Cantor Confusion
in [Math]
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From: G. Frege on 19 Mar 2007 11:23 On Mon, 19 Mar 2007 15:18:59 GMT, "Dik T. Winter" <Dik.Winter(a)cwi.nl> wrote: >>>>> >>>>> By what definition of limit? Pray, for once, provide a definition. >>>>> >>>> Read it: "going on without end". [WM] >>>> >>> That is not a definition. >>> >> That is the only meaningful definition. [WM] >> > The only meaningful definition is a non-definition? > Sure. In /M�ckenheim's World/ this is indeed the case. F. -- E-mail: info<at>simple-line<dot>de
From: mueckenh on 19 Mar 2007 14:10 On 19 Mrz., 16:04, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1174252341.700926.178...(a)e65g2000hsc.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > 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? The addition of fractions is defined by a/b + c/d = (ad + bc)/bd. For this calculation the question "what unit is multiplied by the fractions" does not play a role. You can apply this definition to cakes or cars or nodes. > > > > > 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. Perhaps you have another mental picture in mind. 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. > > > 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. But it is connected to the root node like every other path. The branching offs are countable. > > > 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 number of nodes at each level is finite. Yes. That shows that infinite means always finite but not bounded. The tree disproves every other mental picture. > > > > 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. That number is not visible. Let us take what can be proved: The number of separated paths *is* countable because, within the tree, there *cannot be more than countably many separated paths - at no finite distance from the root! Regards, WM
From: mueckenh on 19 Mar 2007 14:23 On 19 Mrz., 16:18, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1174253026.516861.271...(a)d57g2000hsg.googlegroups.com> mueck...(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 union of all finite numbers gives an infinite number? > > > > 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. Yes. But the path-lengths are natural numbers, and there is no infinite natural number. The axiom of infinity grants the existence of all natural numbers, but not the existence of an an infinite natural number. > Now an > infinite paths is, for instance, the set {(i, 1) | i in N}, which is > a subset of U(T(n)). You try to get an infinite element out of a union of finite elements. That does not work. Further it is not sufficient, because the countable union of countable sets is countable. The set of paths of every finite tree is countable. The set of finite trees is countable. 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. Regards, WM
From: Virgil on 19 Mar 2007 15:31 In article <1174292997.710730.80970(a)n76g2000hsh.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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.) Through every level in a CIBT pass uncountably many paths. Through every node in every level of a CIBT pass uncountably many paths. Consider a mapping from any CIBT to itself which maps the root node to an arbitrary node and maps the left child of any node to the left child of it image node and similarly for right nodes. Then the range of that mapping, though a subset of the original tree is isomorphic to the original with respect to every tree property. So that there are as many paths in the original which contain the image of the root node as in the original altogether. And that number is card(P(N)).
From: Virgil on 19 Mar 2007 15:57
In article <1174327803.614563.300050(a)o5g2000hsb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 19 Mrz., 16:04, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1174252341.700926.178...(a)e65g2000hsc.googlegroups.com> > > mueck...(a)rz.fh-augsburg.de writes: > > > > 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? > > The addition of fractions is defined by a/b + c/d = (ad + bc)/bd. For > this calculation the question "what unit is multiplied by the > fractions" does not play a role. You can apply this definition to > cakes or cars or nodes. Except that two such half cars do not make a whole. Does WM claim that half a root node plus half a left child node equal some node? And if so, which node? And when one supposedly divides a node into halves, what happens when one tries to reunite two left half-nodes or two right-half nodes? If one tried to weld together two left-half cars, would the result be a car? WM's idea of subdividing the indivisible is nonsense. > > > > > > > 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. No one claims more than finitely many in one level, but when one have more that finitely many levels with at least two in every level but the root, there are enough sets of nodes making paths to make an uncountable set of paths. WM gets hung up on his inability to digest Card(N) < Card(P(N)). > 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. (3) is false, since EVERY NODE in every level of a CIBT has uncountably many paths "through" it. > But it is connected to the root node like every other path. The > branching offs are countable. It is only the set of all subsets of those branching offs that limits the number of paths, and that allows unaccountability > > Yes. That shows that infinite means always finite but not bounded. That assumption does not hold in ZF or NBG, and until WM can produce a contradiction within them that does not rely on his assuming that every set is finite, he is SOL. > The > tree disproves every other mental picture. Only to those whose pictures are constrained by unnecessary assumptions. > > > > > > 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. > > That number is not visible. It is to those who know where to look for it. > Let us take what can be proved: The number > of separated paths *is* countable because, within the tree, there > *cannot be more than > countably many separated paths - at no finite distance from the root! What in Nifflheim is a "separated path"? A path in any tree is a maximal chain of nodes in that tree in which each node, with at most one exception, is parent of one and only one other node. A binary tree in which every node has two child nodes can only have paths which have no exceptional "end" nodes, and must necessarily have uncountably many paths. |