Cantor Confusion
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From: Han de Bruijn on 18 Dec 2006 03:46 Jonathan Hoyle wrote [ typo corrected by HdB ]: > [ ... ] Zermelo and Fraenkel > contributed greater rigor to mathematical thinking than Gauss and > Euler. [ ... ] Enough reason to finish this debate, I think. Han de Bruijn
From: Han de Bruijn on 18 Dec 2006 05:04 Virgil wrote: > In article <1166090594.020341.42340(a)80g2000cwy.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > >>I do not understand why you argue that in >>lim{x -> oo} lim{y -> oo} (2x + 3y)/xy = 0 = lim{y -> oo} lim{x -> oo} >>(2x + 3y)/xy >>interchanging limits is not possible. > > Nor do I. > > But for lim{x -> oo} lim{y -> oo} (2x + 3y)/(x + y) = 3 > and lim{y -> oo} lim{x -> oo} (2x + 3y)/(x + y) = 2, > one cannot exchange the order of the limits without changing the value > of the result. This is highly misleading. What's really interesting is the behaviour of the function f(x,y) = (2x + 3y)/(x + y) for large values of x _and_ y. This can be investigated by substituting x = t.cos(phi) , y = t.sin(phi) Giving: f(phi) = (2.cos(phi) + 3.sin(phi))/(cos(phi) + sin(phi)) which turns out to be independent of t for t <> 0 . So for (x,y) -> oo we find a much more complicated behaviour, dependent on the angle phi, being the the direction in which we are "looking". It turns out that the function has several singularities, namely where sin(phi) + cos(phi) = 0, and, as a function of phi, it is increasing everywhere; take e.g. the derivative f'(phi) = 1/(cos(phi) + sin(phi))^2 . For phi = 0 we find indeed a limit = 2 and for phi = pi/2 we find indeed a limit = 3 , but these limits are only quite special cases of the function's "real" behaviour at (oo,oo). Han de Bruijn
From: William Hughes on 18 Dec 2006 07:47 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > William Hughes schrieb: > > > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > William Hughes schrieb: > > > > > > > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > > William Hughes schrieb: > > > > > > > > > > > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > > > > Virgil schrieb: > > > > > > > > > > > > > > > > > > > > > (It is contained in the union of all lines, but the > > > > > > > > > > > > union of all lines is not a line) > > > > > > > > > > > > > > > > > > > > > > That is a void assertion unless you can prove it by showing that > > > > > > > > > > > element by which the union differes from all the lines. > > > > > > > > > > > > > > > > > > > > Not quite. In order to achieve that the diagoal is not in any linem all > > > > > > > > > > that is required is: > > > > > > > > > > Given any line there is an element of the diagonal not in THAT line. > > > > > > > > > > It is not requires that: > > > > > > > > > > There is an element of the diagonal that is not in any line. > > > > > > > > > > > > > > > > > > > > > > > > > > > For linear sets you cannot help yourself by stating that the diagonal > > > > > > > > > differs form line A by element b and from line B by element a, but a is > > > > > > > > > in A and b is in B. This outcome is wrong. > > > > > > > > > > > > > > > > > > Therefore your reasoning "there is an element of the diagonal not in > > > > > > > > > THAT line. It is not required that: There is an element of the diagonal > > > > > > > > > that is not in any line." is inapplicable for linear sets. You see it > > > > > > > > > best if you try to give an example using a finite element a or b. > > > > > > > > > > > > > > > > > > > > > > > > In every finite example the line that contains > > > > > > > > the diagonal is the last line. > > > > > > > > > > > > > > Every example with natural numbers (finite lines) is a finite example. > > > > > > > > > > > > > > > Your claim is that there is a line which contains the diagonal. > > > > > > > > > > > > > > Because a diagonal longer than any line is not a diagonal. > > > > > > > > > > > > > > > Call it L_D. Question: "Is L_D the last line?" > > > > > > > > > > > > > > There is no last line > > > > > > > > > > > > Then, there is a line that comes after L_D. > > > > > > > > > > > > Therefore :L_D does not contain every element > > > > > > that can be shown to exist in the diagonal. > > > > > > > > > > All elements that can be shown to exist in the diagonal can be shown to > > > > > exist in one single line. > > > > > > > > > > > > > Call it L_D > > > > > > > > L_D contains a largest element. n. > > > > > > > > L_D is not the last line, so there is > > > > a line with element n+1, > > > > > > > > Element n+1 can be shown to exist in the diagonal. > > > > > > > > > Element n+1 can be shown to exist in L_D (which is obviously a line > > > containing n+1). > > > > No. L_D is bounded. The largest element of L_D is n. > > L_D does not contain n+1. > > You misinterpret L_D. L_D is that line which contains all numbers > contained in the diagonal. If your L_D does not contain them, then you > have the wrong L_D. Assume that there exists an L_D which contains all the numbers contained in the diagonal. L_D is bounded, therefore there exists a largest element, call it n(L_D). L_D is not the last line therefore there exists a line containing n(L_D) + 1, therefore the diagonal contains n(L_D) + 1, L_D does not contain n(L_D) + 1. Contradiction. Therefore L_D does not exist. - William Hughes
From: William Hughes on 18 Dec 2006 07:56 Han.deBruijn(a)DTO.TUDelft.NL wrote: > William Hughes schreef: > > > Han de Bruijn wrote: > > > > William Hughes wrote: > > > > > > > Han de Bruijn wrote: > > > > > > > >>William Hughes wrote: > > > >> > > > >>>Han de Bruijn wrote: > > > >>> > > > >>>>William Hughes wrote: > > > >>>> > > > >>>>>Han de Bruijn wrote: > > > >>>>> > > > >>>>>>Let's repeat the question. Does there exist more than _one_ concept of > > > >>>>>>infinity? Isn't unbounded the same as infinite = not finite = unlimited > > > >>>>>>= without a limit? Please clarify to us what your "honest" thoughts are. > > > >>>>> > > > >>>>>You are *way* in deficit on clear answers. Try answering > > > >>>>>the following question with yes or no. > > > >>>>> > > > >>>>> Is there a largest natural number? > > > >>>> > > > >>>>No. > > > >>> > > > >>>I there an unbounded set of natural numbers? > > > >> > > > >>Suppose you mean "Is". What does it mean that a set is unbounded? > > > > > > > > An unbounded set of natural numbers is a set of natural > > > > numbers that does not have a largest element. > > > > Please answer yes or no. > > > > > > No. > > > > Does the "potentailly infinite set" of natural numbers > > exist? A potentially infinite set is a function on sets > > that takes on the values true and false. The > > potentially infinite set of natural numbers > > takes on the value true for a set containing only > > natural numbers, and false for any other set. > > (i.e. is there a way of recognizing a set of > > natural numbers?) > > I don't understand this question. > Stop me at the first line you don't understand. [Note if you do not like the term "potentially infinite set" substitue "recognizer".] A "potentially infinite set" is a function on sets. A potentially infinite set takes on the values true and false. The "potentially infinite set of natural numbers", N, is a potentially infinite set such that N(X) is true if X is a set of natural numbers, and false if X is not a set of natural numbers. Does the "potentailly infinite set of natural numbers" exist? - William Hughes
From: mueckenh on 18 Dec 2006 08:47
Virgil schrieb: > > All elements that can be shown to exist in the diagonal can be shown to > > exist in one single line. > > Which line is that? That one which guarantees that a diagonal can contain all the elements which it contains. It is wrong to assume that this could be guaranteed by many lines. Either it is guaranteed be one line or it is not the case at all. > That presumes a last line, which presumption is > unwarranted in ZFC and NBG and most other set theories. It presumes that all elements of the diagonal exist in the EIT. ZFC and NBG and most other set theories which wish to make us believe that all elements of the diagonal do exist although not in one line but distributed over many lines, are obviously wrong. For linear sets it is impossible that an element exists in line m but not in line n > m. Threfore all elements which exists in smaller lines exist in a larger line. The result that only many finite lines contain the complete infinite diagonal shows nothing but a lack of logic thinking and the non-existence of a complete infinite set. > > > > The easiest way to see that would be to name all elements of the > > diagonal. Every element is a natural number and as such a set > > containing all the smaller natural numbers. In fact it is irrelevant > > whether you write > > 1 > > 2 > > 3 > > or > > > > 123. > > > > If the diagonal contains only finite segments, then each one is in a > > line and vice versa each line is in the diagonal. Only if there is an > > infinite segment in the diagonal, then it is not in a line. But in > > order to have the infinite segment, you need an element which is not in > > any line, i.e., an infinite element. > > The idiotic idea that an endless sequence must contain an infinite > object is unwarranted. It is a delusion that such as WM feed their other > delusions upon. > > Consider, for example, the sequence > f(n) = sin(n radians). > It has no last term but every term has a distinctly finite value falling > between -1 and 1, and each term is distinct from all previous terms. Already the sequence (1/n) serves as an example. But the sequence (n) cannot satisfy it because n = 1+1+1+...+1 (1 added n-times) and reaching infinity in the "number of numbers" n = |{1,2,3,...,n}| while not reaching infinity in the number n itself is excluded here. > > Similarly, consider the sequence, g(n) = 1/2^n. According to WM there > must be some n so large that either g(n+1) = g(n) or g(n+1) = 0. No. This example has nothing to do with the observation that between n and n+1 the difference is always 1. Regards, WM |