = 1/2^n and the n-th partial sum is > S_n = sum_{k=1}^n a_k = 2 - 1/2^n for each natural number n. The sum of > the series is S = lim_{n->oo} S_n = 2. Although the symbol "oo" appears > in that last statement, it goes away completely if we replace the > statement by its definition, which is: >>For each epsilon > 0, there exists N > 0 such that >| S_n - 2 | < epsilon for every n > N. >>There are no infinite integers here. The numbers N and n are finite, and >>epsilon is a real number, not an integer. > Thanks. I thought that you would say that, and understand > the analysis. It says that the series sum can become as > close to 2.00.. as you like, but denies the infinity > as a finished thing. You were doing fine up to that last statement. It denies no such thing. In fact, the sequence in question is a mapping f: N -> R, and N is an infinite set (a "finished" infinity, if you like, but that term has no meaning in set theory). > If you have an idealised bouncing ball that loses 1/2 > its height on each bounce and takes 1/2 the time, > how many bounces has it done in 2.00.. seconds? > Hasn't it achieved an actual infinity of bounces as a > finished thing? It bounces infinitely many times. I don't know why you insist on adding the meaningless phrase "as a finished thing". But if you number the bounces sequentially, then each bounce gets a finite number. Mind you, there can be functions defined on larger ordinals (beyond omega), but you haven't mentioned any of those yet. -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. <http://www.mumia2000.org/> |