Cantor Confusion
in [Math]
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From: Virgil on 27 Apr 2007 16:48 In article <1177674968.281638.41520(a)u32g2000prd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 25 Apr., 04:28, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > > Cantor's diagonal can be defined up to any line n of the list. But > > > > > this says *nothing* about an infinite sequence of digits. > > > > > > > > It is defined for *any* line of the list. > > > > > > That does not define it for *all* lines. > > > > It does. > > If it were, than the number of digits were complete. Otherwise there > can be no judgement concerning "all" but at most concerning "each". > But completeness of digits means presence of all, up to the last one > (which does not exist). If you do not agree to my arguing, then you > also must not agree on the idea that single paths exist in the tree. Wrong and wrong and wrong and wrong! > > In the tree we have In the CIBT we have: (1) We have an infinite set of levels, L (2)and for each subset of L a path which branches left at the levels in that subset and right elsewhere, (3) and different subset of L produce different paths, (4) and so there are as many paths as members of P(L) (5) and Card(set of paths) = Card(P(L)) > Card(L) = Aleph_0 > > Wrong. That must only be true if there is a largest natural number. > > Of course. If all natural numbers are available, then there must be a > largest, because the naturals obey trichotomy. Nothing in trichotomy demands a largest. The rationals have trichotomy but no largest. Such phony "logic" is typical of WM's arguments. Please do not mistake > all for each. You can name each natural number (in fact you cannot, > but let's assume, you could), but you cannot name all natural numbers. > All means "completed". Completed means, there is nothing to be added. > In case of a set the elements of which obey trichotomy, there is a > largest one. What version of "trichotomy" does WM claim says that? WM has now descended to outright lies. > > > Consider the following mapping: > > Let's have 'r' a rational number > 0. > > Rewrite 'r' as a continued fraction [r0, r1, ..., rn+1], where the > > ri are natural numbers, except for r0, which can be 0. It is easily > > shown that such a representation is unique. Next map that number to: > > 2^r0.(1 + 2^r1.(1 + ... (1 + 2^rn)...)). So the continued fraction is > > mapped to an integer which has in its binary expansion a specific > > spacing > > of the 1 bits. From this a reverse mapping is obvious. > > Pray tell which rational does not map to an integer, and which integer does > > not map to a unique rational. And also tell me which integer maps to the > > largest rational smaller than 1. > > Your mapping seems to be a unique one but this is only due to your > point of view. > The other possible point of view which is not less justified yields: > > Every path is simultaneously a path bunch, because down to level L(n) > every node belonging to path p is also a node of path p' =/= p. This > holds for every n in N - and others are not available. If we denote > the n-th node belonging to path p by K(p,n) then we have: > forall K(p,n) with n in N thereis p' such that forall m in N with m < > 10^n : K(p',m) = K(p,m). Then there is this simple point of view, which WM is careful to ignore, as it shows the falsity of his own arguments: A CIBT has a set of infinitely many levels, L, every subset of which determines a different path, the one branching left at just those levels. So card(set of paths) = card(P(L)) > card(L)
From: Virgil on 27 Apr 2007 17:01 In article <1177676193.672124.271620(a)b40g2000prd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 27 Apr., 13:56, William Hughes <wpihug...(a)hotmail.com> wrote: > > On Apr 27, 6:40 am, mueck...(a)rz.fh-augsburg.de wrote: > > > > On April 17 Mueckenheim writes > > > > M: Each "point of separation" is a single node > > > > On April 18 Mueckenheim writes > > > > M: a set of separation points can separate a path from all other > > paths. > > You quote this assertion, casually forgetting that it is only valid in > case the path does exist. Given the path exists in the tree, separated > from all other paths, then there must be a set of separation points. > Meanwhile we have found out that this is not the case. Actually, any infinite subset of the set of nodes of a path is enough to separate that path from all others. so that WM's clain is, as usual, contrary to fact. But a simpler way of seeing that WM is wrong is to look at the set of levels of a CIBT. The root is level 1, its child nodes are at level 2, and the children of nodes at any level n, for n e N, are at level n+1 So that the set of all levels, L and the set of all naturals, N, as in ZF and NBG for example, are the same. For each subset, S, of L, let f(S) represent the path which branches left at each member of S and right at each member of L\S. Then if M is the set of paths, f:P(L) --> M is a bijection. Thus there are 'as many' paths in M as subsets of L or as subset of N, so Card(M) > card(N), despite WM's continual denials.
From: Dik T. Winter on 29 Apr 2007 18:25 In article <1177672072.574980.291080(a)n15g2000prd.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 25 Apr., 04:01, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > This does not make a difference. When you remove the stuff between > > parenthesis it makes no sense. There is *no* node that separates a > > specific path from another one. > > Small wonder. There is no specific path in the tree. > > The tree shows us that every path is simultaneously a path bunch, > because down to level L(n) > every node belonging to path p is also a node of path p' =/= p. This > holds for every n in N - and others are not available. If we denote > the n-th node belonging to path p by K(p,n) then we have: > forall K(p,n) with n in N thereis p' such that forall m in N with m < > n^2 : K(p',m) = K(p,m). Yes. So what? Also for all p and p' != p there is an n such that K(p, n) != K(p', n). > > > In order to get two separateed bunches of paths, there is one bunch > > > required which contains all these paths and one node to split the > > > incomin bunch into wo bunches. > > > > As far as I understand from your book, there is no splitting of bunches. > > There are only nodes where bunches terminate. > > A bunch like 0,0 is split into two bunches 0.00 and 0.01 by the due > node. No. The bunch 0.0 terminates and the bunches 0.00 and 0.01 continue. The first edge (going from the root of the tree to the first node) is part of all three bunches. So there is *no* splitting. > > According to the book a > > bunch of paths is a set of paths with a common initial segment (i.e. > > set of nodes). So at each node at least three bunches are involved > > (actually more). > > That is a matter of convention of notation. In any case a node > increases the number of separated path bunches by 1. No. You are confusing bunches with edges. The number of paths does *not* increase at a node. The number of paths remain the same, but some go right and others go left. > > There is the bunch that terminates with the node in > > question, and there are the two bunches that go left and right. All > > three are incoming bunches of the node, and only the latter two are > > outgoing bunches. > > No matter how these are defined, there is the simple mathematical > equation: > One bunch comes in, two bunches go out, the difference is due to the > node splitting the incoming bunch. But that is wrong. The correct statement is "one edge comes in and two edges come out". And indeed, the number of edges is countable. As I already stated, there *is* not a single bunch or path that comes in at a node. All bunches and paths that come out of a node also do come in it. > > > The infinite tree contains all separated paths. Therefore there are as > > > many nodes (minus one) required in the tree. > > > > A proof is still required. > > Simply think: 1 pathbunch goes in, X pathbunches are asserted to be in > the tree. Every node creates one mor pathbunch. X-1 nodes are > required. But there *is* not 1 path bunch that comes in. > > > Should no separated paths exist, then they would also not exist in > > > Cantor's list as entries or as diagonal number. > > > > Makes no sense. > > Cantor's list? I agree. There is no isolated sequence in the list. Your statement that I quoted makes no sense. -- 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 29 Apr 2007 18:45 In article <1177674968.281638.41520(a)u32g2000prd.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 25 Apr., 04:28, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > > Cantor's diagonal can be defined up to any line n of the list. But > > > > > this says *nothing* about an infinite sequence of digits. > > > > > > > > It is defined for *any* line of the list. > > > > > > That does not define it for *all* lines. > > > > It does. > > If it were, than the number of digits were complete. Otherwise there > can be no judgement concerning "all" but at most concerning "each". And in mathematics if something is proven for "each", it is proven for "all. Meaning that if something is proven for "each" element of a set, it is proven for "all" elements of a set. And because the set of all natural numbers exists by the axiom of infinity, this also holds for natural numbers. If something is proven for each natural number, it is proven for all natural numbers (which does *not* mean that it is proven for the set of all natural numbers, because that is not a natural number). > In the tree we have, if K(p,n) denotes the n-th node of path p, > forall K(p,n) with n in N thereis p' such that forall m in N with m =< > n : K(p',m) = K(p,m). Yes, I never did contradict this. On the other hand, for each p and p' != p there is an n such that K(p, n) != K(p', n). > Up to "each" node of p, there is a path p' having the same set of > nodes up to level n. If "each" means "all" in this case, then for path > p there is a path p' =/= p which has the same set of nodes as p, i.e., > which is p' = p. And indeed, for "all" nodes given a path p that goes through that node, there is a path p' that also goes through that node. But that does *not* mean that p and p' have the same set of nodes. > Why do you think, "each" mean "all" in case of Cantor's diagonal but > not in case of the tree? It does, but you have to properly interprete it. You are thinking along the lines of "up to all nodes", but that makes no sense. > > > Cantor's (first diagonalization) method allows to enumerate every > > > rational number. So each rational has its place in the sequence. But > > > this sequence does not enumerate *all rationals". Otherwise there must > > > be the largest rational less than 1. > > > > Wrong. That must only be true if there is a largest natural number. > > Of course. If all natural numbers are available, then there must be a > largest, because the naturals obey trichotomy. Prove that there must be a largest. > Please do not mistake > all for each. You can name each natural number (in fact you cannot, > but let's assume, you could), but you cannot name all natural numbers. We can name all natural numbers, just because that means that we can name each natural number. > All means "completed". Completed means, there is nothing to be added. All means indeed completed. The set of all natural numbers is completed by the axiom of infinity. > In case of a set the elements of which obey trichotomy, there is a > largest one. Prove it. The law of trichotomy only states that given two numbers it is possible to determine that the first is larger, the second is large, or they are equal. It is your simplified finitistic view where that means that there is a larger number. But that is not a mathematical discussion. Within mathematics, it is possible to determine which of the three possibilities is true (although it is likely not possible with the amount of resources we have available). For instance, it is known that the functions pi(x) and Li(x) cross each other infinitely many times, but even the first cross-over point will not be known ever. > > Consider the following mapping: > > Let's have 'r' a rational number > 0. > > Rewrite 'r' as a continued fraction [r0, r1, ..., rn+1], where the > > ri are natural numbers, except for r0, which can be 0. It is easily > > shown that such a representation is unique. Next map that number to: > > 2^r0.(1 + 2^r1.(1 + ... (1 + 2^rn)...)). So the continued fraction is > > mapped to an integer which has in its binary expansion a specific > > spacing of the 1 bits. From this a reverse mapping is obvious. > > Pray tell which rational does not map to an integer, and which integer does > > not map to a unique rational. And also tell me which integer maps to the > > largest rational smaller than 1. > > Your mapping seems to be a unique one but this is only due to your > point of view. In what way is the above mapping *not* a unique one? > The other possible point of view which is not less justified yields: This is not a possible point of view on the mapping above. So it is not even a contradiction. > Every path is simultaneously a path bunch, because down to level L(n) > every node belonging to path p is also a node of path p' =/= p. This > holds for every n in N - and others are not available. If we denote > the n-th node belonging to path p by K(p,n) then we have: > forall K(p,n) with n in N thereis p' such that forall m in N with m < > 10^n : K(p',m) = K(p,m). And for each two unequal paths p and p' there is an n such that K(p, n) != K(p', n). > This also covers decimal expansions and continued fractions and shows > that there is no uniqueness, not even for the representation 0.000... > of the unique number 0. As far as mathematics is concerned, 0.000... is unique (as long as we use the standard definition for that notation). > > > The absence of this rational is as natural as the absence of the set > > > {N,k,f} in Hessenberg's proof. > > > > That is only because you put the cart before the horse. You do not > > understand the conditions. > > That is because barbers are not members of the subsets of natural > numbers? Has no relevance at all. -- 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 30 Apr 2007 08:32
On 27 Apr., 15:37, William Hughes <wpihug...(a)hotmail.com> wrote: > On Apr 27, 8:16 am, mueck...(a)rz.fh-augsburg.de wrote: > > > forall K(p,n) with n in N thereis p' such that forall m in N with m =< > > n : K(p',m) = K(p,m). > > p' may be different for different n's, so I should write p'(n). > I need to show that I can find a single, non changing, > p' that hold for every n. Why do you think to need to show that? Is it a commandment? Or is there any other reason recognizable? In particular, are you afraid that forall K(p,n) with n in N thereis p' such that forall m in N with m =<n : K(p',m) = K(p,m) is false and cheating you? Whether you can find a fixed p' or not: For every node there is such a p'. The necessity to require more than one p' can only result in case (E n in N such that node K(p,n) is in p' and K(p,n) is not in p'') & (E m in N such that node K(p,m) is not in p' and K(p,m) is in p''). But this is impossible. So, what is your concern? Anyhow. The number of path bunches surpassing the number of nodes is zero. If any infinite path p exists, then the bunch {p} does exist too, and {p} is a path bunch. If, on the other hand, {p} does not exist, then it is ridiculous to try to count the number of paths. Regards, WM |