Cantor Confusion
in [Math]
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From: Dik T. Winter on 20 Mar 2007 12:02 In article <1174327803.614563.300050(a)o5g2000hsb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > 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. Second-hand car dealers would like us to believe that. Obviously this can not be applied with undefined entities. Moreover, I do now know how you assign parts of nodes to path. What portion of the first node is assigned to your path? > > 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). Indeed. > (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. Indeed. > (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. No. I would state that at the root node of the tree there start uncountably many paths. Why do you think that number is not uncountable? (Remember, paths are non-terminating.) > > > > 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. Yes, I never contested that the number of branching offs is countable. So what? > > > 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. No. It disproves nothing of the kind. What is shown here is that your reasoning is wrong. > > > 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! And so what, what does that prove about the complete tree? Indeed, at no finite distance from the root. What this *means* is that the number of terminating paths is countable. But there is no finite distance from the root where all non-terminating paths are separated from each other, so the reasoning does not apply to them. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Virgil on 20 Mar 2007 15:46 In article <1174396356.841673.191550(a)n59g2000hsh.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 20 Mrz., 09:30, Virgil <vir...(a)comcast.net> wrote: > > In article <1174372833.140688.156...(a)p15g2000hsd.googlegroups.com>, > > > > > > > > > > > > mueck...(a)rz.fh-augsburg.de wrote: > > > On 19 Mrz., 20:31, Virgil <vir...(a)comcast.net> wrote: > > > > In article <1174292997.710730.80...(a)n76g2000hsh.googlegroups.com>, > > > > > 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 have been separated or will separate in the next level. > > > I fail to see that any of the points WM presents has any relevance > > whatsoever to the issue of whether the number of paths in a CIBT is > > countable or uncountable. > > I am not so much interested in what you may feel relevant to prove > your opinion concerning the multitude of paths. I am only interested > in your boldness to refute the points 1, 2, or 3. WHY are you 'only interested' in my attitude with respect to such irrelevancies? The truth of all three does allow disproof of the uncountability of P(N) WM seems to ignore all the relevant issues that I bring up, so why should I not ignore his irrelevancies. > > > > Does WM agree that EVERY endless sequence of 0's and 1's following that > > header of 0 is an infinite path belonging to the tree T(oo)? > > Every existing path belongs to the complete tree. That carefully avoids answering my question, to which an affirmative answer would justify the uncountability of the set of paths and a negative answer would confirms WM's misrepresentation of the issue. > And at no point of > the tree more than countably many infinite paths have separated. There is also no point at which any infinite path has yet become infinite, but that does not prevent the existence of paths having no last point. And since there is no end of "points", that no more proves the countability of the set of all paths than the finiteness of each natural proves the finiteness of the set of all naturals. No matter how WM squirms, he cannot disprove the uncountability of the set of paths, or the countability of P(N), without invoking axioms to those effects. And there are perfectly good axiom systems which reject WM's axioms.
From: mueckenh on 21 Mar 2007 08:48 On 20 Mrz., 17:02, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1174327803.614563.300...(a)o5g2000hsb.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > What portion of the first node > is assigned to your path? How much does the last term of the geometric series contribute to the value o the series? This is the value which the first node cntributes to he path. > > > 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). > > Indeed. > > > (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. > > Indeed. > > > (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. > > No. I would state that at the root node of the tree there start > uncountably many paths. Why do you think that number is not uncountable? > (Remember, paths are non-terminating.) (3, clarified) 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 have separated. > > > > But it is connected to the root node like every other path. The > > branching offs are countable. > > Yes, I never contested that the number of branching offs is countable. So > what? In the whole tree there can be no more separated paths than nodes. > > > > 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! > > And so what, what does that prove about the complete tree? Indeed, at no > finite distance from the root. What this *means* is that the number of > terminating paths is countable. No. It means that the number of paths which consist only of nodes with natural indexes is countable. > But there is no finite distance from > the root where all non-terminating paths are separated from each other, > so the reasoning does not apply to them. The "distance" is measured by the index of the due node, i.e., by the number n of the corresponding level. "At no finite distance from the root" means at no node which can be enumerated by a natural number. Regards, WM
From: mueckenh on 21 Mar 2007 08:54 On 20 Mrz., 17:07, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1174328592.469787.62...(a)y66g2000hsf.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > 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? > > An infinite set. If you want to call that set a number, go ahead. But you > are confusing yourself using the von Neumann model. In that model, indeed, > the union of all finite integers >= 0 is omega, an infinite number. > > > > > 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 path-length of a non-terminating path is *not* a natural number. How then can it be contained in a set which contains only sets of finite paths? > > > The axiom of infinity grants the existence of all natural numbers, but > > not the existence of an infinite natural number. > > Indeed. why then do you believe that an infinite path (number) is contained in the union of sets of finite paths (numbers). > > > > 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. > > Where is the infinite element? You said: "The path-length of a non-terminating path is *not* a natural number." And you find it in a set of serts of finite paths. > > (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. > > No. From the root node of the tree there start uncountably many paths. But they do never separate into uncountably many paths. (3, clarified) 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: mueckenh on 21 Mar 2007 09:06
On 20 Mrz., 20:46, Virgil <vir...(a)comcast.net> wrote: > In article <1174396356.841673.191...(a)n59g2000hsh.googlegroups.com>, > > > > > 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 have been separated or will separate in the next level. > > > > I fail to see that any of the points WM presents has any relevance > > > whatsoever to the issue of whether the number of paths in a CIBT is > > > countable or uncountable. > > > I am not so much interested in what you may feel relevant to prove > > your opinion concerning the multitude of paths. I am only interested > > in your boldness to refute the points 1, 2, or 3. > > WHY are you 'only interested' in my attitude with respect to such > irrelevancies? Because ever objective observer may form his own opinion on your further claims. >The truth of all three does [not] allow disproof of the > uncountability of P(N) That is obviously impossible to debate with you. > > > Does WM agree that EVERY endless sequence of 0's and 1's following that > > > header of 0 is an infinite path belonging to the tree T(oo)? > > > Every existing path belongs to the complete tree. > > That carefully avoids answering my question, > to which an affirmative answer would justify the uncountability of the > set of paths Perhaps this is the proof of inconsistency of actual infinite and, therefore, the proof of non-existence of those paths? > and a negative answer would confirms WM's misrepresentation > of the issue. > > > And at no point of > > the tree more than countably many infinite paths have separated. > > There is also no point at which any infinite path has yet become > infinite, but that does not prevent the existence of paths having no > last point. For you, it would not even pervent the existence of the non existing last point, I assume. Let us consider only the points (or better nodes) which in fact do exist (because they can be enumerated or indexed by natural numbers): Do you think that the set of those paths, which consist only of existing nodes, is countable? > > And since there is no end of "points", that no more proves the > countability of the set of all paths than the finiteness of each natural > proves the finiteness of the set of all naturals. But you believe that the complete path 0.000... as well as the complete path 0.010101... etc. including their missing ends are contained in the complete tree? Regards, WM |