Cantor Confusion
in [Math]
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From: Virgil on 17 Mar 2007 17:51 In article <1174140129.542070.30270(a)y80g2000hsf.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 16 Mrz., 21:31, Virgil <vir...(a)comcast.net> wrote: > > In article <1174055000.972069.261...(a)y80g2000hsf.googlegroups.com>, > > > > > > > The function of all cross sections, f: n |--> 2^n, is "continuous" in > > > > > the sense that never a jump by more than a factor 2 can occur because > > > > > the nodes of the tree are connected by an untearable network. The > > > > > domain is the same as the range, namely N. That is fact, not by claim > > > > > but by construction of the tree. That's why I constructed it. > > > > > > You constructed the tree to show that 2^{n+1} <= 2*2^n ? Well, that > > > > really must have been fun. Ok, I agree on this. Now we know a property > > > > of the function > > > > > > f : N -> N > > > > n |-> 2^n. > > > > > > This does not tell us anything about 2^aleph_0. > > > > > aleph_0 is not a natural number. > > > > Are you just discovering that? > > Every index of a digit of a real number or of a node, however, is a > natural number. Therefore it is of no interest at all to speculate > about what happens "at aleph_0". That should you try to discover. As WM has speculated on exactly that issue freqeuently, one wonders what has now made him change his mind.
From: Dik T. Winter on 17 Mar 2007 21:02 In article <1174137847.808123.291360(a)l75g2000hse.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 16 Mrz., 16:02, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1174039357.642800.20...(a)d57g2000hsg.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > On 15 Mrz., 16:39, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > > > > A path bundle splits off into two bundles which pass said node. > > > > > Therefore every set of path-bundles in the tree has a finite > > > > > cardinal number > > > > > > > > As long as path-bundles are finite sets of nodes that is right. > > > > > > You should say: As long as path-bundles are sets of nodes that is > > > right. > > > > No, the finite is required here. > > Below you say that there are differences which are not in erms of > nodes. Indeed. There are difference, *not* in terms of nodes. > > > There are only countably many nodes. Unless the number of path-bundles > > > first gets countably infinite it cannot become uncountably infinite. > > > Reason: Continuity of the tree. > > > > Pray define "continuity". > > 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. 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. > > > > > This, in the "limit", may yield an infinite number, but certainly > > > > > not an uncountable number without having an intermediate countably > > > > > infinite number. > > > > > > > > I do not know. What do you understand under "limit"? > > > > > > 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. > > > 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. > > > You seem to imply a difference between the set of all finite paths > > > (which is countable) and the set of infinite paths. > > > > Yes, there is a difference. The difference being that in one set all paths > > are finite and in the other set all paths are infinite. > > > > > In the union of all finite trees there are only finite paths. > > > > Why? The axiom of infinity states different. > > As the tree shows, this statement is self contradictory. 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. > > > 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. > > > The set of all finite paths covers the whole tree with all its nodes. > > > What is the difference, in terms of nodes, between the set of all > > > finite paths and the set of infinite 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. -- 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 17 Mar 2007 21:25 In article <1174138819.038221.295790(a)o5g2000hsb.googlegroups.com> mueckenh(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 calclate 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? > But, more obvious, there can never uncountably many paths exist in the > tree, because there are never uncountably many nodes within oner > level. Again, a statement lacking proof. This is similar to the statement that there can never be uncountably many subsets of N, because there are never uncountably many numbers. 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. > > No. What you need to prove is that that also holds for the *infinite* tree > > together with *infinite* paths. > > It holds as long as there are nodes occupied by paths. Even if there > is not finish. There is no jumping from finite to uncountable. There is no such for finite paths. > > > The cross section of a finite tree (i.e. the number of nodes in its > > > basis) is finite. The union tree U(T(n)) consists of levels all of > > > which are basis levels of finite trees and, therefore, are finite and > > > have a finite cross section. > > > > Yes. > > 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. 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. 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. -- 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 18 Mar 2007 17:12 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? > > > 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 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. > 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? 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. > > > > No. What you need to prove is that that also holds for the *infinite* tree > > > together with *infinite* paths. > > > > It holds as long as there are nodes occupied by paths. Even if there > > is not finish. There is no jumping from finite to uncountable. > > There is no such for finite paths. > > > > > The cross section of a finite tree (i.e. the number of nodes in its > > > > basis) is finite. The union tree U(T(n)) consists of levels all of > > > > which are basis levels of finite trees and, therefore, are finite and > > > > have a finite cross section. > > > > > > Yes. > > > > 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 > 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. > 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! If you require more, than you must introduce an index omega. At a finite n you will fail. Regards, WM
From: mueckenh on 18 Mar 2007 17:23
On 18 Mrz., 02:02, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1174137847.808123.291...(a)l75g2000hse.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > On 16 Mrz., 16:02, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > In article <1174039357.642800.20...(a)d57g2000hsg.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > > On 15 Mrz., 16:39, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > ... > > > > > > A path bundle splits off into two bundles which pass said node. > > > > > > Therefore every set of path-bundles in the tree has a finite > > > > > > cardinal number > > > > > > > > > > As long as path-bundles are finite sets of nodes that is right. > > > > > > > > You should say: As long as path-bundles are sets of nodes that is > > > > right. > > > > > > No, the finite is required here. > > > > Below you say that there are differences which are not in erms of > > nodes. > > Indeed. There are difference, *not* in terms of nodes. > > > > > There are only countably many nodes. Unless the number of path-bundles > > > > first gets countably infinite it cannot become uncountably infinite. > > > > Reason: Continuity of the tree. > > > > > > Pray define "continuity". > > > > 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? > 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? > > > > > > > This, in the "limit", may yield an infinite number, but certainly > > > > > > not an uncountable number without having an intermediate countably > > > > > > infinite number. > > > > > > > > > > I do not know. What do you understand under "limit"? > > > > > > > > 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 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. > > > > > You seem to imply a difference between the set of all finite paths > > > > (which is countable) and the set of infinite paths. > > > > > > Yes, there is a difference. The difference being that in one set all paths > > > are finite and in the other set all paths are infinite. > > > > > > > In the union of all finite trees there are only finite paths. > > > > > > Why? The axiom of infinity states different. > > > > As the tree shows, this statement is self contradictory. > > 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? > > > > > 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! > > > > > The set of all finite paths covers the whole tree with all its nodes. > > > > What is the difference, in terms of nodes, between the set of all > > > > finite paths and the set of infinite 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. Regards, WM |