Cantor Confusion
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From: William Hughes on 13 Dec 2006 18:09 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. The question is: "Is B contained in one of the A_i?". The answer is: "B is contained in one of the A_i if and only if there is a last A_i." >Don't forget, the A_i > form a linear set. It is not possible for B to differ from A_n in > element a but not in element b and from A_m in element b but not in > element a. So A_i will contain all the elements from all the other A_j, if and only if for every A_j, i>=j. Another way to put this is 'Can we replace "all of the A's" by ''a single A_i" ?'. The answer to this question is 'We can make the replacement, if and only if there is a last A_i.' - William Hughes
From: Virgil on 13 Dec 2006 18:14 In article <1166051370.210055.87220(a)73g2000cwn.googlegroups.com>, "William Hughes" <wpihughes(a)hotmail.com> wrote: > 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. > > The question is: "Is B contained in one > of the A_i?". > > The answer is: "B is contained in one of the A_i > if and only if there is a last A_i." > > > >Don't forget, the A_i > > form a linear set. It is not possible for B to differ from A_n in > > element a but not in element b and from A_m in element b but not in > > element a. > > So A_i will contain all the elements from all the other > A_j, if and only if for every A_j, i>=j. > > > Another way to put this is 'Can we replace "all of the A's" > by ''a single A_i" ?'. The answer to this question is > 'We can make the replacement, if and only if there is > a last A_i.' > > > - William Hughes WM assumes (sometimes covertly, sometimes explicitly) that there is a last A_i, which is, of course, false in ZFC or NBG and most other set theories in general use.
From: cbrown on 13 Dec 2006 18:42 mueckenh(a)rz.fh-augsburg.de wrote: > cbrown(a)cbrownsystems.com schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > Dik T. Winter schrieb: > > > > > > > In article <1165923504.410525.226600(a)79g2000cws.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > > > Dik T. Winter schrieb: > > > > ... > > > > > > 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? > > > > > > > > > > I give a surjection on all existing paths. > > > > > > > > But not from the edges. But from parts of edges. > > > > > > > > > > > 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. > > > > > > > > > > Since infinity = infinity became an allowed equation, 1/2 + 1/2 = 1 can > > > > > no longer be true? > > > > > > > > So if the first half of ab edge maps to, say, 1/2 and the other half to, > > > > say, 3/4. To what single path maps the edge itself? > > > > > > Why isn't it sufficient to collect the shares of two edegs for every > > > path? That shows that there are not more paths than edges. Can you > > > explain your objection to factions? > > > > > > > > I previously asked you to "show" how you perform this mapping, given > > that we agree to the following: > > > > (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 > > > > Your claim (which you seem unable to completely articulate) > > Don't mix up lacking understanding of he matter with lacking precision > in stating it. 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". > > > seems to be > > that there are two "full edges" whose "shares" are exactly the shares > > of edges composing a particular path p. But a path p is composed of a > > countable number of shares; so a "full edge" can have at most a > > countable number of shares. > > 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. You claim that you can /prove/, using the 7 statements above upon which we have agreement, that there is a bijection between the reals and and the naturals; and that therefore countable = uncountable. When you say "You are aware of the number 2^omega being a countable one?", I would respond : that's what you're trying to /prove/. You can't just assert it that it's true for other reasons, and claim that therefore you have proven it using your "rational relation". .. > Do you think that lim {n-->oo}1/2^n * 2^n consists of uncountably many > parts? > lim {n->oo}(2^-n*2^n) = 1. 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/. > > Meanwhile, an edge in the original diagram > > consists of an uncountable number of shares. So full edges can only be > > said to be "the same as" edges in the original diagram if you /assume/ > > that countable is the same as uncountable. > > > > But that is what you are trying to prove - you can't simply assume it > > in a proof of it. > > How can 1 + 1/2 + 1/4 + ... +1/2^n + ... yield the value 2 if we add > only countably many shares which are uncountably small? > Why would I think that the real number 1 in the above series is a "share of an edge" in the /same sense/ that you have defined "shares of an edge" as being equinumerous with the real numbers? A "share of an edge" is associated with a particular edge and a particular path. 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. > Now let's argue with aleph0... Your other arguments are not so interesting to me. Cheers - Chas
From: cbrown on 13 Dec 2006 19:00 mueckenh(a)rz.fh-augsburg.de wrote: > cbrown(a)cbrownsystems.com schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > 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. > > > > What? The *range* of a surjection from edges onto paths is /not/ every > > edge; it is every path. The /domain/ of the surjection is every edge. > > (You actually teach Analysis, and you don't know this?!?!) > > Did you never copy and paste a wrong piece? > I corrected it already: The range is "every path". 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". > > > > > 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. Cheers - Chas
From: Dik T. Winter on 13 Dec 2006 22:58
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? 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. > > > 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. > 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. It seems there is a gap between 2^1 and 2^2. Why is that not a problem? Unless you allow sqrt(3) to be a number, but in that context aleph0 is irrelevant. (And, using J. H. Conawys method to define transfinites we also have sqrt(x) for all x in the range aleph0...aleph1.) > This is highly suspicious, if aleph0 is to be a number which is in > trichotomy with the natural numbers, but I can't prove it wrong. How about sqrt(3)? You can not prove it wrong because it is not wrong. > 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. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |