Cantor Confusion
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From: Han de Bruijn on 14 Dec 2006 10:42 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. Han de Bruijn
From: William Hughes on 14 Dec 2006 11:21 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?
From: William Hughes on 14 Dec 2006 12:57 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. - William Hughes
From: William Hughes on 14 Dec 2006 13:07 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: > > > > > > > > > > > > > > > > > A_1 = {1} > > > > > > > > > > A_2 = {1,2} > > > > > > > > > > A_3 = {1,2,3} > > > > > > > > > > > > > > > > > > > > B = {1,2,3} > > > > > > > > > > > > > > > > > > > > then B is contained in the last A_i. If there is no last A_I, then > > > > > > > > > > there is > > > > > > > > > > no A_i that contains B > > > > > > > > > > > > > > > > > > That has nothing to do with "last". > > > > > > > > > > > > > > > > If A_i contains B, then A_i contains any A_j. > > > > > > > > Therefore A_i is "last". > > > > > > > > > > > > > > > > > > > > No comment? > > > > > > > > > > If B contains any A_i then B is the last. > > > > > > > > B cannot be the last A_i, as B is not > > > > one of the A's (B corresponds to the diagnonal, > > > > the A_i correspond to the lines) > > > > > > In which element does it differ from every A_i? > > > > In no element. > > > > We know that B is contained in all of the A_i > > put together. > > How can you put all together, if there is no last one? You will never > be sure that you have all. We know that any element that can be shown to be in B can be shown to be in one of the A_j.. The question is: "Is there an A_D, such that any element that can be shown to be in B is in A_D?" The answer is: "There is an A_D, such that any element that can be shown to be in B is in A_D if and only if there is a last A_i." - William Hughes
From: cbrown on 14 Dec 2006 15:49
mueckenh(a)rz.fh-augsburg.de wrote: > cbrown(a)cbrownsystems.com schrieb: > > > > > > (1) Edges are countable. > > > > (2) Paths can be bijected with the reals. > > > > (3) B(e) is the (uncountable) set of all paths to which edge e belongs. > > > > (4) Shares (of edges in the original diagram) are denoted (p,e) for > > > > some path p and some edge e where e belongs to p. > > > > (5) S(e) is the (uncountable) set of all shares of edge e. > > > > (6) C(p) is the (countable) set of all shares which belong to path p. > > > > (7) 1 + 1/2 + 1/4 + ... +1/2^n + ... = 2 > > > But that is precisely the problem here: you are mixing up two different > > definitions of "part of an edge", because you have not been precise in > > stating what you mean by "part of an edge". > > An edge can be subdivided in x parts. If we collect x parts together, > then we have the same value as if we take one full undivided edge. Where x is... what? A natural number? A rational number? A real number? A cardinal? You have previously also asserted: > Every edge is divided into shares, namely in as many shares as there > are paths to which this edge belongs. Therefore /you/ assert that an edge /can/ be subdivided into "as many" parts as there are paths; which is to say, "as many" parts as there are real numbers (by the bijection (2)). This implies that you assert that the set of all parts of an edge can be bijected with the real numbers; so "x" is the cardinality of the set of real numbers. > > > You are aware of the number 2^omega being a countable one? If so, why > > > then do you speak of an uncountable set of shares? > > > > > > > Here's what I am "aware" of for (1)-(7): there is a bijection between > > paths and reals. There is a bijection between shares of edges and > > paths. Therefore, there is a bijection between reals and shares of > > edges of any particular edge. > > > > By "set X is uncountable" I mean there is a bijection between the > > elements of X and the reals. By "set Y is countable" I mean there is a > > bijection between Y and the naturals. > > Note that I have /defined/ what I mean by "uncountable" in this context. <snip> > No. According to ZFC the ordinal 2^omega is a countable set. Even > omega^omega^omega is a countable set. If your assertion is "2^omega is countable and in bijection with the reals", then we don't /need/ your "rational relation" argument to prove that there is a bijection between naturals and reals. But I am only interested in your "rational relation" argument as a way to /prove/ that there is a bijection between naturals and reals. > > If your question is "does the real number 1 > > consist of uncountably many parts", I must ask: what do you mean by a > > "part" of the real number 1? There are many interpretations of the > > term, each yielding a /different answer/. > > You talked about different parts of an edge. > > Start with 1/3 + 1/3 + 1/3. Do you think that these 3 parts of 1 make > 1? Here is where you make a new definition, which is not compatible with your other definition. Yes, 1/2 + 1/2 = 1. Also, 1/4 + 3/4 = 1; and 1/4 + 1/4 + 1/2 = 1. So does the number 1 have 2 parts, or does it have 3 parts? Or does it have four parts: 1/2, 1/3, 1/4, and 3/4? What does it mean to say "y is a part of 1"? Is 1/sqrt(2) a part of 1, or not? Remember that you have already /asserted/ that another property of "is a part of" is that an edge /can/ be subdivided into as many parts as there are real numbers. So, are you now asserting that it is possible to "make 1" by "adding together" some "value" repeatedly "as many times" as there are real numbers? What "value" would that be? Is it a real number? > > A "share of an edge" is associated with a particular edge and a > > particular path. > > No. In my proof we need only the value, not the individual edge. "No"? If the shares of edges of which you speak are not shares of particular edges in the original diagram, then what relevance do these shares of edges have to the original problem? > Because it is sure from other considerations that any part is used for > one path only., > > > A path is associated with a real number. But the sum > > of the real numbers that correspond to the paths to which edge e > > belongs is not 1; it is instead undefined (the sum does not converge). > > > > > Would you mind to think about that topic again? > > > > > > > Sure. I think that because you fail to make clear definitions, you > > don't even realize that you are using the term "share of an edge" to > > mean two completely different things; and this is the source of your > > mathematical error. > > Please inform you: 2^omega is not an uncountable set. I do not divide > edges in uncountably many parts. Please to inform you: you divide edges into "as many parts" as there are reals. You claim that you can prove using your "rational relation" argument that therefore there is a bijection between the naturals and the reals; and therefore the reals are countable. I have yet to see how you have proven this. If your claim is instead that "2^omega is the cardinality of the reals, and therefore the reals are countable", then that is a /different/ argument than your "rational relation" argument. I am only interested in your "rational relation" argument. Cheers - Chas |