Cantor Confusion
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From: mueckenh on 14 Dec 2006 05:03 Dik T. Winter schrieb: > In article <1166036419.914921.209020(a)j72g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > In article <457ece72(a)news2.lightlink.com> Tony Orlow <tony(a)lightlink.com> writes: > > > > Dik T. Winter wrote: > > > > > In article <457d8cc0$1(a)news2.lightlink.com> Tony Orlow <tony(a)lightlink.com> writes: > > > ... > > > > > > 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. > > > > > > You can assume as much as you want, that does not make it a definition. > > > What *is* a uniform distribution of all natural numbers? > > > > > > > > > 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. > > > > > > Yes, but that does *not* indicate anything about the limit, as I did show > > > below: > > > > > > > > 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? > > > > > > You have: |S_n| = 9n and you think that > > > | lim{n -> oo} S_n | = lim{n -> oo} | S_n | = lim{n -> oo} 9n. > > > but first "lim{n -> oo} S_n" is not defined, and second, when you define > > > it you have to prove that the first part is equal to the second part. > > > > > > > > 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? > > > > > > I do not understand. S_n are sets indexed by n and so form a 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. > > > > > > No. Any S_n with index larger than some n0. > > > > > > > > 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? > > > > > > and if it were not it would not be there by then. (Note that the point > > > where that is the case can be different for each element.) > > > > > > > > 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? > > > > > > Where do I claim that? > > > > > > > > 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? > > > > > > Yes, because the same holds for cardinality. > > > > Could you please explain what same holds for cardinality? > > If you read, it is right above this. I do not understand why you argue that in lim{x -> oo} lim{y -> oo} (2x + 3y)/xy = 0 = lim{y -> oo} lim{x -> oo} (2x + 3y)/xy interchanging limits is not possible. Regards, WM
From: mueckenh on 14 Dec 2006 05:09 Dik T. Winter schrieb: > > You misunderstood. What I explained to Dik is another mapping than that > > discussed by us. It is a mapping of one single edge on the path > > representing 1/3, the next edge is mapped on the path representing pi, > > and so on. I claim that for every real number I can name an edge to be > > mapped on this real number. > > Yes, you state that, and you can. But that does *not* provide a surjection > from the edges to the real numbers. And you stated you had constructed > such a mapping. There is a mapping. This is proven by the fact that there are more edges than paths. Perhaps we cannot display it in the special way you wish. But that is the same with the well ordering of the eals. There you believe in indirect evidence. When did you begin to believe that only constructed objects exist? > > > And noboy can disprove it by constructing a > > diagonal number, because the tree contains them all. > > RIght. The disprove is in the fact that whenever I state a sequence of > numbers that sequence inherently does not contain all real numbers. But the tree does. And the tree does not contain more paths than edges. It is pitty that you cannot combine two thoughts. Regards, WM
From: mueckenh on 14 Dec 2006 05:28 Dik T. Winter schrieb: > In article <1166035958.305158.282870(a)79g2000cws.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > > > > That one which goes left from the root is not engaged, because we need > > > > > > only half of the set of edges. > > > > > > > > > > So your construction is not a surjection. > > > > > > > > Of course it is! It is a surjection from the set of edges onto the set > > > > of paths. > > > > > > Than again. To what number maps the edge that goes left from the root? > > > If it is a surjection, there should be one. > > > > > A surjection onto the paths covers all elements of the *range* = every > > edge. It is not necessary that every edge is mapped on a path, to show > > that there are not less edges than paths. It is only necessary that > > there is no path without edge mapped on it. > > Yes. But you stated that you had constructed a surjection. And if you can > not state to what number a particular edge maps you have *not* constructed > a sufjection. So you now state that you did *not* construct a surjection? I proved the possibility of a surjection. > But again your statement "it is only necessary that there is no path without > edge mapped to it" is blatantly wrong. Consider the infinite set of decimal > numbers. Map a digit (from 0 to 9) to a number when it has somewhere in > its expansion that digit. If there are more numbers a digit maps to, use > shares. By this reasoning (which is your reasoning) there is a surjection > of the set of digits 0 to 9 to the decimal numbers. You have not understood, deplorably. The shares of every edge I use add to 1 edge. And every share is mapped on one path only. And every path gets as many shares to restate two full edges on its own. According to your proposal, every decimal number gets only some shares of the digits 1 to 9, but not as many shares as one digit is divided into. So not one full digit is mapped on a real number. > > > > > It is not even necessary to know what a mapping is in order to see that > > > > you cannot construct (or define) a path representing a number of [0,1] > > > > which is not in the tree. > > > > > > You had said that you had constructed a surjection. A surjection is a > > > mapping with specific properties. It is necessary to know what a mapping > > > is in order to construct a surjection, which you did not do. And I have > > > *no* idea what you mean with the second half of your sentence. > > > > If someone makes an asserted list of all real numbers and hands it out > > to Cantor, then Cantor can construct a diagonal number which is not in > > the list, contradicting the assertion. > > Right. > > > If someone makes a binary tree of all real numbers and hands it out to > > Cantor, then Cantor cannot construct a diagonal number which is not in > > the tree, so no contradiction of completeness is possible. > > You assume that some way to disprove one assertion also will work to > disprove another assertion. No I simply state that the diagonal proof fails in case of the tree. > > > The problem with Cantor's theorem aleph0 < 2^aleph0 is that the > > function f(x) = 2^x must be discontinuous: > > For natural x = n we have 2^n < aleph0, > > for the least infinite x = aleph0 we have 2^aleph > alep0. > > So there is a gap. The values between aleph0 and 2^aleph0 cannot be > > taken by the function 2^x. > > Yes? What is the problem with that? There are enough naturals that > cannot be taken by that function either. We can simply count the edges. They are countable. So we an count the biginnings of separated parts of paths (because every edge is a beginnng of the separated part of a path, notwithstanding which it may be). Therefore the beginnings of separated parts of the paths are countable. In the whole tree we have an uncountable set of separated parts of paths. The tree is continuous. Therefore your arguing fails. It seems there is a gap > between 2^1 and 2^2. Why is that not a problem? There is no gap in looking at the beginnings of separated parts of paths. If always counting from the left side, we come to the result 3 edges 0. / \ / and so on. Every number of beginning paths is there. > > > By means of the binary tree I can exclude any discontinuity. Therefore > > the number of branches isolated at a certain level n grows continuosly > > from 1 to its maximum value. If this was 2^aleph0, then the number of > > branches could not but call on aleph0 or at least at a value between > > aleph0 and 2^aleph0. To show that this is impossible is achieved by the > > big advantage of the tree: its undoubted continuity. > > Again, this makes no sense to me. Look at the edges. Do you deny that every edge is the beginning of that part of a path where it is separated from other paths? Do you believe that paths beginn to run separated wihout any edge being inolved? Regards, WM
From: mueckenh on 14 Dec 2006 05:35 cbrown(a)cbrownsystems.com schrieb: > And the domain is "every edge". Therefore, you claim to have > constructed a function T such that for every edge e, T(e) is a path. > And yet you seemed to deny this by saying: "That one which goes left > from the root is not engaged". The domain in my original proof (the rational relation with 1 + 1/2 + 1/4 + ...edges) is the set of all edges. But in fact not all edges are required. We can waste as (finitely) many as we want to remain enough to be mapped on every path. > > > > > > > > It is not necessary that every edge is mapped on a path, to show > > > > that there are not less edges than paths. It is only necessary that > > > > there is no path without edge mapped on it. > > > > > > > > > > That is poorly stated. The mapping that maps the first edge to the left > > > to every path provides us with a mapping such that "there is no path > > > without edge mapped to it"; but this mapping does not prove that the > > > cardinality of the set of paths <= the cardinality ({a single edge}) = > > > 1. > > > > You misunderstood. What I explained to Dik is another mapping than that > > discussed by us. It is a mapping of one single edge on the path > > representing 1/3, the next edge is mapped on the path representing pi, > > and so on. > > That sounds like a mapping from edges to paths to me. I give you an > edge, you give me a path. This can easily be accomplished: map the nth > edge to the nth path. The question is: is there a /surjection/? > > > I claim that for every real number I can name an edge to be > > mapped on this real number. > > And that sounds instead like a mapping from paths to edges. I give you > a real number, you give me a path. My point was: if that's what you > claim to be able to do, do you also claim that this mapping paths -> > edges is an /injection/? If not, it's easy to provide a non-injective > function by simply mapping every path to the same edge. The mapping paths to edges is an injection. The mapping edges (or a subset of them) to paths in a surjection. Regards, WM
From: mueckenh on 14 Dec 2006 05:39
Virgil schrieb: > WM deceives himself over the number of lines versus the number of > elements in any one line. The number of lines is not finite but the > number of elements in any one line is finite. Each line differs by 1 element from the preceding line. If the number of lines is actually infinite then the number of differences must be actually infinite too. Therefore the assumption of an actual infinity of lines without an actual infinity of elements is a first grade insanity. Here only potental infinity can be accepted. But that would disprove set theory in other places, for instance in Cantor's diagonal proof. > > > Then we need a line with infinitely many > > elements. > > WM needs a transfusion of sanity. > > > > I think you should consider it as a fact, that an actually infinite set > > of finite numbers is nonsense. > > While any physical infiniteness maybe impossible, until someone can show > that ZFC or NBG is internally contradictory, actually infinite sets will > continue to flourish within them. > If someone claims to be Napoleon I will not be able to cure him. But I hope to help all those who are at the point but have not yet fallen into this sickness. > > 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. > > It is to me, and to anyone who does not assume a priori that infinite > sets are impossible. Then our positions are clear, and there is no further reason to discuss this topic. It was useful, however, because many young students of mathematics can read now what they most probably have not yet known: The intellectual tortures to be suffered when assuming transfinite set theory to be correct. Regards, WM |