Cantor Confusion
in [Math]
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From: Virgil on 18 Mar 2007 17:45 In article <1174252341.700926.178590(a)e65g2000hsc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 18 Mrz., 02:25, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1174138819.038221.295...(a)o5g2000hsb.googlegroups.com> > > mueck...(a)rz.fh-augsburg.de writes: > > > > > On 16 Mrz., 16:15, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > In article <1174041538.385009.91...(a)d57g2000hsg.googlegroups.com> > > > > mueck...(a)rz.fh-augsburg.de writes: > > ... > > > > > It is very easy to define such paths. The following, for instance, > > > > > is > > > > > a path: > > > > > 1/2 > > > > > 1/4 > > > > > 1/8 > > > > > 1/16 > > > > > ... > > > > > As it does never end, it is an infinite path. Do you agree? Do you > > > > > claim that there are uncountably many of such paths? > > > > > > > > Yes, and yes, and that is easily proven by a slight adaption of > > > > Cantor's > > > > proof. > > > > > > And it is easily proven that there are as many paths as nodes in the > > > tree. You need only calculate the sum of the fractions. > > > > What fractions are you talking about? Again fractional nodes? How do > > you define them? A node is by definition a point where edges connect. > > What is half a node? > > Again, pretend to be stupid? Unlike WM, who does not need to pretend? > > > > > > But, more obvious, there can never uncountably many paths exist in the > > > tree, because there are never uncountably many nodes within one > > > level. > > > > Again, a statement lacking proof. > > There are only countably many nodes in the whole tree. So there cannot > be more in one level. But as the number of paths corresponds more with the number of sets of nodes than with the number of nodes, WM's claim still requires, and lacks, and will forever lack, proof. > > > This is similar to the statement that > > there can never be uncountably many subsets of N, because there are never > > uncountably many numbers. > > No. The tree has a unique structure which cleans up with this > superstition. Since both the uncountability of P(N) and of the set of all infinite paths in a CIBT have been validly proved, and WM is incapable of producing counterproofs, both because he is incapable of producing any proofs and because such counterproofs do not exist, WM's religious beliefs d not override logically valid proofs which contradict his faith. > > > A path is a set of nodes, so it is a subset of > > the set of nodes, and so is an element of the power set. Now paths are > > restricted sets of nodes, so you have to *prove* that that restriction > > is sufficient to make the set of paths countable. I can tell you > > beforehand that you will fail. Let us consider infinite paths only. > > Count levels in the tree starting at 1. For each level state whether > > the path is going left ('w') or right ('m'). An infinite path is an > > infinite sequence of such symbols. Now apply Cantor's diagonal proof. > > Do you think that this proof creates a path which is not in the tree? As it doesn't need to in order to get uncountably many paths, why do you think so? > > On the other hand we know: There are only countably many nodes in the > whole tree. So there cannot be more in one level. Therefore here is no > level of the tree where uncountably many paths could exist. If you > insis in their existence, then they must exist outside of the tree. There is no level (decimal place) at which more that a finite umber of digits can occur, yet the set of all endless decimals is uncountable. > > > > > > But you seem to claim, though you do not explicitly state it (as far > > > as I remember): The number of paths / paths-bundles becomes > > > uncountable before the paths leave the tree. > > > > I do not know. I have no idea how you define that division. > > There is no division indicated but it is stressed that paths are also > path-bundles What "path-bundle" which is only one path? And why would any such thing be relevant to the uncountability of the set of all paths in the CIBT? > > > What I see > > in the infinite tree is: > > (1) For each level there are path-bundles, that could also be defined as > > finite paths. > > (2) There are paths, without any dependence on path-bundles or levels, > > they are all infinite in length. > > They are not caught within the outermost left and the outermost right > paths of the tree, namely 0.000... and 0.111...? Between these two > paths only a countable set of paths is possible. Claimed often, proven never. On the other hand our claim that a CIBY must have uncountably many paths has been proved often and refuted never. > > > > The number of paths in (2) is uncountable. The number of path-bundles in > > (1) is countable. So however you define that definition, there is nothing > > about "becomes uncountable". The number of paths *is* uncountable. It > > does not suddenly become uncountable. > > The number of paths *is* countable because, within the tree, i.e. > between the paths 0.000... and 0.111..., there *cannot be more than > countably many paths - at no finite distance from the root! WM's paths, being limited to finite distances from the root, are not infinite paths as for any finite distance from the root an infinite path has nodes further from the root. WM is again conflating the finiteness of each node in an infinite path with the finiteness of the set of nodes in that path. > > If you require more, than you must introduce an index omega. At a > finite n you will fail. WM again claims that the set of naturals, as index set for the nodes of an infinite path, in order not to be finite, must contain an infinite member. That is his inescapable tangle with quantifier dyslexia confusing him again.
From: Virgil on 18 Mar 2007 18:48 In article <1174253026.516861.271800(a)d57g2000hsg.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 18 Mrz., 02:02, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > The function Card(paths) over n cannot jump from finite to > > > uncountable. > > > > It does not when you consider finite paths. There is another jump, from > > finite paths to infinite paths. > > Where does that happen? When one jumps from finite trees to infinite trees, of course! Finite trees have finite paths, infinite trees have infinite paths, and there is a jump from finite to infinite. > > > > But you state: "the number of paths of every set of finite paths is > > finite". > > Again, you are contradicting yourself. > > No. I take only the silly speech of infinitely many finite numbers and > apply it to infinitely many finite paths. What is silly is someone who insists that every set is finite arguing about what happens when some sets are not finite. So > > > > > you > > > > > agree that in the union of all finite trees the number of paths is > > > > > countable? > > > > > > > > No, because in the union there are also infinite paths. Or do you now > > > > think that 0.010101... is *not* a path in that union? > > > > > > Is it in the union of all finite paths? > > > However, as long as it consists of nodes it has only countably many co- > > > paths. > > > > Prove it, please. All your proofs until now contain at one stage or > > another a statement, without proof, that is in direct contradiction > > with the axiom of infinity. > > Do you claim now that there are uncountably many nodes in the tree? If > not then you can be sure that there are only countably many in one > level. In every level! Wm keeps looking for things which are countable and refusing to look at those things in infinite trees (in which all paths are endless binary sequences) which are not. The number of positions in an endless binary sequence of members of {w,m} is countable but the number of such sequences is not. > > The tree is a static structure. Every node for which is another node > is already contained therein. Every path which does exist, exists in > the tree, i.e., in the union of all finite trees. Nevertheless, within > the tree there is no uncountable set of paths. Then WM's pseudo-tree is not a CIBT.
From: mueckenh on 19 Mar 2007 04:29 On 18 Mrz., 22:45, Virgil <vir...(a)comcast.net> wrote: > In article <1174252341.700926.178...(a)e65g2000hsc.googlegroups.com>, > > > WM's paths, being limited to finite distances from the root, are not > infinite paths as for any finite distance from the root an infinite path > has nodes further from the root. Can we agree upon the following facts? (1) The infinite paths 0.000... and 0.111... are completely contained in the complete tree T(oo), i.e., every node belonging to one of these paths is a node belonging to the tree T(oo). (2) The set of nodes of the tree T(oo) is countable. In particular, the number of nodes belonging to one level L(n) is countable for every n. (3) There is no node in the tree (including every node of the paths 0.000... and 0.111...) which belongs to a level where uncountably many paths arrive at or start off from or cross through. (I am not sure whether the prepositions here are used correctly, but I hope you will understand.) Regards, WM
From: Dik T. Winter on 19 Mar 2007 11:04 In article <1174252341.700926.178590(a)e65g2000hsc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 18 Mrz., 02:25, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1174138819.038221.295...(a)o5g2000hsb.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > > > On 16 Mrz., 16:15, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > In article <1174041538.385009.91...(a)d57g2000hsg.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > ... > > > > > It is very easy to define such paths. The following, for instance, is > > > > > a path: > > > > > 1/2 > > > > > 1/4 > > > > > 1/8 > > > > > 1/16 > > > > > ... > > > > > As it does never end, it is an infinite path. Do you agree? Do you > > > > > claim that there are uncountably many of such paths? > > > > > > > > Yes, and yes, and that is easily proven by a slight adaption of Cantor's > > > > proof. > > > > > > And it is easily proven that there are as many paths as nodes in the > > > tree. You need only calculate the sum of the fractions. > > > > What fractions are you talking about? Again fractional nodes? How do > > you define them? A node is by definition a point where edges connect. > > What is half a node? > > Again, pretend to be stupid? No, waiting for you to finally give proper definitions of the things you are using. So, again, what fractions are you talking about? How do you add such fractions? > > > But, more obvious, there can never uncountably many paths exist in the > > > tree, because there are never uncountably many nodes within one > > > level. > > > > Again, a statement lacking proof. > > There are only countably many nodes in the whole tree. So there cannot > be more in one level. This is not a proof. As there are uncountably many paths *at every level*, this is not an argument. This is only an argument where you consider finite paths: at each level there are finitely many paths terminating at that level. So the total number of finite paths in the complete tree is countable. This says nothing about the infinite paths. > > This is similar to the statement that > > there can never be uncountably many subsets of N, because there are never > > uncountably many numbers. > > No. The tree has a unique structure which cleans up with this > superstition. Oh. Prove it. > > A path is a set of nodes, so it is a subset of > > the set of nodes, and so is an element of the power set. Now paths are > > restricted sets of nodes, so you have to *prove* that that restriction > > is sufficient to make the set of paths countable. I can tell you > > beforehand that you will fail. Let us consider infinite paths only. > > Count levels in the tree starting at 1. For each level state whether > > the path is going left ('w') or right ('m'). An infinite path is an > > infinite sequence of such symbols. Now apply Cantor's diagonal proof. > > Do you think that this proof creates a path which is not in the tree? No. It creates a path that is in the tree, but is not in the given enumeration of paths. > On the other hand we know: There are only countably many nodes in the > whole tree. So there cannot be more in one level. Therefore here is no > level of the tree where uncountably many paths could exist. If you > insis in their existence, then they must exist outside of the tree. As the number of infinite paths that goes through a level is the same for each level, I do not think this argument is extremely convincing. Try again. By your reasoning there is no level of the tree were infinitely many paths could exist because the nuber of nodes at each level is finite. > > > This is true as long as the paths occupy nodes. > > > > > > But you seem to claim, though you do not explicitly state it (as far > > > as I remember): The number of paths / paths-bundles becomes > > > uncountable before the paths leave the tree. > > > > I do not know. I have no idea how you define that division. > > There is no division indicated but it is stressed that paths are also > path-bundles Yes, you may stress that, but it is better to focus on finite paths and infinite paths. But whatever the case, both the number of paths and the number of path-bundles is fixed, so there is no question about the quotient (however you do define it) becomes something. > > What I see > > in the infinite tree is: > > (1) For each level there are path-bundles, that could also be defined as > > finite paths. > > (2) There are paths, without any dependence on path-bundles or levels, > > they are all infinite in length. > > They are not caught within the outermost left and the outermost right > paths of the tree, namely 0.000... and 0.111...? Between these two > paths only a countable set of paths is possible. Prove that. (It is false.) > > The number of paths in (2) is uncountable. The number of path-bundles in > > (1) is countable. So however you define that definition, there is nothing > > about "becomes uncountable". The number of paths *is* uncountable. It > > does not suddenly become uncountable. > > The number of paths *is* countable because, within the tree, i.e. > between the paths 0.000... and 0.111..., there *cannot be more than > countably many paths - at no finite distance from the root! You state it again. at no finite distance from the root. But you are wrong. At each finite distance from the root there are uncountably many paths that go through that level. That number is the same for each level. What you can consider is the number of finite paths that terminate at that level, but in that case you are considering terminating paths only. > If you require more, than you must introduce an index omega. At a > finite n you will fail. Again: each two paths differ at some index n does *not* mean that there is an index n where all paths do differ. -- 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 19 Mar 2007 11:18
In article <1174253026.516861.271800(a)d57g2000hsg.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 18 Mrz., 02:02, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > > The function Card(paths) over n cannot jump from finite to > > > uncountable. > > > > It does not when you consider finite paths. There is another jump, from > > finite paths to infinite paths. > > Where does that happen? When you consider the union of all finite trees. > > The set of *finite* paths in the complete > > tree is countable. The set of *infinite* paths in the complete tree is > > not countable. As in each finite tree there are only finite paths, even > > your continuity considerations can not give any information about the > > infinite paths. > > How can they enter a union of finite trees U(T(n)) or a union of all > paths of all finite trees? By the definition of that union. Let us consider a node indexed by a pair of numbers (i, j), where i indicates the level and j indicates the node at that level. Let's have i go from 1 and j from 1 to 2^(i-1). The tree T(n) is the set of all nodes with i <= n. A terminating path in the tree T(n) is the set of nodes {(k, l)} where k ranges from 1 to n and each k occurs only once, and l is within the bounds with some additional requirements: if (k1, l1) and (k2, l2) are in a path, and k2 = k1 + 1 then l2 is either l1 * 2 - 1 or l1 * 2. This is a complete definition of finite paths and finite trees. U(T(n)) is by definition the set of pairs that contains all pairs contained in all of the T(n). So it is the infinite set {(i, j) | i in N and 1 < j < 2^(i - 1)}. The existence of this set is granted by the axiom of infinity. Now an infinite paths is, for instance, the set {(i, 1) | i in N}, which is a subset of U(T(n)). > > > > > In this case nothing else but: "going on without end". > > > > > But I will agree to any other definition which you might supply. > > > > > The limit of U[k<n]{1,2,3,...,k} is the union of all finite initial > > > > > segments for example. > > > > > > > > By what definition of limit? Pray, for once, provide a definition. > > > > > > Read it: "going on without end". > > > > That is not a definition. > > That is the only meaningful definition. The only meaningful definition is a non-definition? > > > > > The number of paths of every set of finite paths is finite. > > > > > > > > Eh? So there are not even infinitely many paths? Why not? > > > > > > There are. Infinite is "going on without end". > > > > But you state: "the number of paths of every set of finite paths is > > finite". Again, you are contradicting yourself. > > No. I take only the silly speech of infinitely many finite numbers and > apply it to infinitely many finite paths. In what way? You state "the number of paths of every set of finite paths is finite". Applying that to your analogy with finite numbers, I get: "the number of numbers of every set of finite numbers is finite". This is indeed very silly speech, because it states that there are finitely many finite numbers. > > It does not show anything of that kind. You define above: "infinite is > > 'going on without end'", but that definition in itself is already a > > negation of the axiom of infinity, so you can not use it in a proof > > that purports to show that that axiom is self contradictory. Obviously > > that axiom is in contradiction with its negation. So, unless you can > > prove its negation in the context of that axiom, you have shown nothing. > > Is "infinite is going on without end" in contradiction with the axiom > of infinity? If I do understand what you mean with it, yes. But as you do not yet provide a definition of that terminology I am not entirely sure. > > > > No, because in the union there are also infinite paths. Or do you now > > > > think that 0.010101... is *not* a path in that union? > > > > > > Is it in the union of all finite paths? > > > However, as long as it consists of nodes it has only countably many co- > > > paths. > > > > Prove it, please. All your proofs until now contain at one stage or > > another a statement, without proof, that is in direct contradiction > > with the axiom of infinity. > > Do you claim now that there are uncountably many nodes in the tree? If > not then you can be sure that there are only countably many in one > level. In every level! Yes. But *paths*, remember: *paths*. Why do you always think you have to convince me that there are countably many nodes when I ask you about the number of paths? > > > > Not in terms of nodes, why should there be a difference in terms of > > > > nodes? > > > > > > What else can distinguish paths? Soul? Feeling? Belief? > > > > For every two paths there is a node where they are different. This does > > *not* mean that there is a node where all paths are different. Back to > > that again. > > The tree is a static structure. Every node for which is another node > is already contained therein. Every path which does exist, exists in > the tree, i.e., in the union of all finite trees. Nevertheless, within > the tree there is no uncountable set of paths. Again, stated without proof. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |