Cantor Confusion
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From: mueckenh on 7 Feb 2007 07:48 On 6 Feb., 21:15, Virgil <vir...(a)comcast.net> wrote: > In article <1170756166.580698.67...(a)p10g2000cwp.googlegroups.com>, > > mueck...(a)rz.fh-augsburg.de wrote: > > On 4 Feb., 20:28, Virgil <vir...(a)comcast.net> wrote: > > > > > > Not so. Induction can only prove every set of naturals which is bounded > > > > > above by a natural is finite, but that does not, according to the axiom > > > > > of infinity, exhaust the possible sets of naturals. > > > > > There is no natural number in this set which is not covered by > > > > induction. So, which number is missing to exhaust N? > > It is not numbers but sets of naturals of which I spoke. The empty set of natural numbers? I am only interested in sets of natural numbers which contain at least one natural number and contain nothing but natural numbers. They are covered by induction. Regards, WM
From: mueckenh on 7 Feb 2007 07:51 On 6 Feb., 23:07, Virgil <vir...(a)comcast.net> wrote: > > The basic way to establish IV c V is to use the numbers in their basic > > form IIII c IIIII. (Numbers *are* their representations.) > > Numerals are no more numbers than names are people. Wrong! People can exist and do exist without names. Numbers cannot. Regards, WM
From: mueckenh on 7 Feb 2007 08:06 On 7 Feb., 02:52, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1170761231.100893.322...(a)v33g2000cwv.googlegroups.com> mueck....(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 ... > > > > 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? > 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. > > > > > > > > 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 natürliche Zahl" means "1 in N". > > But N is not yet defined in (1). You give meaning beyond what is stated. If you use "natürliche Zahl" or "N" does not matter. You use something not yet defined (and never defined by Peano). > > > 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. > > > 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. > > > > 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 DEFI NITION. It is the criterion of existence. > 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! > > > > > > Your existence is not a mathematical existence. > > > > > > > > This form of existence is the only possible existence. > > > > > > As you do not define your form of existence, > > > > Read chapter 10 of my book. > > I still state that that is *not* methametical existence, but I am not yet > at that chapter. > > > > it is impossible to talk about > > > it, at least mathematically. > > > > Present mathematics has nothing in common with existence. > > 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. > > > > > I gave two definitions: Peano and that with "+1" which is very close > > > > to Dedekind's attitude. (I don't know whether he actually created it, > > > > but I know that he would have liked it with "+1" as a primitive). But > > > > the axioms do not establish any existence, in particular not when you > > > > apply Dedekind's definition of what a number is. > > > > > > Existence is a mathematical thing when you can establish it by axioms > > > or through theorems based on axioms. Anything else is merely phylosophy. > > > > 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. Note, Euclid did not talk about infinitely many prime numbers. There he was very correct. Regards, WM
From: Tez on 7 Feb 2007 08:20 Federico Ferreres Solana wrote: [snip] > FRACT --> NATURALS > 0.0 --> 0 > 0.1 --> 1 > 0.2 --> 2 > 0.3 --> 3 > .. --> ... > 0.9 --> 9 > 0.01 --> 10 > 0.11 --> 11 > 0.21 --> 12 > (etc) Then I take it you think the above is bijection from the real numbers to the naturals? Which natural number does the real number 1/3 map to? -Tez
From: mueckenh on 7 Feb 2007 08:22
On 7 Feb., 04:05, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1170762893.276254.14...(a)s48g2000cws.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > On 5 Feb., 06:04, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > In article <1170606249.100767.104...(a)j27g2000cwj.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > > > > > On 3 Feb., 04:12, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > In article <1170413742.648825.136...(a)l53g2000cwa.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > > > > So you say: > > > > > > The union of finite trees contains an infinite tree. > > > > > > The union of finite chains contains an infinite chain. > > > > > > The union of finite paths contains an infinite path. > > > > > > > > > > Indeed. > > > > > > > > > How can a union of finite elements contain an infinite element? > > > > > > Where in the above is there a union of finite elements? > > > > In all three statements. > > 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. > > > > Unions are about *sets*, not about elements. > > > > 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. > > > > > The union of finite sets of finite paths > > > > {p(0)} U {p(1),q(1)} U {p(2), q(2), r(2),s(2)} U ... > > > > is the set of all finite paths > > > > {p(0), p(1), q(1), p(2), q(2), r(2),s(2) ... } > > > > > > Indeed. > > > > > > > which obviously contains the union of finite paths of the special form > > > > p(0), p(1), p(2), .... > > > > as a subset. > > > > > > Wrong. There is no obviously here. > > > > It can be proven. {p(0), p(1), q(1), p(2), q(2), r(2),s(2) ... } > > contains {p(0), p(1), p(2), ... } > > p(0) U p(1) U p(2) U ... != {p(0), p(1), p(2), ...} > > As the p(i) are sets of nodes, on the left hand we have a set of nodes, on > the right hand we have a set of sets of nodes. The set on the right is > a subset, the set on the left is not. A set of nodes can not be a subset > of a set of sets of nodes. To abuse notation a bit, the best you could > state is: > > p(0) U p(1) U p(2) U ... = U{p(0), p(1), p(2), ...} Yes, I forgot the U. > > where U{a, b, c} is defined as U{U{a, b}, c}, which can be done using the > axiom of pairing and the axiom of union. But there is a huge difference > between: > {a, b, c, ...} > and > U{a, b, c, ...}. Note it how you like. The important thing is what the tree does. > > > > 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? > > > {p(0), p(1), q(1), p(2), q(2), r(2),s(2) ... } contains {p(0), p(1), > > p(2), ... } > > As a subset, but not as an element. That does't matter. It is contained! > > > > In what way does > > > {p(0), p(1), q(1), p(2), q(2), r(2),s(2) ... } > > > contain: > > > p(0) U p(1) U p(2) U ...? > > > > In the way the tree contains them as sets of nodes: p(0) U p(1) U > > p(2) U ... = {p(0), p(1), p(2), ... } = p(oo) is a subset of the set > > of paths contained in the tree. > > 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). > 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! > > > > > If the union p(0), p(1), p(2), ....contains p(oo) then the union > > > > p(0), p(1), q(1), p(2), q(2), r(2),s(2), ... > > > > cannot miss it. > > > > > > What *sets* of paths are you using? > > > > 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). > > Correct. And a set of ndes can be a path. > > And in most cases is not. No problem. > > > 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. > > > 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 union of the P(i) is the set of all paths. It includes the set of > > > > all paths of the form p(n). Why should it not contain p(oo) if the set > > > > of all paths of the form p(n) alone contains p(oo)? > > > > > > I think we are talking at cross purposes. And are misunderstanding each > > > other about what the meaning is of the term "the set of all paths p(n)". > > > > I think P(i) is the set of all paths of the finite tree with i levels. > > p(n) is a path like 0.000...0 with n zeros behind the point. The union > > of all p(n), i.e., of all nodes 0, is p(oo) = 0.000... > > 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. > > > That set is a tree but contains the paths as subsets. > > Yes. But it is not a path. What is this remark good for? > > 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. Regards, WM |