Cantor Confusion
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From: William Hughes on 8 Feb 2007 07:37 On Feb 8, 3:07 am, mueck...(a)rz.fh-augsburg.de wrote: > On 7 Feb., 16:02, "William Hughes" <wpihug...(a)hotmail.com> wrote: > > > > > On Feb 7, 9:29 am, mueck...(a)rz.fh-augsburg.de wrote: > > > > For all for finite natural numbers > > > > > n = |{2,4,6,...,2n}| < 2n > > > > > Now take the limit as n-->oo (the < becomes <= in a limit, > > > > No. n < 2n is always true for natural numbers (which property excludes > > > 0). There is no reason to assume that 1/2 becomes 1 in "the limit". > > > > > lim[n-->oo] n = oo, lim[n-->oo] 2n = oo) > > > > > oo <= oo > > > > > There is no contradiction. > > > > Wrong for all finite numbers. > > > And since the limit is not a finite number > > Every number which can appear in this inequality is a finite number. > If something can appear in place of n which is not a finite number, > then you agree that N contains infinite numbers. No. The thing that may be able to appear in this inequality is the cardinality of E. Either -The cardinality of E does not exist or -The cardiality of E is not a finite number. In neither case does N contain infinite numbers. In neither case is the cardinality of E a finite number. In neither case is there a contradiction. >But that is wrong for > any set N, be it actually or potentially infinite. > Since I do no make such a claim, this statment is irrelevent. > > the fact that there > > is a contradiction for all finite numbers does not mean that > > there is a contradiction for the limit. > > Potentially infinite means here only that the value represented by n > can become as large as we like. > Recall the statment we are discussing is -E contains numbers which are larger than the cardinal number of You note: "Potentially infinite means here only that the value represented by n can become as large as we like" It follows immediately that either -The cardinality of E does not exist or -The cardiality of E is not a finite number. The fact that potentially infinite means that the set E can become as large as we like is very relevent. > > > > > > > Can you answer the following question yes or no? > > > > > Is the potentially infinite set of finite even numbers > > > > a set of finite even numbers? > > > > Yes, I can. The answer is yes. Probably set theory gives another > > > answer. > > > Let E be the potentially infinite set of finite even numbers > > > You have now made three claims > > > -every set of finite even numbers contains numbers which > > are larger than the cardinal number of the set > > True. > > > - E is a set of finite even numbers > > Yes, but not a complete or finished set as set theory requires. > > > - the statment "E contains numbers which are larger > > than the cardinal number of the E" is false > > No. The answer is not so simple. > The statement can be proved correct for an actually existing set E > with a cardinal number |E| which is a number that can be compared by > size with natural numbers. > The statement is false for a potentially infinite set E because such > sets do not have cardinal numbers. > In other words the statement A: every set of finite even numbers contains numbers which are larger than the cardinal number of the set is false. The statement B: every set of finite even numbers with a finite cardinal contains numbers which are larger than the cardinal number of the set The statement B is true, and of little interest. In particular, statement B does not mean that extending the concept of cardinality to sets that cannot have a natural number as a cardinal leads to a contradiction. As has been pointed out before, if you wish to claim that only sets with finite cardinality exist, then you are free to do so. However, saying that E does not have a cardinal does not change the properties of E. E still has a sparrow (recall a sparrow is an equivalence class under the equivalence relation equitransform which generalizes the concept of bijection to include potentially infinite sets). This sparrow can be compared with other sparrow's including the sparrows of finite sets, which are just the finite cardinals. There is nothing contradictory about defining a sparrow. Extending the concept of cardinality to include potentially infinite sets does not lead to a contradiction. - William Hughes
From: Dik T. Winter on 8 Feb 2007 07:48 In article <1170853587.867792.9070(a)v33g2000cwv.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 7 Feb., 02:52, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1170761231.100893.322...(a)v33g2000cwv.googlegroups.com> mueck.= > .=2E(a)rz.fh-augsburg.de writes: > > > > > On 5 Feb., 05:10, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > > > > What *is* IIII. You never have defined it. You really do not like > > > > definitions, as they pin down the real meaning. > > > > > > IIII is a primitive. Everybody knows it - even without definition by > > > Peano or Dedekind. That means, we do not need Peano to know small > > > natural numbers. > > > > Perhaps not. But in axiomatic mathematics it is best to define it. > > Unless you want to do non-axiomatic mathematics, like Cantor. The > > problem with non-axiomatic mathematics is that proofs are so difficult > > as nothing is defined and people can talk at cross-purposes. It is > > quite possible that I have quite different ideas about IIII than you > > have. And indeed, that is the case. I do not see it as the number 4, > > but as a sequence of four strokes. In my opinion the successor to > > that would be four strokes with a fifth stroke starting at the bottom > > left and ending at the top right. > > Why not. IIIII can be abbreviated by 5 or V or your proposal or ... Again, still no definition. So IIIIII can be abbreviated 5I, I5, VI or IV? And IIIIIIIIII can be abbreviated 55 or VV? > > > > When I asked you about what basic > > > > way, III c IV c V, you answered that I had to continue with IIII, > > > > IIIII, etc. > > > > > > The basic way to establish IV c V is to use the numbers in their basic > > > form IIII c IIIII. (Numbers *are* their representations.) > > > > No. Numbers are abstract entities. Their representations are > > concretisations of those abstract entities. > > Where and what are these entities? Abstract entities. > > are many different representations possible, and they may even be > > contradictionary when you mix representations. If you see the greek > > letter "delta" do you associate that immediately with the number 4? > > And if you see the letter "lambda" do you associate it with the number > > 30? Nevertheless, they *are* representations. > > You have to use some agrrement if you want to use abbreviations. For > IIII you need no agreement. Oh. I think one is needed. > > > > > Why do you say N is wrong in (2) but not in (1)? > > > > > > > > Where in (1) is N? I do not see N at all. > > > > > > "1 ist eine nat=FCrliche Zahl" means "1 in N". > > > > But N is not yet defined in (1). You give meaning beyond what is stated. > > If you use "nat�rliche Zahl" or "N" does not matter. You use something > not yet defined (and never defined by Peano). You apparently do not know how a recursive definition works. (1), (2) (when properly corrected) define the natural numbers. (3) defines the set of natural numbers. > > > The property "being a > > > natural number" implies the existence of N. > > > > How can that be the case if N is not yet mentioned or even defined? > > N is only an abbrevation of set of natural numbers. An undefined abbreviation. WHen doing definitions you should do it the correct way. See: We are going to define natural numbers: (1) 1 is a natural number (2) if a is a natural number, the successor of a is also a natural number These two together define the natural numbers. > > > Of course the requirement > > > to decide whether n is in N is more circular than the statement that 1 > > > is in N. > > > > There is no such requirement at all. As long as you do not define N, you > > can not use N. Pray read your book were it is done proper. At the end it > > is stated that "a in N" means "a is a natural number" (but not that > > "a is a natural number" means "a in N"). > > That is MERELY an abbreviation. An UNDEFINED abbreviation. > > > > You stated that when I asked you for a definition. So what is > > > > happening here? > > > > > > You will have read the definitions in my book. "3 is the set of all > > > sets of 3 elements" is explained in chapter 10. > > > > Not yet, I am in chapter 9 now. But that is a ridiculous definition, as > > that is extremely circular. And with that definition you will not be > > able to show that 2 is a subset of 3. > > IT IS NOT A DEFINITION. It is the criterion of existence. So when I ask for a definition you refuse to give a definition? > > Yes, those are representations. There are umpteen many ways to represent > > numbers. But the representations are just that, representations. Not > > numbers. Just as sqrt(2) is a representation of the square root of 2. > > Where does sqrt(2) exist? Axioms do not guarantee existence! In the set of real numbers, set of algebraic numbers. But as each and every natural number, it is an abstract entity. > > You are focussed on your very personal definition of existence. And I > > think you mean physical existence (without reading your chapter 10 at > > all). But in that case, the Euclidean straight line has also no > > physical existance. > > Correct. There is no infinite line existing. All we assume is a line > longer than any line we have measured yet. Such lines also do not have physical existence. Each and every physical line has a width smaller than anything measured yet. And I do not think that physical lines are really straight either. > > > There is something called reality and another thing called matheology. > > > Both are disjoint. > > > > With your reasoning, Euclidean geometry is matheology. > > Only if we forget that the adjective "infinite" is merely an > approximation. And straight, and width zero. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: William Hughes on 8 Feb 2007 08:10 On Feb 8, 2:42 am, mueck...(a)rz.fh-augsburg.de wrote: > On 7 Feb., 14:35, "William Hughes" <wpihug...(a)hotmail.com> wrote: > > > > The potentially infinite set of even numbers is *constructed* by its > > > segments > > > > {2,4,6,...,2n} > > > > Every time we increase n by 1 we increase 2n by 2. This cannot be > > > avoided. Therefore it is impossible to have for finite natural numbers > > > > lim[n-->oo] |{2,4,6,...,2n}| > 2n > > > The limit of finite natural numbers is not a finite natural number. > > There is no limit. And since something that does not exist is not a finite natural number the statement -The limit of finite natural numbers is not a finite natural number. is true. > The notation lim[n-->oo] says nothing but that > there is *no* limit. The size of n is not bounded. One can of course choose whether or not lim[n-->oo] n exists. However, whether or not it exists it will not be a finite natural number. Statements that are true for all finite natural numbers may or may not be true for things that are not finite natural numbers. - William Hughes
From: Dik T. Winter on 8 Feb 2007 08:20 In article <1170854542.370817.5020(a)a34g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 7 Feb., 04:05, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > No. A finite tree, chain and path are all sets of nodes (by your own > > definition). The elements in the union are *not* the trees, chains > > or paths, but the nodes. > > Does not matter. The nodes are finite elements. The sets of nodes are > finite too. The sets are finite, unless you unite *all* trees, or *all* chains, or *all* paths. In that case the union is the set containing *all* nodes. > > > Every element is a set. There are only sets in ZFC. > > > > In some models of ZFC. You can not prove that every element is a set based > > on ZFC only. > > It is simply a definition. Yes, you can not prove it. So a statement like "every element is a set in ZFC" is false. > > > > The union of sets of elements does > > > > *not* contain the union of elements. > > > > > > The tree contains it. > > > > Yes. So what? > > Is your argument about infinite sets of finite sets of nodes which > cannot contain infinite paths now removed? No. I still argue that p(0) U p(1) U p(2) U ... *is* p(oo), it does not contain p(oo). Moreover, when P(i) is the set of paths at a finite level, P(0) U P(1) U P(2) U ... does *not* contain p(oo). Because, for this union to contain p(oo) at least one of the constituent sets P(i) *must* contain p(oo). But the infinite tree *does* contain p(oo). In short P(oo) is *not* a subset of P(0) U P(1) U P(2) ... . That is my objection against your argument. When you unite sets of paths you do *not* unite paths. > > How do you find that > > {p(0), p(1), p(2), ...} = p(oo)? > > on the left there is a set of sets of nodes, on the right there is a set > > of nodes. How can they be equal? > > Because the tree contains the union U{p(0), p(1), p(2), ...} = p(oo). You are still maintaining that {p(0), p(1), p(2), ...} = p(oo)? So that a set of sets of nodes is a set of nodes? Or a set of paths is a path? > > I agree that > > p(0) U p(1) U p(2) U ... = p(oo), > > (allowing finite paths in p(oo)), but neither is equal to > > {p(0), p(1), p(2), ...} > > It is already not true for the finite case, so I wonder why it must be > > true in the infinite case. > > The tree makes the union! How can a set of paths be a path, when a path is defined as a set of nodes? > > > The set of paths contained in the finite tree T(0), the set of paths > > > contained in the finite tree T(1), ..., the set of paths contained in > > > the finite tree T(n), and so on. > > > > In what way does the union of p(0), p(1), p(2), ..., contain p(oo)? It > > does not contain it as an element, but it *is* the union. > > It is the union. Just in the same way T(1), T(2), T(3), ... is the > union T(oo). Pray define "contain". Up to now I have seen you use it in two ways: (1) a contains b if b is an element of a (2) a contains b if b is a (possibly improper) subset of a > > > Every path is a subsets of the union, for example {0, 2, 6} c {0, 1, > > > 2, 3, 4, 5, 6}. Which path do you miss? > > > > None. So every path is a subset of the union of the union of the sets of > > finite paths. That is *not* what you did state. You stated that every > > path is an element of the union of the sets of finite paths. > > 1) The union of finite trees contains the union of finite paths. > 2) The union of finite trees is the whole tree. > 3) The whole tree contains all paths. > 4) The union of finite paths contains all paths. > 5) This union is countable. The union of finite paths is indeed a countable set of nodes. It contains all paths as subsets. But the set of subsets of that set of nodes is the powerset, which is *not* countable. And as the set of all paths is a subset of that powerset you can not prove this way that that set is countable. (And (1), (2) and (3) are not really needed, they merely serve as obfuscation.) > > > > Yes. Infinite unions can give finite things. But that is not a > > > > contradiction. > > > > > > It shows clearly that not the number of elements is decisive but their > > > sizes. > > > > Prove it. > > The infinite union of sets {1} is a finite set. The infinite union of the sets {1}, {2}, {3}, is not a finite set. > > So P(oo) is not the union of all sets of finite paths? But P(oo) is the > > set of the union of all finite paths? There is a difference. > > It is or is not in the tree. That is important. But you want to show that P(oo) is countable by showing that it is the union of finite sets. But if it is the set of the unions of finite paths, it is *not* the union of finite sets. It is a set of uniions. > > > That set is a tree but contains the paths as subsets. > > > > Yes. But it is not a path. > > What is this remark good for? Because P(oo) can not be the set of unions of finite paths. Because most unions of finite paths are not paths, and so are not in P(oo). > > Yes. So when we take the union of finite sets of paths, and take the > > union again of the resulting set, we get a set that has the intended > > paths as subsets. But you claim that the intended path is an element > > of the union of finite sets of paths. There is quite some difference. > > The intended path is a subset of the union of the union of finite sets > > of paths. Not anything more. > > That is enough. In the union of ll finte trees here are all paths. No. You want that P(oo) is the countable union of finite sets. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Franziska Neugebauer on 8 Feb 2007 08:43
mueckenh(a)rz.fh-augsburg.de wrote: > On 8 Feb., 11:02, Franziska Neugebauer <Franziska- > Neugeba...(a)neugeb.dnsalias.net> wrote: > >> This is not true. A tree *then* is the structure plus the set of >> paths and/or nodes. You persistently try to vaporize the essential >> constituents of a tree until it fits your preconception. > > What is wrong with my use of trees, except that it is inconvenient if > one is a bit clumsy in adapting new ideas? ,----[ WN in <1170852009.005792.107360(a)p10g2000cwp.googlegroups.com> ] | The tree is a set of nodes. [(TISON)] `---- (This is not Virgil's definition that a tree is a set of paths. This is not a definition of tree at all since tree structure is not taken into account. A set of nodes is simply a set of nodes.) But let us /assume/ that a tree is a set of nodes. Then we read ,----[ contd. ] | It contains paths as subsets. `---- This is wrong and remains wrong under your definitions of path: ,----[ WM in <1170923642.541518.218750(a)h3g2000cwc.googlegroups.com> ] | A path is an ordered set of nodes. [(*)] `---- Assume that the tree is the set of nodes S and there is a path p' = (S', <). Your claim (*) is p' c S -> A x(x e p' -> x e S) -> (S' e p' -> S' e S) & (< e p' -> < e S) -> S' e S & < e S -> false & false -> false Hence your "new ideas" are plain wrong. > This structure is given once and for all. Hence we do not need to > mention it further. Fallacy of composition. >> > Froget this complicated picture. I does not lead ahead. >> >> It leads again leads to the insight, that you posit what does not >> hold. > > If you mention the structure in any step, the description gets > complicated. "As simple as possible, but not simpler." (Einstein) > The result does not differ from the case where the > structure is given once and for all. Given once and for all and forgotten therafter? Fallacy of composition. >> > A path is an ordered set of nodes. >> >> In this picture a path is an ordered pair (S, <) of a set of nodes S >> and an order relation <. > > The path also has some nodes. It has, but it *is* *not* some nodes or a set of nodes. > And we need only this aspect here. Fallacy of composition. >> > But we need not use this order. >> >> To show that you are wrong even in this picture we *need* this order: > > But we don't need it for our conclusion. Your conclusion is a non-sequitur. > Therefore we do not use this picture. Show what is wrong with the > conclusion that the nodes of a path are a subset of the set of nodes > of the tree. 1. If I read "the nodes of a path are a ..." as "every node of a path is a ..." you claim A p A n(n e p -> n c SON(T)) According to (TISON) SON(T) = T and we get A p A n(n e p -> n c T). -> A p ((S e p -> S e T) & (< e p -> < e T)) | assume some p' -> false (s. above) 2. If I read "the nodes of a path are a ..." as "the set of the nodes of a path is a ..." you claim A p (SON(p) c SON(T)) | T is a set of nodes -> A p (SON(p) c T) | S is the set of nodes of p=(S, <) -> A p (S(p) c T) This is not decidable without explicitly examining the structure of the tree. F. N. -- xyz |