Cantor Confusion
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From: Dik T. Winter on 10 Dec 2006 21:01 In article <1165761763.908889.34550(a)80g2000cwy.googlegroups.com> Han.deBruijn(a)DTO.TUDelft.NL writes: .... > Let P(a) be the probability that an arbitrary natural is divisible by > a fixed natural a. Then P(a) = 1/a . Forbidden by set theory. No. Not specifically forbidden by set theory. Forbidden because there are no appropriate definitions for the words you are using (they are not used conforming to standard definitions, so you better supply definitions). In probability theory (as is commonly use) you have to define how you *select* your arbitrary natural. You have not done so, so probability theory does not have an answer. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 10 Dec 2006 21:07 In article <1165775585.061252.183930(a)n67g2000cwd.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > 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. You apparently have not read the definition in Merriam-Webster. And apparently never have heard about loanwords from one language to another with a change of meaning. In short, the "dummkopf" above is not German but English. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: David Marcus on 10 Dec 2006 21:18 Virgil wrote: > In article <1165778054.244969.226800(a)j72g2000cwa.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > 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. I think WM has posted several proofs. The problem is that none of them are valid. > Ergo, we have every right to reject his assertion. Ditto. -- David Marcus
From: Dik T. Winter on 10 Dec 2006 21:19 In article <1165778477.143980.296870(a)80g2000cwy.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > > > Pray reread what I wrote: "the nodes can be made to represent numbers in > > > > your tree". That is an easy exercise, I even did show it. The same for > > > > the edges, I did show that too. So actually the nodes and edges also > > > > represent numbers in some way. > > > > > > In the same way as the first few digits of a real number represents a > > > number. 3.1, 3.14, and so on represent numbers in some way. But that is > > > not at all important or interesting for the tree argument. > > > > So why did you state that I erronously believed that nodes represent > > numbers? > > Because you erroneously did. I did not. Reread what is stated above. > > > > > Not at all! I represent numbers by standard binary notations. > > > > > > > > It is using the limits where you are doing something non-standard. > > > > > > I do nothing. The tree cares that even in the limit the number of paths > > > cannot become uncountable. 2^n remains the cardinal number of a > > > countale set, even in the limit n --> oo. That's why I devised the > > > tree! > > > > And it is exactly that what is wrong. For each finite n 2^n is the > > cardinal number of a countable set (even of a finite set), that does > > not make something like that also true in the limit. > > What limit are you talking about? My statement is true for each finite > level n. There are no others. I was talking about the limit you just wrote about. > > It is easy > > enough to construct a bijection between the natural numbers and the > > edges, because the edges are countable. > > After all you have grasped that too? Until now you denied. Not at all, I only had some misapprehentions about the way you constructed your tree. > > Contrary to what you write > > elsewhere, you have *not* constructed a surjection from the edges to > > the paths. If you think you did that present us with an edge that > > maps to 1/3, and show how the mapping is constructed. > > Please give me all the bits of 1/3. Then I will show the bijection. I can't so you can not show a bijection. But I did not even *ask* for a bijection. You said you had constructed a surjection from the edges to the paths. To prove that show us a single edge that maps to 1/3 and also tell us how the construction of the *surjection* works. > > It is easy enough to construct a surjection from the edges to the > > binary rational numbers with terminating expansion. But indeed, WM > > does not show a surjection at all. > > Indeed you are unable to understand fractions? I understand them well enough. But fractions have nothing to do with this at all. If you state that part of an edge maps to some number and another part of that same edge maps to some other number that may be fine. But that does *not* define a surjection from edges to numbers. To define such a thing you should be able to tell the single number to which a specific edge maps. That is what a surjection is (and indeed is also what a mapping is). So you have not shown a surjection. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: David Marcus on 10 Dec 2006 21:22
cbrown(a)cbrownsystems.com wrote: > mueckenh(a)rz.fh-augsburg.de wrote: > > cbrown(a)cbrownsystems.com schrieb: > > > > > > Because your argument is of the form: "There exists a bijection f > > > between the naturals and the edges. > > > > This is evident... > > I'm glad you agree. > > > ... and has nothing to do with shares. > > On the contrary; we will use f to prove that there is a bijection > between the shares composing any given path p and N: each path p is > composed of a countable number of shares of edges. > > > > > > There exists a bijection h between > > > the paths and the reals. > > > > Again tat is evident ... > > I'm glad you agree that h exists. > > ... and has nothing to do with shares. > > On the contrary; we will use h to prove that there is a bijection > between shares of an edge and R: each edge is composed of an > uncountable number of shares of edges. > > > > > > 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. > > "... 1/2 + 1/2 = 1 and some stuff that like" is hardly a mathematical > definition, sophisticated or otherwise. > > Either you have a surjection from edges in the original diagram to > paths, or you don't. If you do, you have proven your claim that paths > are both countable and uncountable. If you don't, your claim that edges > are equinumerous with paths must be proved some other way. > > <snip> > > > Every edge is divided into shares, namely in as many shares as there > > are paths to which this edge belongs. > > Sweet Mary! Finally, something resembling a definition! > > Let B(e) be the set of paths to which the edge e "belongs". > > Then the set of shares of an edge e is a set of elements called > "shares", such that there exists a bijection from these shares to B(e). > > I suppose that if a share s is a share of edge a and s is also a share > of edge b, then edge a = edge b; is that correct? I.e., a "share of an > edge" is always "marked" as a share of some /specific/ edge e in the > original diagram, yes? > > If that is the case, I would suggest that we define "shares of edges" > as members of the set (paths X edges). We then define "the set of > shares of the edge e in the original diagram" as: > > S(e) = {(p,e) : p in B(e)}. > > There is an obvious bijection from S(e) <-> B(e); so as required every > edge is "divided" into "as many shares as there are paths" to which e > "belongs". And if s is a share of edge a, it is not a share of edge b, > unless a = b; so a "share of an edge" is uniquely associated with an > edge in the original diagram. > > By these definitions, if edge e belongs to path p, then the share of > edge e that belongs to path p is (p,e). Every share of an edge e (p,e) > belongs to some path p in B(e). Every path p is "composed" of shares > (p,e) where e belongs to p. > > To anticipate, let us also define "the composition of path p" as > > C(p) = {(p,e) : p in B(e)} > > So C(p) contains a unique "share of an edge" for each edge e that > "belongs to" the path p. > > Do you agree with the above definitions of B, shares, S and C? > > Now, B(e) = {p : e "belongs to" p}. We haven't defined "belongs to" > explicitly, but if you mean the "obvious" thing, then the set of paths > to which edge e belongs can be bijected with the set of all paths > (informally, by appending some unique finite length prefix associated > with the edge e to all paths). > > Next, recall that we have assumed a bijection h : paths <-> R. > Therefore, by composition, we have a bijection B(e) <-> paths <-> R for > every edge e; and thus a bijection S(e) <-> R for every edge e; in > other words, every edge is "divided" into an uncountable number of > "shares". > > Clearly, the cardinality of C(p) is the same as the cardinality of some > subset of the edges in the original diagram; and so card(C(p)) <= > card(edges). Equally clearly, the cardinality of C(p) is at least > countable ("every path is infinite"). Therefore, by the bijection f, it > follows that there is a bijection C(p) <-> N; in other words, every > path is composed of a countable number of shares of edges. > > <snip> > > > > > But it is no necessary to divide an edge at all, because ... > > Let's stick to one argument at a time, shall we? Or have you given up > on trying to prove your "rational relation" argument? > > You now have the following functions whose existence we have > established: > > (1) A bijection f between the naturals N and the edges. > > (2) A bijection h between the paths and the reals, R. > > (3) A function B : edges /in the original diagram/ -> sets of paths, > where B(e) = { p : edge e "belongs to" path p}, and card(B(e)) = > card(paths) = card(R). > > (4) The set of all "shares" is a subset of (paths X edges); i.e., "s is > a share" implies s is of the form (p,e) for some path p and edge e in > the original diagram, and where e belongs to p. > > (5) A function S: edges /in the original diagram/ -> sets of shares, > where S(e) = {(p,e) : p in B(e)}; and card(S(e)) = card(R) for all > edges e. > > (6) A function C: paths -> sets of share, where C(p) = {(p,e) : p in > B(e)}; and card(C(p)) = card(N). > > (7) A function g : (paths X edges X N) -> R, with the property that lim > n->oo sum(over all edges) g(p,e,n) = 2. > > In English, these 7 statements correspond to: > > (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 > > Now, how do we use the above functions to produce a surjection T : > edges /in the original diagram/ -> paths? That's a good question! > You have previously mentioned "full edges" as distinct from "edges in > the original diagram". How do they fit into the picture? What is the > role of the function g in your argument? -- David Marcus |