Cantor Confusion
in [Math]
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From: Dik T. Winter on 4 Jun 2007 23:19 In article <1180979008.019427.261130(a)n4g2000hsb.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > On 26 Mai, 04:28, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > > 3)Is their theorem formal enough? 1.3 Theorem Let L be an > > > infinite cardinal, let K_A (A < L) be nonzero cardinal numbers, and > > > let > > > K = sup {K_A | (A < L)}. Then Sum {A < L} K_A = L * sup {K_A | (A < > > > L)} > > > > And *where* in that formal statement is there a sequence? > > Can't you see simplest facts any longer? It seems you cannot see them. > Sum {A < L} K_A for L = aleph_0 is the sum of all cardinals less than > aleph_0. This sum can also be written by 1 + 2 + 3 + .... You may write it such, but there is *no* sequencing in *their* statement. Take L = aleph_1. Where is the sequencing? > > Just as I said: > > No. Not just as you said. It is really funny to see how you try to > veil your errors. Oh. > > 1 + 2 + 3 + ... > > is an informal notation for: > > sum{i in N} > > or > > sum{i < aleph_0} > > no sequence at all. Just summation over a set. > > Yes, summation over the ordered set of natural numbers. In general > this is written as a series. Their statement is *not* about natural numbers. Try to read. It is about cardinal numbers. > Hracek and Jech even state the identity on p. 188: > "It is not very difficult to evaluate infinite sums. For example, > consider > 1 + 2 + 3 + ... + n + ... (n in N). > It is easy to see that this sum, SUM{n in N}n is equal to aleph_0." > Can you read: 1 + 2 + 3 + ... + n + ... = this sum = SUM{n in N}n? Yes, that is an informal statement. Can you distinguish formal definitions from informal statements? > But probably you will not acknowledge these authors, because they also > use, as I did, the term "adic", which you opposed to as not being > common use. Then you were as wrong as you are here. Oh, that was a long time ago. I will look how they use the term "adic", if it is in the index, otherwise provide a reference. > > > > > > But definitions *can* lead to > > > > > > another result. > > > > > > > > > > For instance? > > > > > > > > Definition: sum{i in N} i = 0. > > > > > > Wrong definition. > > > > What is wrong about it? > > Simply the fact that the equality is not an equality. Can you prove that? Stating something is a fact is no proof. > > Indeed. That has been proven. But if you want to add the set of natural > > numbers, using addition for natural numbers, you have a lot more to do than > > just state it. In that book they add sets of *cardinal* numbers, and > > give a definition which leads to a result that is a *cardinal* number. > > Assuming you can replace "cardinal" by "natural" in that proof leads to > > wrong conclusions. > > I do not replace anything but use the notation they use, namely the > symbols 1, 2, 3, ... and n in N. By the way: Every finite cardinal is > a natural. And only finite cardinals are added. Yes, and the sum is not a finite cardinal. > > > > > 1 + 2 + 3 + ... + n + ... (n in N) > > > > > > > > > > They write, it is easy to see that this sum is equal to aleph_0. > > > > > > > > And they also write that they are doing cardinal arithmetic. > > > All natural numbers are cardinal numbers! > > > > Yup. But not all cardinal numbers are natural numbers. > > That is completely irrelevant as long as only natural numbers are > summed up. No, it is not irrelevant. They are adding cardinal numbers, leading to a cardinal number. Whether that sum is also a natural number is something you assume, again without proof. > > Yes. This does *not* prove my definition wrong. Because in arithmetic > > on natural numbers every *finite* sum is a natural number. Arithmetic > > on natural numbers tells us nothing about *infinite* sums. > > Now I understand why you (and some others) are so unable to follow > logic arguments, for instance concerning the tree. You think > "infinite" is some holy grail? Even the infinite sum of natural > numbers cannot be less than 1. Again stated without proof. As for natural numbers there is *no* standard definition of infinite sums, you have to provide your definition because you can state anything about it. > > > 1+2+3+...+X = X(X+1)/2 > > > > > > for every X in N. Proof by induction left as an exercise. > > > > Yes. So not for infinite X. Unless you first prove that there *is* an > > infinite X in N. You cannot use this as proof that there is indeed > > an infinite X. > > Wrong again. Hrbacek and Jech proved that the right hand side is an > infinite number, so X on the right is infinite. This does not imply > the same symbol on the left being infinite? How wrong can you get? They define: 1+2+3+... = aleph_0 see? No X. Their sequence does not terminate, but you have problems with non-terminating sequences. -- 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 4 Jun 2007 23:28 In article <1180979700.783883.14940(a)g4g2000hsf.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > On 29 Mai, 04:25, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > > The diagonal number is a limit. Without having the limit in a line, it > > > is not in the diagonal. > > > > Makes no sense. Can you put it in mathematical terms? The diagonal number > > is *not* a limit. The only limit involved is the proof that it is also a > > real number. > > Every real number is a limit. Cp. Cauchy. Nope, when you want to get formal, every real number is an equivalence class. > > So, there is a bijection between lines and *finite* initial segments. > > Nobody contested that. It is *not* a bijection between lines and the > > single *infinite* initial segment of the diagonal number, because > > nowhere in your mapping that segment does occur. > > Nowhere an infinite diagonal does occur either. Again stating the negation of the axion of infinity. > Do you agree that a bijection of the initial segments of the diagonal > with the initial segments of the diagonal is possible and can be > extended to include the complete diagonal mapped on itself? Yes. > And the bijection of the initial segments of the diagonal and the > numbers enumerating the lines can also be extended to include the > complete diagonal and all natural numbers? Yes. Suppose D_i it the initial segment of the diagonal upto i digits, consider the following mapping: f: D_i -> i D -> N > But the bijection of the initial segments of the diagonal and the line > entries cannot be extended to include the complete diagonal? Of course not, because N is not a line entry. > Remember: > The triangle is equilateral by definition. Without an infinite base > there is no infinite height. Remember there are both infinite base and infinite height. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: WM on 5 Jun 2007 01:55 On 5 Jun., 05:19, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > Sum {A < L} K_A for L = aleph_0 is the sum of all cardinals less than > > aleph_0. This sum can also be written by 1 + 2 + 3 + .... > > You may write it such, but there is *no* sequencing in *their* statement. > Take L = aleph_1. Where is the sequencing? And I talked about aleph_0 being the sum of all natual numbers. They take aleph_0. > > > > Just as I said: > > > > No. Not just as you said. It is really funny to see how you try to > > veil your errors. > > Oh. > > > > 1 + 2 + 3 + ... > > > is an informal notation for: > > > sum{i in N} > > > or > > > sum{i < aleph_0} > > > no sequence at all. Just summation over a set. > > > > Yes, summation over the ordered set of natural numbers. In general > > this is written as a series. > > Their statement is *not* about natural numbers. Try to read. It is > about cardinal numbers. It is about 1 and 2 and 3 and so on. > > > Hracek and Jech even state the identity on p. 188: > > "It is not very difficult to evaluate infinite sums. For example, > > consider > > 1 + 2 + 3 + ... + n + ... (n in N). > > It is easy to see that this sum, SUM{n in N}n is equal to aleph_0." > > Can you read: 1 + 2 + 3 + ... + n + ... = this sum = SUM{n in N}n? > > Yes, that is an informal statement. Can you distinguish formal definitions > from informal statements? Not in the way you believe you could. > > > But probably you will not acknowledge these authors, because they also > > use, as I did, the term "adic", which you opposed to as not being > > common use. Then you were as wrong as you are here. > > Oh, that was a long time ago. I will look how they use the term "adic", > if it is in the index, otherwise provide a reference. p.98:" We conclude this section with a few remarks on decimal (or, in general, p-adic) expansions of rationals." > > > > > > > > But definitions *can* lead to > > > > > > > another result. > > > > > > > > > > > > For instance? > > > > > > > > > > Definition: sum{i in N} i = 0. > > > > > > > > Wrong definition. > > > > > > What is wrong about it? > > > > Simply the fact that the equality is not an equality. > > Can you prove that? Stating something is a fact is no proof. I am glad to see the arguing which is typical for set theory. Here even you should be able to recognize the nonsense involved in your statement. SUM{n>0}1/2^n is not larger than 1. Do you agree? This is correct without any axiom, just by observing that always the remaining is divided by 2 and one half of it is removed. SUM{n>0}1/n is not less than 1. Do you agree? Yes, even for the whole infinite set of natural numbers this holds. We can prove it. SUM{n>0}n is not less than 1. This is true because for every natural number n we have n >= 1/n. So your Definition: sum{i in N} i = 0 is obviously wrong. The same holds for Definition |{2,4,6,...}| = aleph_0. We can argue with the same limits as above. For every set {2,4,6,...,2n} we have larger numbers than n = | {2,4,6,...,2n}|. Taking he limit as in SUM{n>0}1/2^n or SUM{n>0}1/n we get the result |{2,4,6,...}| = aleph_0 cannot be true without an infinite natural number (which dos not exist). Regards, WM
From: WM on 5 Jun 2007 16:51 On 4 Jun., 20:49, Virgil <vir...(a)comcast.net> wrote: > In article <1180966798.913696.131...(a)n4g2000hsb.googlegroups.com>, > > > > Mathematicians seem to be very short sighted. If they see a great many > > of numbers, they quickly assume that here are infinitely many. It is > > simply a ridiculous claim. There are not more than 10^100 numbers, and > > there are many natural numbers less that 10^10^100 which do not exist > > and by no means can be brought to existence. > > There is a critical difference between the notions of physical > existence and that of mathematical existence. > Physically, there is no such thing as a triangle, or a line, or a point, > but mathematically all of these "exist". The points of a triangle or a circle exist, if their coordinates, at least in principle, can be approximated with arbitrarily small error. But in many cases they cannot. In particular some natural numbers less than 10^10^100 and most of the natural numbers larger than 10^10^100 cannot be approximated to better than +/- 100. And this fact means, that those numbers do not exist. > > So that these, like WM, who would limit mathematical existence by the > limits of physical existence, are mathematically clueless. > Mathematics is inevitably about ideals that need not and usually cannot > be matched physically. To object to that is to misunderstand mathematics. You intermingle mathematics with religion or fine art. Once upon a time, mathematics used to be the most precise science (or art). Now it has degenerated to the fruitless occupation of some somnabule dreamers who loudly shout that they are the majority. > > > I need not a bijection in order to show that a set is countable. > > You need to show that there exists a surjection from the naturals to the > set in question, or an injection in the reverse direction, , as one or > the other is required by the definition. This is shown by the fact that no path can be isolated without a node. Obviously there is an injection from the set of all separated paths into the countable set of nodes. > > > Remember the set of constructible numbers. There is no bijection with > > N - there is no list. I need merely the simple fact that there can be > > no more separated paths than separations in the tree. > > This idiot notion of "separations", at least as pushed by WM, makes a > number of false assumptions about what separates what in a CIBT. All paths which are contained in the CIBT have separated at the level omega (if omega exists). A path which has not separated at omega will not separate at omega + 1 either. > > The truth is, in any CIBT: > Any finite set of nodes, if it acheives any sort of separation at all, > only separates one uncountable set of paths from another uncountable set > of paths, each set being easily bijectable with the set of all nodes. > > It takes an infinite set of nodes in any one path to separate it from > all other paths. > > Yes. And this infinite set of nodes is a subset of the countable set of all nodes. > > Wrong. For every path there is a node where it separates from all > > paths considered up to a certain level. > > But "all paths up to a certain level" never includes all other paths > unless there is a LAST level, which does not occur in CIBTs. Does omega exist? > > > > > > That is false. > > > That means that there must be as many nodes as paths for every finite > > segment of the tree, but not for the whole tree? You consider ghost > > paths, which do not exist. > > In WM's world, there may be ghost trees, but in ZF and NBG, there are > actual CIBTs which have properties that WM is apparently not capable of > analysing. I am not willing to believe in the fruitless and unprecise imaginations of intellectual clowns. > > WM is still trying to argue that properties dependent on the finiteness > of a tree must also be properties of infinite trees. Properties that are dependent on the completeness should be realized in the complete binary tree. > > > > > Every node is the end of a finite path. > > Except that in CIBTs there are no finite paths at all. There are paths > of finite subtrees, but they are not paths of the CIBT itself. Every finite path segments end at a node. And there is nothing in the tree which could support an infinite path. > > > There is simply nothing left > > to prove the existence of the other (even uncountably many) paths. > > In finite trees, of course not, but finite trees are not infinite. But they are complete? What do you mean by that term? Regards, WM
From: Virgil on 5 Jun 2007 17:49
In article <1181076666.933401.301080(a)p47g2000hsd.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 4 Jun., 20:49, Virgil <vir...(a)comcast.net> wrote: > > In article <1180966798.913696.131...(a)n4g2000hsb.googlegroups.com>, > > > > > > > Mathematicians seem to be very short sighted. If they see a great many > > > of numbers, they quickly assume that here are infinitely many. It is > > > simply a ridiculous claim. There are not more than 10^100 numbers, and > > > there are many natural numbers less that 10^10^100 which do not exist > > > and by no means can be brought to existence. > > > > There is a critical difference between the notions of physical > > existence and that of mathematical existence. > > Physically, there is no such thing as a triangle, or a line, or a point, > > but mathematically all of these "exist". > > The points of a triangle or a circle exist, if their coordinates, at > least in principle, can be approximated with arbitrarily small error. They have no physical existence, because their coordinates also have no physical existence. The only sort of existence that any mathematical objects have is mathematical existence. > But in many cases they cannot. In particular some natural numbers less > than 10^10^100 and most of the natural numbers larger than 10^10^100 > cannot be approximated to better than +/- 100. And this fact means, > that those numbers do not exist. Then 1 and 2 do not exist either, at least in WM's world. > > > > > So that these, like WM, who would limit mathematical existence by the > > limits of physical existence, are mathematically clueless. > > Mathematics is inevitably about ideals that need not and usually cannot > > be matched physically. To object to that is to misunderstand mathematics. > > You intermingle mathematics with religion or fine art. Once upon a > time, mathematics used to be the most precise science (or art). Now it > has degenerated to the fruitless occupation of some somnabule dreamers > who loudly shout that they are the majority. WM seems to want to embed mathematics in politics fantasy fiction by eliminating logic and proofs and establishing truth solely by shouts the loudest. Those of us who favor logic and proofs as a more reliable source of truth, reject his fantasies. > > > > > I need not a bijection in order to show that a set is countable. > > > > You need to show that there exists a surjection from the naturals to the > > set in question, or an injection in the reverse direction, , as one or > > the other is required by the definition. > > This is shown by the fact that no path can be isolated without a node. No path in an infinite tree can be isolated by one node either, it takes a set of infinitely many nodes to isolate any one path in an infinite tree from all other paths, as has been proven several times here. That WM does not understand either the proof or its consequences marks his as mathematically incompetent. And the mathematically incompetent, like WM, are incapable of governing mathematics. > Obviously there is an injection from the set of all separated paths > into the countable set of nodes. It is not obvious to me that WM's "seprated paths" means anything relevant to CIBTs. To "separate" one such path from all others by nodes requires a set of infinitely many nodes. WM's occasional claim that one node suffices is obviously false for infinite trees, as such a node would have to be a "last" node, and infinite paths don't have "last" nodes. > > > > > Remember the set of constructible numbers. There is no bijection with > > > N - there is no list. I need merely the simple fact that there can be > > > no more separated paths than separations in the tree. > > > > This idiot notion of "separations", at least as pushed by WM, makes a > > number of false assumptions about what separates what in a CIBT. > > All paths which are contained in the CIBT have separated at the level > omega (if omega exists). While "omega" exists, there is no omega level. Levels are members of N, and omega in not a member of N. > A path which has not separated at omega will > not separate at omega + 1 either. Paths in finite trees are "separate" only at their leaf nodes. Paths in infinite trees do not have leaf nodes, so cannot be "separate" at any one node, however any infinite subset of nodes of such a path is sufficient to 'separate' it from all other paths. > > > > > The truth is, in any CIBT: > > Any finite set of nodes, if it acheives any sort of separation at all, > > only separates one uncountable set of paths from another uncountable set > > of paths, each set being easily bijectable with the set of all nodes. > > > > It takes an infinite set of nodes in any one path to separate it from > > all other paths. > > > > > Yes. And this infinite set of nodes is a subset of the countable set > of all nodes. Quite so, but there are uncountably many subsets of any infinite set. > > > > > Wrong. For every path there is a node where it separates from all > > > paths considered up to a certain level. > > > > But "all paths up to a certain level" never includes all other paths > > unless there is a LAST level, which does not occur in CIBTs. > > Does omega exist? Does WM claim that omega is a member of omega? Unless he does, his question is irrelevant. And if he does , his "set theory" runs smack into Russell's paradox. > > > > > > > > > > > That is false. > > > > > That means that there must be as many nodes as paths for every finite > > > segment of the tree, but not for the whole tree? You consider ghost > > > paths, which do not exist. > > > > In WM's world, there may be ghost trees, but in ZF and NBG, there are > > actual CIBTs which have properties that WM is apparently not capable of > > analysing. > > I am not willing to believe in the fruitless and unprecise > imaginations of intellectual clowns. Then don't ever look carefully at your own garbage. > > > > WM is still trying to argue that properties dependent on the finiteness > > of a tree must also be properties of infinite trees. > > Properties that are dependent on the completeness should be realized > in the complete binary tree. They are in our CIBTs, but not in WM's. The properties we claim all have logically satisfactory proofs in ZF and NBG and the like, whereas there is no system yet presented here in which WM's anything like logically satisfactory proofs. > > > > > > > > > > Every node is the end of a finite path. > > > > Except that in CIBTs there are no finite paths at all. There are paths > > of finite subtrees, but they are not paths of the CIBT itself. > > Every finite path segments end at a node. And there is nothing in the > tree which could support an infinite path. Then WM's "infinite trees" are not infinite trees, and all this time he has been blathering about properties of what he does not have. > > > > > There is simply nothing left > > > to prove the existence of the other (even uncountably many) paths. > > > > In finite trees, of course not, but finite trees are not infinite. > > But they are complete? What do you mean by that term? "Infinite" means "not finite". For trees, an infinite binary tree is like the set of all functions from N to {0,1}, each such function being a path in that tree, and a node the intersection ,as sets of ordered pairs, of a set of functions, and thus is, itself, a function. |