Cantor Confusion
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From: William Hughes on 10 Dec 2006 13:51 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > Dik T. Winter schrieb: > > > > > > > In article <1165492756.322548.255040(a)j72g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > > > Dik T. Winter schrieb: > > > > > > In article <1165421463.339178.48680(a)j44g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > > ... > > > > > > The set in the quantifiers you are using is not finite either. The > > > > > > quantifier are not over a single line, but over the set of natural > > > > > > numbers. > > > > > > > > > > For finite natural numbers, we have finite lines only. It is not the > > > > > question of a single line. Every line is finite. Therefore there is no > > > > > line where quantifier reversal could not be applied. > > > > > > > > But the quantifier reversal is *not* applied to individual lines, it > > > > is applied to the set of natural numbers. > > > > > > No. > > > > > > > The reversal given was > > > > For every natural number n there exists a line L(n), such that > > every natural number m <= n is an element of L(n) > > > > There exists a line L, such that for every natural number n, > > every natural number m<=n, is contained in L. > > > > Note the movement of the phrase "every natural number". > > > > Please provide an alternate formulation that does not > > involve the set of natural numbers. > > The set N can be involved and the quatifier can be changed as noted > above as long as it is asured that: > 1) every line has a finite number of elements > 2) there is no element of the diagonal outside of every line. > > If you disagree please provide a counter example (with a finite line). The problem is not making a statment for each finite line. The problem is combining all these statments (one for each natural number n) into a single statement. - William Hughes
From: mueckenh on 10 Dec 2006 14:06 Virgil schrieb: > In article <1165488738.492139.68600(a)73g2000cwn.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Bob Kolker schrieb: > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > > > One of many examples: The set {2,4,6,...,2n} has a cardinal number less > > > > than some numbers in the set. This does not change when n grows (yes, > > > > it can grow!) over all upper bounds. Therefore the assertion that the > > > > set of all even natural numbers has a cardinal number gretaer than any > > > > even number is false. > > > > > > The set of natural numbers has a cardinal greater then any set > > > > > > {1, 2, ... , n} for any integer n. > > > > Wrong. > > It is WM who is wrong Wrong. > > > The set of all natural numbers contains only natural numbers. > > These number count themselves by > > |{1, 2, ... , n}| = n. > > For every {1,2,3,...,n}, there is an n+1 greater than n. Correct. The set is potentially infinite. There is no number aleph_0 or omega counting all natural numbers. ========================= > Repeating a falsehood does not make it any less false. Why then did you repeat always that here are no balls in the vase at time 0? Why do you repeat without reasonable arguments that there are more paths than edges in the binary tree, although no path springs off without its own egde? ========================= > Why he maintains this this somehow bijects the individual branches with > paths, is not clear. Because every branch consists of two edges, one for the each of the resulting paths. ========================= > It is not the contents of a single line but the set of all lines that is > being quantified, so that the finiteness of lines is irrelevant. Wrong. We know from every line that it is finite. therefore it is uninteresting which line we consider. ========================== > Such faith in the illogical is charming, but doomed. > ZF, or NBG, or something like them, will be around long after we are all > gone. Of course, in the way as the old science astrology remains being around here, never betrayed by the majority - but this majority is a quantity, not a quality. =========================== > I did not claim to have personally shaken hands with each of them, as > Ramanujan was rumored to have done, but I know where they live. You think you know where some of them live. And you know them as well as you know the integer [pi*10^10^100] ============================= > > You cannot imagine the integer [pi*10^10^100]. > Why not, you just did! No. I did not. But I now understand why you think that you could know natural numbers. You simply think "number" and imply that this covers all numbers. You simply think "pi" and assert that you imagine the number pi. You simply think "ZFC is free of contradictions" and so you have proved that ZFC is free of contradictions. ================================== > Sqrt(2) has a directly approachable decimal numerical address. Yes. But there are only countably many addresses. Regards, WM
From: William Hughes on 10 Dec 2006 14:08 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > <statements which make it clear that certain > > things which were though to be settled are not settled> > > > > Terminology: If we say that X exists > > then we can use X in a proof. > > That already depends on what you understand by "a set exists". I > suspect that you understand that all its elements exist. If a set exists X we can use the statement "X exists therefore " in a proof. We cannot use the statment "The elements of X exist ". > > > > > > On Dec 4 I wrote: > > > > You now agree that a potentially infinite set can have > > a cardinal number and that this cardinal is not > > a natural number. > > > > As your latest post points out, this is not (or > > no longer) true. > > I never agreed. oo is not a number, so it is not a cardinal number, it > is at most a "cardinal number". > > > > > Stop me when I make a statement you disagree with > > We can then discuss this statement before proceding. > > > > -a potentially infinite set exists (this leaves > > open the question of whether the elements > > of a potentially infinite set exist.) > > > > -if we are given x and a potentially > > infinite set we can determine whether > > x is an element of the potentially infinite set. > > Correct. But we cannot determine every element of the set. > > > > > > > -a bijection can exist between two potentially > > infinite sets > > That again depends on what you understand by "to exist". I suspect that > you understand that all its elements exist. Even the bijection between > positive and negative integers: f(n) = -n is not complete, because > domain and range are not complete. > Therei is not claim that the bijection is "complete". The claim is that a bijection can exist. > > > > -given two potentially infinite sets A and B > > the question "Is there a bijection between > > A and B?" has an answer which exists. > > This answer again depends on what you understand by "to exist". That in a proof I can use the statement " A bijection between A and B exists, therefore ..." > > > > > -a cardinal number is an equivalence class on > > sets with respect to the equivalence relation > > bijection > > Yes, that is correct. But it does not satisfy the order-relation > "greater than" with natural cardinals. Since no order relation has been defined, this is vacuously true. (An extension is trivial, but we will not do that here). Note however it does make sense to talk about equals and not equals, which is all I ever do. > > > > > -the equivalence relation bijection can be extended > > to include potentially infinite sets > > > > -given a potentially infinite set A, the set C > > of ordered pairs (a,a) exists, where C has > > the property > > > > if > > a is an element of A > > then > > (a,a) is an element of C > > > > Call C the identity function. C is a bijection > > on A. > > That again depends on what you understand by "to exist". The identity > function f(n) = n does not provide the existence of all natural > numbers. We see it by the fact that otherwise finite (every number > including the limit) = infinite (the limit of initial segments of > natural numbers). > Completely irrelevent. The proof of existence of the identiy function on A does not require knowing that the elements of A exist. > > > > -A belongs to an equivalence > > class with respect to the equivalence relation > > bijection > > > > -A has a cardinal number > > > > -the cardinal number of A is not a natural number > > neither it is larger than every (or even any) natural number. However, it is not equal to any natural number, that is all that was claimed. > > > > > -given two sets of natural numbers E and F where E is a > > potentially > > infinite set, and F has a largest element. there does > > not exist a bijection between E and F > > > > -the diagonal is the potentially infinite set of natural > > numbers. > > > > -every line L has a largest number > > > > -there is no bijection between the diagonal and a line L > > There is no complete diagonal. Never claimed. It is not necessary to show that a potentially infinite set is complete to define a bijection on it. (see above) So, if we condsider your objections to be clarifications of what I said (note that in no case did you actually disagree with what was said), Do you agree with the above statments? - William Hughes
From: Mike Kelly on 10 Dec 2006 14:11 Tony Orlow wrote: > stephen(a)nomail.com wrote: > > Han.deBruijn(a)dto.tudelft.nl wrote: > >> stephen(a)nomail.com schreef: > > Do you or do you not wish to abolish any mathematics > > that involves infinity? If you are perfectly content > > to let others freely explore whatever they wish, then > > why are you so aggressive? > > > > I believe Han agreed that, if he saw a more satisfying treatment of > infinite sets, he'd be open to it. Of course, he's been rather resistant > to my alternatives, but it seems everyone is, for one reason or another. Because they're impossible for anybody but you to understand and of no apparent utility anyhow. > In any case, I think his objections are specific enough to warrant > some attention. "Calculus XOR Probability" was about the loss of > additive probability measure, when you have an infinite set of equally > likely possibilities, as a result of the notion of aleph_0 elements, and > its standard inverse, if there is such a thing, of 0% probability each. > No sum of 0's can be anything but 0. Is it unreasonable to want to > preserve additive measure within probability over an infinite set? I > don't think so, and the answer to that issue was obviously to allow some > infinitesimal probability for each natural. Doesn't work. Even in systems with infinitesimals, a countably additive union of sets with measure zero has measure zero. There is no uniform probability distribution over the natural numbers. Even if you "allow" infinitesimals. >In that case it can easily follow that the probability the n/3 e N is 1/3. Only by vigorous handwaving by people who really have no idea what they are talking about. >Set theory "disagrees". Interpret that as you wish. Measure theory and probability theory do too, more to the point. -- mike.
From: mueckenh on 10 Dec 2006 14:14
cbrown(a)cbrownsystems.com schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > cbrown(a)cbrownsystems.com schrieb: > > > > > > > So now you return to claiming that your mapping T /is/ a surjection > > > from edges onto paths. Perhaps there is some confusion regarding "what > > > is the domain/range of T?". > > > > > > To clarify, let e be any edge; say the first branch to the left in your > > > original diagram. According to your above statement, e is in the domain > > > of T. Which path is T(e)? > > > > You know that the two edges mapped on a path consist of shares of many > > edges. Why do you put your question? Why should I name a special edge? > > Because your argument is of the form: "There exists a bijection f > between the naturals and the edges. This is evident and has nothing to do with shares. > There exists a bijection h between > the paths and the reals. Again tat is evident and has nothing to do with shares. > There exists a surjection T between edges and > paths. Therefore, the composition h o T o f is a surjection of the > naturals onto the reals; contradiction." > > I assume you know what "a surjection T from edges onto paths" is, yes? > It doesn't map 1/2 of an edge to 1/4 of a path! It is a function that > maps each edge /in the original diagram/ to a path; so that every path > is in the image of the function. Of course. But what I do is somewhat more sophisticated. It requires to kow that 1/2 + 1/2 = 1 and some stuff that like. > > Since you refuse to describe what you mean by "shares" in a > mathematical sense, I can't assign any particular meaning to the > assertion "there are enough shares". How do you define this phrase so > that we can construct a surjection T (edges /in the original diagram/ > -> paths)? > Every edge is divided into shares, namely in as many shares as there are paths to which this edge belongs. > > > > > > > Everything in this tree is countable. > > Could you instead answer my questions regarding S(e)? What is the set > of shares associated with edge e, where e is an edge in the original > diagram? You claim it is a set; so one assumes it has members. The set of shares into which each edge is divided is an infinite set. One share for every path to which this edge belongs. But it is no necessary to divide an edge at all, because it is equally well possible to to map one full edge on every path. In order to disprove this assertion you have only two ways: (1) You have to find a path which does not split from another path by an edge. This is impossible. (2) You have to construct a diagonal number, which in the tree is impossible too. Regards, WM |