Cantor Confusion
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From: MoeBlee on 24 Jan 2007 13:52 On Jan 24, 4:37 am, mueck...(a)rz.fh-augsburg.de wrote: > On 23 Jan., 18:17, "MoeBlee" <jazzm...(a)hotmail.com> wrote: > > No, we do NOT have to suppose the list is of all real numbers. Yes, I wrote that. > But you have to suppose that all enumerated lines and columns are > there. So you have to suppose the finished presence of an actually > infinishable set N. That is nonsense. No, I didn't write that; you did. And your claim is irrelevent to the set theoretical proof. We don't have predicate symbols for 'finished', 'presence', and 'actually'. You are welcome to try to devise, IN SET THEORY, set theoretical definitions, but until you do, you've not said anything that bears upon the set theoretical proof that there is no function from the set of natural numbers onto the set of real numbers. MoeBlee
From: Virgil on 24 Jan 2007 14:44 In article <1169626572.550075.102890(a)l53g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > > > > You have to choose whether you want to see a tree as > > > > a) the standard tree, T = (M, E) > > b) the "Virgil"-tree. T = (set of paths having certain properties). > > > > Which definition do you choose? > > I defined the trees which I use by nodes and edges. > > 0. > / \ > 0 1 > / \ / \ > 0 1 0 1 > | | | | > 0 1 0 1 > | | | | > 0 1 0 1 > . . . . > . . . . > . . . . > > > This is a tree of type weeping willow, T(1), with edges 0 and 1. It is > easy to see how the general definition of T(n) looks like but it is > tedious to write it down here. However, who is unable to understand it, > will be unable to follow my thoughts and should refrain from trying so. > > > The corresponding cut tree T(1) is > > 0. > / \ > 0 1 Why not 0. / \ 0 1 / \ / \ 0 1 0 1 ? That is a complete binary tree of depth 2 within you WW tree. And your Weeping Willow Trees are essentially the same as my eventually constant trees, to which you objected. > > We see that the union of two trees is the larger one. We see further > that the union is the set theoretic union. And by enumerating the nodes > > 0 > 1 2 > 3 4 5 6 > ... > > we prove that the union of all nodes and, threfore, of all levels and > of all finite trees does exist in the original set theoretic meaning. But since in each WW tree every path is eventually constant, in the union of all such trees every path is still eventually constant, and no path in that union can have both infinitely many 0 branchings and infinitely many 1 branchings. Thus WM's "union" of WW trees is not the complete binary tree he claims, and the countability of WW paths does not imply the countability of the set of apths of the complete infintie binary tree. So WM fails again! > > > > If so then the set of paths of G is cardinally greater than the > > std-union of sets of paths of the trees in the set of all finite trees > > V*. > > The set of paths of a tree is solely given by its nodes and edges. Same > nodes and edges result in same paths. Everything else is nonsense. Except that the paths in WM's union of WW trees omit "most" paths of the complete infinite nary tree. > > > > What is a segment of paths? > > A finite initial segment of a path is a finite part of a path from > level 0 to level n. > > > > Since the union of V* is the std-union we know that that U V* > > contains only "finite" paths. > > Correct. > > Hence 0.[10] is _not_ in UV*. > > Right words but wrong conclusion. On the contrary, it is WM who comes to the wrong conclusion based on his religious faith that what he believes is Gospel. > > I include some other posts of you because their contents are very > similar to the former one, and therefore, fit together > > > > > You have misunderstood the induction principle. It is not made for > > "counting over to the infinite". > > It is valid for all existing natural numbers. Counting over to the > infinite is nonsense. Counting occurs in the finite. Then why does WM keep arguing otherwise? > > > > If tree-union was the std-union of ZFC then by the axiom of union we can > > state that U V* is defined. But that requires a commitment to Virgil's > > definition of trees, which is not the standard-meaning of trees. > > Virgil dreams of different sets of paths in one and the same tree. I > would not believe in his ideas. WM's Weeping willow trees and their union are exactly the same as my "eventually constant" paths and their union. In either case every path in every tree in the union is eventually constant. > > The union of all finite trees is the complete infinite tree. The mathematical union of two or more different trees is not a tree at all. A suitably defined conjunction of a nested sequence of trees can be a tree, but it is, at best, a pseudo-union, not an actual union. > > Wrong. If A c B or B c A, then A U B is a tree, namely B or A, > repsectively. If A = (M_A,E_A) = {M_A, {M_A,E_A}}, E_A \subset M_A x M_A, and B = (M_B,E_B) = {M_B, {M_B,E_B}}, E_B \subset M_B x M_B, Where M denotes a set of nodes and EW a set of edges then A U B = {M_A, {M_A,E_A} ,M_B, {M_B,E_B}} which is not an ordered pair at all, much less a tree. If one chooses to define an operation of conjunction on nested rooted binary trees by A \conj B + (M_A,E_A) \conj (M_B,E_B) = ( M_A u M_B, E_A U E_B ) only then will WM have the conjunction of two trees being a tree. > > I did not say what was first. (There is no time in mathematics.) I said > what is fact. It is possible to define the union of two finite numbers > by the maximum of both. This can be extended to the union of all > numbers. > So it is with finite trees too. Not so. With two natural numbers, the usual definitions always have one natural as a subset of the other, but this is not the case with trees. It is quite possible to have two rooted binary trees with neither a subtree of the other, or even being isomorphic to a subtree of the other. So that any conjunction of two such trees will be different from either of them. Note that for any set of 7 distinguishable nodes, there are some 7! = 5040 different sets of edges which will make those nodes into a level 2 rooted binary tree, at least if one can always distinguish between branching left and branching right, and some 28 distinct trees even if one can't. > > > The union of two > > sets a and b is defined > > > > a U b = { x | x e a v x e b } > > > > first. Thereafter for two distinct ordinals a and b "larger" is simply > > defined as > > > > a < b := a c b > > Because the number n is by definition is n = {0, 1, 2, ..., n-1}. This > leads unavoidably to n U m = {0, 1, 2, ..., m-1} for m > n. And it > leads to Un = {0, 1, 2, ...} = N. > > Exactly the same situation we have in the tree for the union of two or > all distinct trees. Except that a tree is not defined as simple set of nodes or set of edges, it is a concatenation of two sets, and unions of such catenations are not the same as the catenations of the unions.
From: Andy Smith on 24 Jan 2007 14:48 stephen(a)nomail.com writes (snip) >>> >> Thanks for the clarification. I had previously mistakenly been using the >> expression "actually infinite" to mean the infinite sequence limit - a >> "completed infinity" as Aristotle wouldn't have approved of. > >What do you mean by the infinite sequence limit? The limit of an >infinite sequence diverges, does not exist, or it is finite. Perhaps I should have said the "limit of the infinite sequence" e.g. the limit of 1/2,3/4,7/8,... is 1. > >> But, anyway, you require an infinite number of bits to define the set of >> reals and so any indexing scheme also requires an infinite number of >> bits, and so any corresponding index is not a natural number. Crank, >> crank... > >What do you mean by an indexing scheme? Each real number can be >represented by an infinite number of bits. Each of those bits >has a finite index. > Maybe you missed the earlier posts. To summarise, I made the mistake of saying, it is "obvious" that the natural numbers can't address/index/map the reals. This was based on : 1) any systmatic numbering scheme for the reals is the same underpermutations (unless, somehow, the order of indexing can affect matters) 2) A systematic ordering/indexing scheme is to look at the reals as defined by the their bits in [0,1]. So we set up natural numbers with bits in correspondence with the bits of the reals - so the first bit of a real .b0 is a0 in the corresponding naturals, and proceed in bit order through the reals, an<->bn; so after e.g. 3 bits, a real .011 becomes 110. This defines a unique and 1:1 systematic mapping between the natural numbers and the reals in [0,1]. 3) at first sight, so what is the problem - we can increase n indefinitely; n is always finite. But, then we observe that there are reals with a genuinely infinite bit sequence. So under that systematic mapping, .0111... becomes ...1110 . An infinite number of bits is required to specify each real; in contrast the natural numbers are always finite. Aristotle wouldn't have liked the discussion, but he would have said that the natural numbers are potentially infinite, wheras the reals are actually infinite. You need an infinity of bits to specify a transcendental point. 4) So, you can't map the reals to the natural numbers - it is a difference in category. The address space of the reals is too big to be represented by the natural numbers. Anyway, following Moeblee's kind advice, I am buggering off to read some books. Bye -- Andy Smith
From: Franziska Neugebauer on 24 Jan 2007 15:05 mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: >> You have to choose whether you want to see a tree as >> >> a) the standard tree, T = (M, E) >> b) the "Virgil"-tree. T = (set of paths having certain properties). >> >> Which definition do you choose? > > I defined the trees which I use by nodes and edges. [...] ,----[ WM in <1169545581.873125.47200(a)q2g2000cwa.googlegroups.com> ] | It is defined. It is the tree which contains the root node and if it | contains the tree with n levels, then it contains the tree with n+1 | levels. `---- >> If so then the set of paths of G is cardinally greater than the >> std-union of sets of paths of the trees in the set of all finite >> trees V*. > > The set of paths of a tree is solely given by its nodes and edges. In Virgil's trees a tree is given by a set of paths. > Same nodes and edges result in same paths. Everything else is > nonsense. Using Virgil's trees we have the result that the induced std-tree (M(P), E(P)) may be the same if different sets of paths P are given. [...] >> Since the union of V* is the std-union we know that that U V* >> contains only "finite" paths. > > Correct. This is correct _only_ for Virgil's trees. For std-trees there the std-union is not a tree and for David's tree-union the tree-union is not the std-union and the tree-union of all finite trees is the binary tree G. > Hence 0.[10] is _not_ in UV*. Using Virgil's trees this is correct. Using David's tree-union this is not correct. > Right words but wrong conclusion. You should say: If 0.010101... > exists, then it is in a tree which contains all nodes and, therefore, > all alternations 0 to 1 and 1 to 0. Your observation shows that > 0.010101... does not exist in such a tree (the union of all finite > trees), so it does not exist at all. 1. 0.[01] is assumed to exist. It is an infinite sequence. This has already been explained to you many times. 2. "To exist in a tree" is unmathematical wording. You probably mean "is member of the set of paths of the tree". 3. 0.[01] is not a member of Virgil's UV*. 4. 0.[01] is a member the set of paths of David's UV*. 5. 0.[01] is a member the set of paths of G. > I include some other posts of you because their contents are very > similar to the former one, and therefore, fit together Please when copying paste the MID also! What you do paste now is bad practice since you cut the context. So I will restore the context. ,----[ <45b5ec2c$0$97243$892e7fe2(a)authen.yellow.readfreenews.net> ] | >> Again: Your notations | >> | >> T(1) U T(2) U ... | >> | >> and | >> | >> U {T(i) | i e N } | >> | >> are undefined. | > | > You are in error. The union of the trees T(n) and T(n+1) is defined. | > n is a natural number. Therefore the union of all finite trees is | > defined. | | You have misunderstood the induction principle. It is not made for | "counting over to the infinite". `---- >> You have misunderstood the induction principle. It is not made for >> "counting over to the infinite". > > It is valid for all existing natural numbers. Counting over to the > infinite is nonsense. Counting occurs in the finite. What you do _is_ "counting over to the infinite" when you claim that induction proves existence of U V* where in fact is does only prove existence of UV(n) A n e N. V* is not a finite set, i.e. a set of the form V(n). Hence induction does not "reach" V*. >> If tree-union was the std-union of ZFC then by the axiom of union we >> can state that U V* is defined. But that requires a commitment to >> Virgil's definition of trees, which is not the standard-meaning of >> trees. > > Virgil dreams of different sets of paths in one and the same tree. I > would not believe in his ideas. Virgil has states that there are at least two different sets of path induce the _same_ std-tree G. This is no dream or believe. This is proven fact. >> > If you try to construct the tree with n levels, do you fail at some >> > number of levels? No. Therefore the union is defined for every n. >> >> But not for V*. Recall: V* = { T(n) | n e omega } is the infinite set >> of all finite binary trees. > > It is a set which can be formed if the set of all n exists. In ZFC omega is the set of all natural numbers and it exists. If you assume omega not to exist you can hardly show a contradiction in ZFC. >> > The union of all finite trees. >> >> So you implicitly commit to Virgil's trees? I.e. a tree is a set of >> paths. > > The union of all finite trees is the complete infinite tree. 1. G? 2. Do you commit yourself to std-trees, Virgil's trees? 3. Do you commit yourself to std-union or David's tree-union? > It contains the set of all paths How do any of your trees "contain" the set of all paths? Can you write it down in common formal set theoretical language? > which are the unions of initial segments (finite paths of finite > trees) because the paths are sets of nodes and proper subsets of the > trees. This set is the same set of paths which > is in the complete infinite tree T. This paragraphs makes no sense to me at all. >> Only under Virgil's definition this union is identical to the >> standard-set-theoretical notion of union. If so UV* is not identical >> to the full set of paths of the infinite binary tree G given by me in >> a elsewhere. Virgil's UV* only comprises finite paths. > > Every UV* comprises only finite paths and comprises all infinite paths > existing. David's UV* does not "comprise" paths at all. Nonetheless the his UV* equals G and hence the set of paths induces by G is comprises even the infinite paths. Your first attempt to specify the "union of all finite trees" does not define union at all. >> > Further, both Virgil and William understand that the union defined >> > by me is identical to the complete tree T as far as nodes and edges >> > and levels are concerned. >> >> Nodes, edges and levels. Shaken or stirred? > > The levels are identical. > The nodes are idebtical. > The edges are identical. > > Correct or not? Define tree and "union of all finite trees" formally or commit yourself to one of the available definitions then we can continue to write about trees. In my opinion everything is already entirely clear except your position. >> > They merely doubt the identity of path due to some inexplicable >> > religious belief in a death religion. >> >> I must admit that great part of the discussion is to subtle for you. > > You are right. I will never understand how one can be so blind to see > more paths in the tree T than in the tree T. You have the Third Eye. So no wonder that you see things we can't. In the following quote you have fallaciously cut the context: ,----[ context (the not superseded one) ] | >> No, it does not. The usual set theoretical union is defined as | >> | >> A U B := { x | x e A v x e B } | >> | >> If A and B are trees {M_A, E_A} and {M_B, E_B} then the union is `---- So A and B are supposed to be trees! >> >> A U B = { x | x e {M_A, E_A} v x e {M_B, E_B} } >> >> = { M_A, E_A, M_B, E_B } >> >> >> >> This union is not a tree. >> > >> > unless M_A, E_A and M_B, E_B have same elements at all levels they >> > have in common. >> >> Of course. This in this special case (a = a U a) you are right. >> If A /= B then the A U B is not a tree. > > Wrong. If A c B or B c A, then A U B is a tree, namely B or A, > repsectively. There are no two trees A and B (using std-Kuratowski-pairs:) A := {{M_A}, {M_A, E_A}} B := {{M_B}, {M_B, E_B}} where A c B or B c A. Trees necessarily have cardinality 2. Hence your remark is entirely pointless. >> >> > The nodes can be enumerated in various ways. Here is a very >> >> > simple method: >> >> > 1 >> >> > 11, 12 >> >> > 21, 22, 23, 24, >> >> >> >> A fomal version of this "numbering" I have given in >> >> 45b483f8$0$97267$892e7fe2(a)authen.yellow.readfreenews.net > > Fine. Then you should know it. >> >> >> >> > ... >> >> Please take a decision: > > I did already. If you cut more of the context and paste from more different sources ---maybe you could shuffle the lines around--- it would be even more challenging to communicate with you. >> I would prefer David Marcus' definition. We can than examine the sets >> of paths separately from the trees. > > I do not read this kind of writers. You see the truths, I know. >> > The union of two natural numbers is defined to be the larger one. >> > This is a set theoretic union. >> >> 1. You are in error about what is defined first in ZFC. > > I did not say what was first. (There is no time in mathematics.) When I mean "first" I do not mean time but dependence. In ZFC '<' depends on 'c' not the other way round. > I said what is fact. It is possible to define the union of two finite > numbers by the maximum of both. It is not the aim of any axiomatic theory to rely on what you see. Maybe someone can explain the rules of the game to you. [...] >>From this follows that the union of all levels L(n) exists. >>From this it follows that the union of all finite trees exists. >> > >> > >> > What about the union of levels? >> > What about the union of initial segments of one path? >> > >> >> Again. U V* is not defined. >> > >> > Again: I defined it so that everybody can construct the finite >> > union of finite trees. >> >> You have not defined UV* (the union of actually _all_ finite trees). >> >> > This construction does not come to an end. >> >> In ZFC there is no time and no processes. You must be writing on >> something entirely different. Have you scrapped your plan to show a >> contradiction in contemporary set theory? > > "Infinite" means "not ending", "unending" or "not coming to an end". > "Construction" is a notion that belongs to current mathematics. > "A construction which does not come to an end" is simply the verbal > description of "an infinite construction". So you _have_ scrapped your plan to show a contradiction _in_ contemporary set theory? F. N. -- xyz
From: Franziska Neugebauer on 24 Jan 2007 15:16
mueckenh(a)rz.fh-augsburg.de wrote: > On 23 Jan., 13:35, Franziska Neugebauer > <Franziska-Neugeba...(a)neugeb.dnsalias.net> wrote: > >> ** The std-union of the set of sets of paths of every finite binary >> std-tree is not identical to the set of paths of the infinite >> binary std-tree G. ** > > Summary > > 1) Every complete infinite binary tree T (containing all nodes and > edges) contains all paths. No we cannot say so. Please define "contains" first. > 2) The union tree T(oo) of all finite trees is well defined (as I have > shown elsewhere) and yields the complete infinite binary tree > containing all nodes and edges: T = T(oo). So you commit to David's trees (std-trees + tree-union). I will identify T(oo) with my tree G defined elsewhere. > 3) The union of all finite trees includes the union of all nodes and, > with it, the union of all such subsets which are paths (because every > path is a well defined subset of the set of nodes if the structure of > the tree is well defined). We cannot say so. *In* std-trees there no paths. The set of paths is induced by the tree. > 4) The set of paths in T(oo) is a subset of the countable set of > finite sets of all paths in the finite trees. No. The set of paths of T(oo)/G is the full set of paths. > 5) A countable union of countable sets is a countable set (according > to ZF with AC). This is true. > ==> The set of all path is countable. (==> The real numbers are > countable.) Non sequitur. Equivocation fallacy. > Going on, we can say: > > 6) T(oo) = T contains only finite paths. No! The set of paths of T(oo)/G is complete, i.e. contains really all paths. > 7) T(oo) = T contains all paths including all infinite paths. Right. > ==> There are no infinite paths. (There are no irrational numbers.) Non sequitur. And wrong. > Nothing further remains to say. True. F. N. -- xyz |