Cantor Confusion
in [Math]
|
Prev: Pi berechnen: Ramanujan oder BBP
Next: Group Theory
From: mueckenh on 11 Dec 2006 16:38 Virgil schrieb: > > > > 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. > > > > > > > > If you disagree please provide a counter example (with a finite line). > > > > > > The problem is not making a statment for each > > > finite line. The problem is combining all these statments > > > (one for each natural number n) into a single statement. > > > > > > Why should there be a problem as long as *all* lines are finite? > > Because what is being quantified is not line lenght, but line number, of > which there are infinitely many. Wrong. Quantified is over the finite number of elements of a line. It plays no role which line, because each one is finite. If you disagree please provide a counter example (with a finite line). > > > Do you have a counter example where quantifier reversal in any finite > > line / set was prohibited? > > There are lots of them for infinite sets, which is what is involved in > this case. We are just investigting this case. >The statement is about every line, of which there are > infinitely many, not every position in any one line. > > It is the case that for every line there are elements of the diagonal > not in that line. And it is the case that for every element of the diagonal there is a finite line. Regards, WM
From: mueckenh on 11 Dec 2006 16:40 Eckard Blumschein schrieb: > On 12/10/2006 8:16 PM, mueckenh(a)rz.fh-augsburg.de wrote: > > Eckard Blumschein schrieb: > > > > > >> > Then why do you disagree with Cantor's results? > >> > >> This is a good question. > > > > I disagree with those of his results which are in error. Cantor knew > > how to distinguish potential from actual infinity, that is more than > > most discussers here can achieve, > > This is true. > > but he was not infallible. > > I would like to apologize for not realizing that you wrote Gauss > brackets. I confess, it surprised me somewhat. But now it is ok. Regards, WM
From: Dik T. Winter on 11 Dec 2006 21:21 In article <457d8cc0$1(a)news2.lightlink.com> Tony Orlow <tony(a)lightlink.com> writes: > Dik T. Winter wrote: > > 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. > > Why does that matter? It does matter because if you do not properly define your problem, mathematics is not able to give an answer. > This is the same thing as your stupid ball and > vase trick. Why do you need to label anything, or know what you're > choosing from the infinite set? Because that is part of the problem setting. Giving that setten will allow mathematics to model the question and give an answer. And it is bad to think that because for a sequence of sets holds that lim{n -> oo} |S_n| = k with some particular value of k, that also | lim{n -> oo} S_n | = k because the latter statement contains something that has not been defined in mathematics. But even when we define it, it is not certain that it holds. Given the following (I think reasonable) definition: lim{n -> oo} S_n = S if: (1) for every element a in S there is an n0 such that a is in each of the sets S_n with n > n0 (2) for every element a not in S there is an n0 such that a is not in each of the sets S_n with n > n0. So from some particular point an element either remains in the sets in the sequence or remains out of the sets. With this definition (when we look at the rationals) we have that lim{n -> oo} [0, 1/n] = [0] and so: lim{n -> oo} | [0, 1/n] | = aleph0 != 1 = | lim{n -> oo} [0, 1/n] | (I am talking standard mathematics here). So taking cardinality and limits can not be interchanged except in some particular cases. But that is not unprecedented in mathematics. limits and integrals can also not be interchanged except in particular cases. And so can the interchange is not in general passoble if one of the things you interchange is a limit. Even interchanging limits is not in general possible. Consider: lim{x -> oo} lim{y -> oo} (2x + 3y)/xy -- 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 11 Dec 2006 21:30 In article <1165872695.605284.98210(a)f1g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > > 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. > > Reread what you wrote on 12 November "What is the node number of 1/3?" > and "Do you really think the node 1/3 is finitely far from the root in > the tree?" > > Reread what you wrote on 28 November: "But as I did show in another > article, when you assign bits to nodes, you can show that the nodes > represent numbers,..." I did not state that nodes represent numbers, I stated that it can be shown that nodes represent numbers. And I gave an easy proof. That you apparently could not follow that proof tells me a lot about your mathematical reasoning. > > > > 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, > > Look here: > > Dik T. Winter, Datum: Fr 3 Nov. 2006 16:18 > > You answered my assertion: > > The > > mapping N --> {edges} has been established such that every edge knows > > its number. > > You are wrong. Yes, that was when I though you were thinking about a different construction. > > > 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. > > Representations (= paths) which do not exist cannot be part of a > bijection. So you can not show a bijection (in your opinion), but nevertheless you state that you have given a surjection. Do you not think you are contradicting yourself a bit? > > 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. > > I showed you a sum of shares of two edges. Since when does mathematics > deny the vailidity of > 1 + 1/2+ 1/4 + ... = 2 ? > > > > 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. > > No? Even if there are enough parts to gather more than a whole edge to > be mapped on every path? No. Each part of an edge may map to a single path, that does *not* give a map from each edge to a single path. > > 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. > > You cannot add fractions to get a whole edge? I can. And I see that each edge maps, when you do this, to a plethora of paths, so that is *not* a surjection from edges to paths. The crucial thing in a surjection (and indeed for a mapping) from A to B is that each element of A maps to a *single* element of B. So, let me ask a different question. To which single path does the edge that goes left from the root map. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: mueckenh on 12 Dec 2006 05:22
William Hughes schrieb: > > Why should there be a problem as long as *all* lines are finite? > > The problem is not with any one line, it is with > the attempt to combine all lines into one. > The fact that *all* lines are finite does not > mean there is a last line. But it means that every line is finite. Therefore for every line we can reverse quantifiers. I do not understand why you always talk about how many lines there are. > > > Do you have a counter example where quantifier reversal in any finite > > line / set was prohibited? > > No. For any natural number n > the statement > > > For every m<=n, there exists a line L(m) > such that m is in L(m) > > can be reversed to form > > There exists a line L(n) such that for every m <=n, > m in in L(n) > > However the L(n) is different in > every pair. You can combine all lines into > > There exists a line L such that for every natural number n > n is in L > > if and only if there is a last line. > The finiteness of all lines would then imply that there are only finitely many lines (= natural numbers). Or the finiteness requirement of all lines must be dropped. Then we need a line with infinitely many elements. And in fact, if all elements of the set N actually do exist, then one of these two alternatives must apply. The second one cannot apply by the definition of natural number (= line). Therefore, only the first alternative remains. There is no actually existing infinite set N. I think you should consider it as a fact, that an actually infinite set of finite numbers is nonsense. The only way to maintain this claim is to assert that a diagonal can have more elements than any line. That alternative is not acceptable. Regards, WM |