Cantor Confusion
in [Math]
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From: mueckenh on 19 Feb 2007 08:01 On 19 Feb., 00:57, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1171789678.561730.262...(a)t69g2000cwt.googlegroups.com> mueck...(a)rz.fh-augsburg.de > > The union of sets of alle finite paths is the union of all finite > > trees, and this union is the complete tree T(oo) if the union of all > > initial segmens of N is N. > > Wrong. The union of sets of paths is *not* the tree. The union of sets of paths is one set of paths. The union of this set of paths is a set of nodes. > According to your > notation above, the tree is the union of paths. The tree is a set of nodes which is the union of paths which is the union of the union of a set of paths. Here is the complete relation for the tree T(1) 0 1 2 T(1) = {0,1,2} T(1) = p(0) U (p(1) U q(1) = {0} U {0,1} U {0,2} P(0) = {p(0)} = {{0}} P(1) = {p(1), q(1)} = {{0,1}, {0,2}} P(0) U P(1) = {p(0), p(1), q(1)} = {{0}} U {{0,1}, {0,2}} = {{0}, {0,1}, {0,2}} U(P(0) U P(1)) = {0,1,2} = T(1) T(1) = U P(1) = The paths when repesented as node-values are denoted by binary numbers p(1) = 0.0 p(2) = 0.1 > There is a huge > difference between the union of sets of paths (which is a set of paths) > and the union of paths (which is a set of nodes). Of course there is a difference. But the tree is the union of the set of paths. > And the union of > sets of things that do not have property A can not be a set that contains > an element that *has* property A. This is fundamental. > > > This is the fundamental problem of set theory. The union of all finite > > segments of N is and must be the same as the union of all natural > > numbers. But while the latter union is infinite by defintion and by > > axiom, the union of all finite segments cannot be the infinite segment > > N (because N is not contained in this union). > > It need not be and is not in that union. That is your funamental > misunderstanding. It *is* that union. Moreover, the union of all finite > numbers (using von Neumann representation) *is* N, it does not contain N. The union of segments {1}, {1,2}, {1,2,3}, ... of the set N of natural numbers does not contain N = {1,2,3,...}. But if all these segments are subsets of a set R then also N is a subset of this set R. Now replace R by T. > > > "The infinite union of finite numbers" is illogic but it has been > > swallowed by most of us (yes, I did also swallow it until deeper > > thinking lead me the correct way). > > What is the illogic? > > > But the infinite union of finite segments is obviously incapable of > > yielding an infinite path, as you say above: "there can be no infinite > > path in that union because none of the constituent sets contains an > > infinite path." > > There is not infinite path as an element *in* that union, that union *is* > the infinite path. Pray look at the difference. And this path belongs to the tree. > > > > > In this union P(0) U P(1) U ... there is the path p(0) as an element > > > > and there is the path p(1) as an element and so on. Therefor the set > > > > of all the paths p(0), p(1), p(2), ... is a subset of P(0) U P(1) > > > > U ... > > > > > > The set of all the *finite* paths. > > > > Correct. Even more precisely: The infinite set of finite paths p(0), > > p(1), p(2), ... is a subset of the set of all finite paths p(i), q(j), > > r(k), .... The union of the infinite set of finite paths p(0), p(1), > > p(2), ... is the infinite path p(oo), if the union of all finite > > segments of N is N. > > But that one is correct. But in P(0) U P(1) U P(2) U ... you are *not* > uniting path, you are uniting sets of paths. Yes. You get a set of paths. The paths of this set can be united to get a set of nodes, namely the tree. > T(2) = {0, 1, 2, 3, 4, 5, 6} > so T(2) != P(0) U P(1) U P(2). T(2) = U(P(0) U P(1) U P(2)). Regards, WM
From: mueckenh on 19 Feb 2007 08:03 On 19 Feb., 01:02, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1171789749.778902.224...(a)h3g2000cwc.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > On 16 Feb., 16:36, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > ... > > > > > In none of them I do immediately see the number three. What I see > > > > > is that there are three of something, not the number three. > > > > > > > > What is common to all? > > > > > > I stated that already just above. > > > > Yes. You said "three of something". That is exactly the meaning of the > > number three. > > Not in my opinion, and I think not in mathematics. If the meaning of > 3 is "three of something" how than do we calculate "three of something" > times "three of something"? Or "nine of something" divided by > "three of something"? Or more concrete, if I divide nine apples by > three apples what is the result? three of something, where the "something" here means the unit. Regards, WM
From: mueckenh on 19 Feb 2007 08:08 On 19 Feb., 01:31, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1171790940.416072.43...(a)k78g2000cwa.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > On 18 Feb., 02:51, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > In article <1171702351.590890.177...(a)h3g2000cwc.googlegroups.com> mueck..= > ....@rz.fh-augsburg.de writes: > > > > > On 16 Feb., 15:58, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > > This is not a representation of 3 other than in a perverted > > > > > > system, which calls 0 the first number, 1 the second and so > > > > > > on. Of course {{{{}}}}, or better and easier {{{{, denotes the > > > > > > fourth number which is 4 and not 3. > > > > > > > > > > Can you tell me a form of set theory where 0 is *not* the first > > > > > ordinal or cardinal number? If so, how many elements does the > > > > > empty set have in such a system? > > > > > > > > 0 may be the first (or better the zeroest) ordinal or cardinal number > > > > (if you wish to have the empty set in the theory). Nevertheless it is > > > > not the first natural number and not a natural number at all. > > > > > > But you said "perverted system which calls 0 the first number". The only > > > place where it is called the first *natural* number is in Bourbaki and its > > > > In all set theory 0 is called the first ordinal number, but in fact it > > is the zeroth one. Why do you start counting ordinals with 0 but start > > counting ordinally with 1? > > So your "natural number" above was a red herring? A human being in its > first year has the age 0. Before completing his first year, the being has the number of years which comes before 1. The first number drawn in lottery may be the 7. The first is that ordinal number which we start with. It is the first ordinal number. 0 is not commonly accepted as an ordinal number. That is why usually we speak of first, even set theorists do so. But however this may be, 0 is not a natural number. > A human being in its twentieth year has the > age 19. What is so strange about doing the same with ordinal numbers? > The twentieth century encompasses the years 1901-2000. There are all > kinds of places where the linguistic ordinal number does not match the > mathematical ordinal number. (Note that the above holds for Western > cultures, there are cultures where different conventions do apply. > Like those cultures where the first year of a ruler starts a year > after taking the throne, or that state where year 35 of the rule is > actually the 29-th year.) > > > > But following your reasoning, {{{}}} is the third number, which is 3. > > > BTW, I can quote you as saying: > > > > > > > {{{}}} > > > > > > It was page 93 of my book. > > > > > I have seen it earlier than that. > > > > By the way, above is only number 2 given. > > > So earlier you said it is 2. What is it? > > > > 2 is the number called in set theory which I referred to here and in > > chapter 7 of my book (I hope without too many errors). > > 3 is the number called in any reasonable system. > > So you are trying to obfuscate the issue. > > > > If my house contains no dogs, in what way does the set of dogs in my house > > > not exist in reality? > > > > There is no dog and no set of dogs, because it would be the same set > > of cats. Your dog would be your cat. That is impossible. They would > > not live together, let alone form the same (identical) set. > > In my house the set of dogs is the same as the set of cats. And indeed, > my dog is my cat. But that is all pretty unmathematical. Indeed. > > > > If you think {} to be an unnatural set, so be it (that is not mathematics, > > > because there is no mathematical definition of natural set). > > > > There not even a mathematical definition of an unnatural set. > > Indeed. There is only a definition of set. Not even that. > > > > > > But if > > > somebody asks how many coins I have in my purse, he is asking for the > > > cardinality of the number of coins in my purse. And I can correctly > > > answer 0 at some times. In that case the set is the set of coins in > > > my purse, and that can be empty (and is quite often in reality). > > > According to current definitions and axioms in set theory, {} *is* a > > > set. > > > > That is known to me but will not prevent me from criticising it. > > Well, I do know you do not like the axiom of infinity, but the other > axioms of ZF imply the existence (and uniqueness) of the empty set. > -- Of course. There is usually the axiom of its existence. Regards, WM
From: mueckenh on 19 Feb 2007 08:10 On 19 Feb., 01:35, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1171816407.391676.216...(a)v33g2000cwv.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > ... > > The limit {1,2,3,...} = N does exist according to set theory. Why > > should the limit {1}, {1,2}, {1,2,3}, ... = N not exist? > > There you go again. Set theory does *not* use limits. I know of *no* > definition that gives the limit of a sequence of sets. So pray write > down here how *you* define the limit of a sequence of sets. We need not get involved into a discussion about limits of sets or the etymology of the limit ordinal number. Consider whether the following definitions of N in your opinion are correct or not. N = {n | n e N} = {0,1, 2, 3, ...} N = U {{0,1,2,...,n} | n e N} = U {{0}, {0,1}, {0,1,2}, ...} If they are acceptable, then consider whether a set which contains all sets of the form {0,1,2,...,n} also contains N. And if every set which contains all sets of the form {0,1,2,...,n} contains N, why does the union of finite trees T(n) not contain an infinite path? Regards, WM
From: mueckenh on 19 Feb 2007 08:16
On 19 Feb., 02:56, Michael Press <rub...(a)pacbell.net> wrote: > In article > <1171790147.558130.297...(a)j27g2000cwj.googlegroups.com> > , > > > > > > mueck...(a)rz.fh-augsburg.de wrote: > > On 17 Feb., 03:05, Michael Press <rub...(a)pacbell.net> wrote: > > > In article > > > > <1171468278.284645.273...(a)v33g2000cwv.googlegroups.com>, mueck...(a)rz.fh-augsburg.de wrote: > > > > On 13 Feb., 21:17, Virgil <vir...(a)comcast.net> wrote: > > > > > In article <1171364856.226197.135...(a)l53g2000cwa.googlegroups.com>, > > > > > mueck...(a)rz.fh-augsburg.de wrote: > > > > [...] > > > > > > > What about all existing sets with 3 objects, i.e., the fundamenal set > > > > > > of 3? > > > > > > What fundamental set of 3 does WM refer to? > > > > > Olease read before writing. The set of all existing sets with 3 > > > > objects. > > > > How many sets with three elements are there? > > > No idea. Less than 10^100. But that is unimportant. Important is that > > there is at least one set with 3 elements. > > It is important to me. Why? However, I can't help you. > > Show me a set with three elements. Here it is: lll Regards, WM |