Cantor Confusion
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From: Franziska Neugebauer on 19 Dec 2006 11:54 mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: >> mueckenh(a)rz.fh-augsburg.de wrote: >> >> > Franziska Neugebauer schrieb: >> >> mueckenh(a)rz.fh-augsburg.de wrote: >> >> > William Hughes schrieb: >> >> [...] >> >> >> Assume that there exists an L_D which contains all the numbers >> >> >> contained in the diagonal. L_D is bounded, >> >> > >> >> > If an unbounded diagonal exists, then obviously an unbounded >> >> > line >> >> > must exist. >> >> > [(*)] > >> 6. You shall refocus on a proof of (*). > > Every element of the diagonal is n some line. Do you agree? > Every element of the diagonal is a finite natural number, by > definition. > > Up to the finite natural number n: Every element =< n of the diagonal > is contained in line n. Induction shows this is valid for any finite > natural number. That does not prove (*) since the whole diagonal is _not_ bounded by any such n. > There are only finite numbers in the diagonal. QED. Modus M�ckenheim? > If you cannot understand this proof, then simply try harder. I suppose I will not "understand" this "proof" even if you'll beat me with your new hard-covered concoction. > If you don't succeed but nevertheless are convinced of the contrary, > remember: N is a well ordered set. So there must be a least number n > of the subset for which my statement is not true. Try to find it. The burden of proof is upon you who crafted the claim (*). F. N. -- xyz
From: William Hughes on 19 Dec 2006 11:57 mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Franziska Neugebauer schrieb: > > >> mueckenh(a)rz.fh-augsburg.de wrote: > > >> > William Hughes schrieb: > > >> [...] > > >> >> Assume that there exists an L_D which contains all the numbers > > >> >> contained in the diagonal. L_D is bounded, > > >> > > > >> > If an unbounded diagonal exists, then obviously an unbounded line > > >> > must exist. [(*)] > > > 6. You shall refocus on a proof of (*). > > Every element of the diagonal is n some line. Do you agree? > Every element of the diagonal is a finite natural number, by > definition. > > Up to the finite natural number n: Every element =< n of the diagonal > is contained in line n. Induction shows this is valid for any finite > natural number. > > There are only finite numbers in the diagonal. QED. > > If you cannot understand this proof, then simply try harder. If you > don't succeed but nevertheless are convinced of the contrary, remember: > N is a well ordered set. So there must be a least number n of the > subset for which my statement is not true. Try to find it. You have correctly show that: A: For every natural number n, there is a line L(n), such that n is contained in L(n). You need to prove B: There is a single line, L_D, such that every natural number n is contained in L_D. A and B are not the same statement. You have not given a proof of B. (B follows from (A + linear ordering) iff there is a last line) On the other hand a simple proof of (not B) has been given. This proof assumes that all the elements of the diagonal are finite, and does not assume that every element of the diagonal exists. - William Hughes
From: Virgil on 19 Dec 2006 15:39 In article <29569$45879e6a$82a1e228$23181(a)news1.tudelft.nl>, Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> wrote: > Virgil wrote: > > > In article <1a591$458667c1$82a1e228$22650(a)news1.tudelft.nl>, > > Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> wrote: > > > >>Virgil wrote: > >> > >>>In article <1166090594.020341.42340(a)80g2000cwy.googlegroups.com>, > >>> mueckenh(a)rz.fh-augsburg.de wrote: > > >>>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 is misleading about a true and relevant statement? > > Both double limits exist but they have different values. > > > > The issue is whether such double limits are always reversible, > > and the answer, as demonstrated by the example above, is "NO". > > The issue is that your limits are actually an ill-posed problem. Does HdB claim that any of the limits mentioned does not exist according to standard delta-epsilonics? If not then the question of equality is quite well-posed.
From: Virgil on 19 Dec 2006 15:45 In article <1166526723.021088.294160(a)n67g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > > The Earth is round. That can be proved and has been proved - contrary > to the consistency of set theory. Whether the Earth is "round" depends on what one means by round. It is certainly not perfectly spherical. The earth has been proved not flat. ZFC has not been proved inconsistent.
From: Virgil on 19 Dec 2006 15:53
In article <1166526857.570806.262360(a)79g2000cws.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > WM presupposes things that do contradict ZFC. > > I presuppose the existence of irrational numbers as infinite > paths in the tree which differ from other paths, i.e., which exist as > individual entities. Actually, there is no natural bijection between the set of reals and the the paths of an infinite binary tree, as, under the most natural form of mapping between them, for certain rational reals there are corresponding two paths. However as this only occurs for countably many rationals and countably many paths, out of uncountably many, it is fixable. > > I conclude that, if their cardinality is to be 2^aleph0, then they > cannot exist and their cardinality is zero. WM often concluded things he cannot prove which others can prove false. > > > > > > > > > Your edges are not > > > > being broken into 2 "shares" each; they are being divided into an > > > > infinite number of shares, and that is very different from 2 shares! > > > > > > For this sake we have mathematics, I mean really correct mathematics, > > > not based on dubious infinite sets, which states that 1 + 1/2 + 1/4 + > > > ... = 2. > > > > That is not an equality but a hidden limit statement: > > Lim_{n --> oo} Sum[k=0..n: 1/2^k] = 2 > > Set theory veils even the most simple facts. We need not a limit in > order to calculate > 1 + 1/2 + 1/4 + ...+ 1/2^n + ... >= 2 - 1/2^n. And that is sufficient. Without a defined limit process, its not sufficient. |