Cantor Confusion
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From: Virgil on 10 Dec 2006 15:29 In article <1165775417.113389.137470(a)80g2000cwy.googlegroups.com>, 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. Note also: 3) for every line, there are elements of the diagonal outside that line. 4) there is a natural bijection between the set of last elements of lines ( one last element per line) and the set of elements of the diagonal (in which the last element of line n corresponds to the nth element of the diagonal). 5) there is no last line. 6) the diagonal does not have a last element. But does WM still wish to claim that "there exists a natural n such that for all naturals m, n > m" and "for all naturals m there exists a natural n such that n > m" are logically equivalent?
From: Virgil on 10 Dec 2006 15:33 In article <1165775585.061252.183930(a)n67g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Bob Kolker schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > > > > You cannot imagine the integer [pi*10^10^100]. > > > > That is not an integer, dummkopf. It is an irrational real number. > > In any case "Dummkopf" starts with a capital letter, and is applied by > uneducated people for uneducated people. i.e., such with little > knowledge or experience. In English usage one is not expected to capitalize nouns other than proper names, so it appears as if Wm is choosing to adopt "Dummkopf" a his proper name. > > By the way, have you never seen expressions like [1.8] = 1, involving > Gauss brackets? Do you know Gauss? Never met him personally. Not all his notatains have the rule of law. > > > Regards, WM
From: Virgil on 10 Dec 2006 15:43 In article <1165775686.611151.88330(a)79g2000cws.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > stephen(a)nomail.com schrieb: > > > David Marcus <DavidMarcus(a)alumdotmit.edu> wrote: > > > It is such an odd belief. Why use a set for something that a function is > > > naturally for? I don't really understand why cranks insist on using sets > > > for everything, while at the same time insisting that sets are useless > > > or illogical or whatever. > > > > I do not think it is that odd. In everyday usage, the word "set" is used > > to denote something that changes. But then again, so is the word "number". > > The number of people in a room may change, but that does not imply that > > a specific number, such as 5, changes. > > Only cranks may not know that a number is a set. In everyday usage, outside of certain set theories, numbers are not required to be sets. So WM is labeling the majority of humans as cranks. > Only cranks may not > know that functions may be defined on sets. That same majority may have no idea what a mathematical function is. > > > Most people seem to have an abstract > > enough concept of number that the common usage does not confuse them. > > However > > they do not apply this abstraction to sets, so if someone says the set of > > people > > in the room changes, they think a specific set changes. > > A number, according to set theory, is a set. Numbers are only required to be sets in such set theories in which everything is a set. There are set theories without any such requirement. > So what is appropriate for > numbers is also appropriate for sets. Even in those set theories in which every number is a set, not every set is a number, so there are things appropriate to numbers that are not appropriate to those other kinds of sets.
From: Virgil on 10 Dec 2006 16:01 In article <1165777585.721007.267970(a)16g2000cwy.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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. Yup, 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. Perhaps not in WM's philosophy, but there are more things... > > 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? Because that is what the logic requires. > 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? Because I have constricted an injection from the set of branches to the set of paths, and constructed and posted several proofs that there cannot be any injection from the set of paths to the set of branches (or to the set of nodes). > > > 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. WM is confusing branches (edges) with branch points (nodes). > > ========================= > > > 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. Then all lines must, in WM's eyes, be identical. The finiteness of lines does not mean the identity of all lines any more than the finiteness of decimal representations of naturals means such equality of such representations. > > ========================== > > > > 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. If Wm wishes to preserve his mathematical equivalent of astrology, he will not eradicate our mathematical equivalent of astronomy by doing so. > > =========================== > > > > 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. If you say that you did not write "[pi*10^10^100]", then perhaps you should lock your computer when not using it yourself, as someone did it in your name. > > ================================== > > > Sqrt(2) has a directly approachable decimal numerical address. > > Yes. But there are only countably many addresses. But that does not identify which numbers have them.
From: Virgil on 10 Dec 2006 16:20
In article <1165778054.244969.226800(a)j72g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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. WM has nothing that does not require existence of an infinite sequence of different nodes or different branches for each path, so his alleged bijection is not on branches or nodes but on infinite sequences of them. > > > > 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. As every branch "belongs" to (is a part of) infinitely many paths, that means every branch must be infinitely subdivided. > > > > > > > > > Everything in this tree is countable. Only in finite trees. > > > > 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. Then construct such a maping and present it for inspectin. > In order to > disprove this assertion you have only two ways: WM misses the point again. WE do not have to disprove anything about his assertion. If WM wishes to assert something and have his claim honored, HE must provide a proof in support of that assertion. WM has not done so. Ergo, we have every right to reject his assertion. In mathematics, assertions without proofs are called conjectures, or assumptions. Even Fermat's Last "Theorem" was not really a theorem, but merely a conjecture, until Wiles, et al, nailed it down. So WM's "tree theorem" is really only WM's "tree conjecture". But the counter-theorem that there are uncountably many branches IS a theorem, as it has been proved independently by several people. And when the negation of a conjecture is a theorem, the conjecture loses out. |