Cantor Confusion
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From: William Hughes on 12 Dec 2006 09:46 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: > > > > > > > > > > > > > > > > This is the argument you present. > > > > > > > > > > > > > > > > > > > > 1) The set of lines is also the potentially infinite set of natural > > > > > > > numbers. > > > > > > > 2) Every element of the diagonal is in at least one line. > > > > > > > 3) Every initial segment of the diagonal is (in) a line. > > > > > > > > > > > > No, there is one initial segment of the diagonal that is > > > > > > not in a line. > > > > > > > > > > Then name the element of the diagonal please, which supports your > > > > > claim. > > > > > > > > > > > > There is one initial segment of the diagonal which is not defined > > > > by a single element. > > > > > > > > The potentially infinite sequence > > > > > > > > {1,2,3 ...} > > > > > > > > is an initial segment of the diagonal, but it is not in a line, nor > > > > does it have a largest element. > > > > > > If it is not in any line then there must be at least one element of it > > > which is not in any line. > > > > No. Consider the trivial > > > > A_1 = {1} > > A_2 = {2} > > A_3 = {3} > > > > B = {1,2,3} > > > > There is no element of B which is not in one of the A's, but B is not > > in any of the A's. We need something more. > > And we have it: We are dealing with linear segments. Your set A_3 does > not satisfy this important requirement: A_1 is not a subset of A_3. > > This kind of inappropriate arguing occurs again and again, somtimes > with dancers and sometimes with lovers. But it dos not hit the point! > > Don't you wonder why it is always necessary to fail to argue with > linear segments? I can tell you: Because for linear segments this > aguing in invalid. > > Your statement that the union of all natural numbers is more than any > natural number is simply wrong. And you could know it by the fact, that > you cannot name an element of the union which is in no natural number. > According to extensionality then there is no difference. > > > > If we have the > > situation where A_i contains A_j, for i>j > > > > 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". >It has all to do with "every A is finite". No. From the statement "every A is finite" we cannot conclude that there exists and A_i that contains B. > > > > Which of the following three do you claim > > (note that none of them require assuming that > > the set of natural numbers actually exists). > > > > - {1,2,3 ...} is not an initial segment > > If no actual existence is assumed, what is the meaning of "..."? Which of the following three do you claim (note that none of them require assuming that the set of natural numbers actually exists). - The potentially infinite sequence of natural numers is not an initial segment - there is a natural number which is not an element of the potentially infinite sequence of natural numbers - there is a line for which you cannot find a natural number which is not an element of the line - William Hughes
From: William Hughes on 12 Dec 2006 09:56 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > Do you now claim > > > > the natural numbers are not a potentially infinite set? > > N is a potentially infinite set. > > > > the natural numbers do not exist? > > More than enoug do exist. (More than we will ever need could be brought > to existence.) -does a potentially infinite set exist? (note that this is not the same question as "do all the elelments of a potentially infinite set exist) - William Hughes
From: William Hughes on 12 Dec 2006 10:05 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > 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. > > > > Because there are three steps needed to reverse > > the quantifiers: > > > > 1. break up the original statement, giving > > on statement per line > > > > 2. reverse the quantifiers in each statment > > > > 3. recombine the statements to make a single > > statement. > > > > Step 3 (which you have not shown to be possible) depends on > > how many lines there are and whether there is a last line. > > There is only *one* important statement: > > "Every member of the diagonal is a member of at least one line." > > It is not necessary to break up anything and to recombine anything. Yes it is. A simple statement that it is not is insufficient. You have stated and I have agreed that if you have a statement about a single line you can reverse the quantifiers. However, you have not demontrated that this can be used to reverse the quantifiers in a statement about all lines without first breaking up the statement, reversing, then recombining . Until you demonstrate that this is possible, your simple statement carries no weight. - William Hughes
From: Tony Orlow on 12 Dec 2006 10:43 Dik T. Winter wrote: > 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. It's sufficiently defined if one assumes that there is a uniform probability distribution. > > > 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. > That problem has a clear answer with or without the labels: the sum diverges as f(n)=9n. The labels are confounding, not clarifying. > 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. I'm not sure what that statement is supposed to say. Can uoi give an example? 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 So, what, S_n is supposed to be an initial segment of the sequence? > 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. In (2), it sounds like a would not exist in ANY S_n if it's not in S. > So from some particular point an element either remains in the sets in > the sequence or remains out of the sets. You mean, at some point you can tell whether a given element a is in S, because if it were, it would be there by then? > > With this definition (when we look at the rationals) we have that > lim{n -> oo} [0, 1/n] = [0] Okay that interval degenerates to 0.... > and so: > lim{n -> oo} | [0, 1/n] | = aleph0 != 1 = | lim{n -> oo} [0, 1/n] | > (I am talking standard mathematics here). Are the |'s supposed to denote set size? If so, how can you claim that [0,0] contains aleph_0 elements? > > 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 True, but is it relevant?
From: Tony Orlow on 12 Dec 2006 10:49
Dik T. Winter wrote: > 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, consider this. If the number of levels in the tree is countable, then every path in the tree is finite, and marked by a specific edge where the value arises. Is that not true? In order to allow infinitely long strings, you must have an uncountable number of levels in the tree, in which case 1/3 will exist with a specific edge, infinitely far from the root node. Can you have infinite paths/strings with a countable number of levels/bit positions? TOny |