Cantor Confusion
in [Math]
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From: Dik T. Winter on 16 Feb 2007 10:36 In article <1171626693.139060.112450(a)v45g2000cwv.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 16 Feb., 02:06, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > > But it is > > > correct that there are more obvious and less obvious numbers: > > > ... > > > > An ellipsis. > > > > > ||| > > > > A sequence of bars. > > > > > {{{ > > > > A sequence of braces. > > > > > abc > > > > The first letters in some of the alphabets that use the standard order. > > > > 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. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Andy Smith on 16 Feb 2007 11:30 In message <JDKBA5.MwM(a)cwi.nl>, Dik T. Winter <Dik.Winter(a)cwi.nl> writes >In article <1171626163.912450.149230(a)m58g2000cwm.googlegroups.com> >mueckenh(a)rz.fh-augsburg.de writes: > > On 16 Feb., 02:38, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > In article <1171575875.054198.56...(a)q2g2000cwa.googlegroups.com> > > >mueck...(a)rz.fh-augsburg.de writes: > > > > > > > On 15 Feb., 14:03, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > I think we should first clarify the notation before continuing: > > > > P(n) is the set of all paths of the tree T(n) with n levels. > > P(0) is a set with one path, namely p(0) = {0.} > > P(1) is a set with two paths, namely p(1) = {0.0} and q(1) = {0.1} > > P(2) is a set with four paths, namely p(2) = {0.00}, and some others. > >Using that representation for paths is misleading, As you state that >each path is a set of nodes, and if that is the case, with that notation >0.00 is a node, and each path is a set containing a single node. > > > The union of all sets of finite paths is P(0) U P(1) U ... is the set > > of all paths. > >Why? It is the set of all *finite* paths, because there can be no >infinite path in that union because none of the constituent sets >contains an infinite path. > > > 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. > > > The tree forms unions from sets of paths. > > The tree T(2) is the union of the sets of paths P(0) U P(1) U P(2). > > But the tree T(2) does not contain the paths of P(0) and P(1), namely > > p(0) and p(1) and q(1). > >So it is *not* that union. If T(2) = P(0) U P(1) U P(2) it should >contain all paths of P(0), P(1) and P(2). > > > Similar the tree T(oo) is the infinite union P(0) U P(1) U P(2) U ... > >In a similar way, it is *not*. > > > Form the subset p(0), p(1), p(2), ... of this union the tree forms > > the union p(0) U p(1), U p(2) U ... = p(oo). > > > > So the tree contains the union of the paths p(0), p(1), p(2), ... > > which according to you is p(oo). This is the outmost left path of the > > union tree. > >The last part is right. > > > > > > I never stated that p(oo) = {p(0)} U {p(1)} U ...; that can > > > > >the > > > > > case, because on the left hand side we have a set of nodes, > > > > >and on the > > > > > right hand side we have a set of paths. And a path is not a node. > > > > > Moreover, {p(0)} U {p(1)} U ... is not the union of a subset of > > > > > P(0) U P(1) U ...; it *is* a subset. > > > > > > > > {p(0)} U {p(1)} U ... is a subset of the union of P(0) U P(1) U .... > >You are dishonest. You snipped part of my article (without any indication), >making it appear that my refutation below was to your statement above. It >was not, the refutation was in a paragraph you snipped. Also the statement >by you to which the response below belongs did you snip (or do you not see >the difference between the two statemens by you, both wrong?). >Re-inserted your statement here: > > > > > p(0) U p(1) U ... is a subset of P(0) U P(1) U ... > > > > > Wrong. p(0) U p(1) U ... is a set of nodes. > > > Every path is a set of nodes. > > You said p(0) U p(1) U ... = p(oo). Of course this is a set of nodes. > > The paths of this union, p(0), p(1), ... are elements of the union > > P(0) U P(1) U ... . > >Right. > > > Therefore these elements form a subset of P(0) U P(1) U ... . > >Right. The set of these elements form a subset of that union. But >their *union* does not form a subset of that union. > > > > P(0) U P(1) U ... is a set > > > of paths. Again, how can a set of nodes be a subset of a set of paths > > > (which is a set of sets of nodes)? > > > > One path can be a subset of a set of paths as a singleton or it can be > > an element of the set of paths. > >It can *not* be a subset of a set of paths, a subset of a set of paths >is a set of paths, possibly a singleton, in that case it is a set >containing one path, not the path itself. If it is in the set of paths >it is as an element, *not* as a subset. > > > > Do you not know the difference between a set of sets of nodes and a set > > > of nodes? This is the same as stating: > > > {{1}, {1, 2}, {1, 2, 3}} subset U {{1}, {1, 2}, {1, 2, 3}} = {1, 2, 3} >This was about your first statement. > > > and at the same time: > > > {1, 2, 3} subset {{1}, {2, 3}}. >And this about your second statement. > > > Please keep a clear head about what the *elements* of the unions are. > >Apparently not, because you maintain that a path can be a subset of a >set of paths. Stating in fact: > {1, 2, 3} subset {{1}, {1, 2}, {1, 2, 3}} >which is clearly false. > > > Please see above. The union of paths cotains elements p(n). > >Keep a clear head about what the elements *are*. A union of paths is a >set of nodes, so the elements are nodes, *not* paths. If you are talking >about the paths that are united, they are *not* called elements of the >union. You may call the "constituent paths" if you wish. > > > The tree > > forms unions from these elements. The tree T(oo) does not contain any > > finite paths, but the unions of finite paths. > >Yes, what is the relevance. > > > > > > > > This subset is unioned by the tree. It yields p(oo). > > > > P(oo) is composed of the sets p(oo), q(oo), r(oo). > > > > > > The tree is not interesting in this. > > > > Why do you think did I choose the tree? > >I do not know, but it is not interesting in this. > > > > If the finite sets of paths contain only finite paths, their union can > > > *not* contain an infinite path. Because when taking that union you are > > > uniting sets of paths, not paths. That it is a subset of the union of > > > that union does not matter at all. The union of that union is simply > > > a set of nodes, *not* a set of paths. > > > > See above. > >What is the relevance of what you wrote above? The threes do *not* define >the unions of the finite sets of paths. > > > > > The tree contains a path which *is* the union of the p(i), namely > > > > p(oo). The tree, i.e., the union of finite trees is or establishes the > > > > paths q(oo), r(oo), ... too. Therefore it establishes all elements of > > > > the set P(oo). > > > > > > Perhaps, depending on what you mean with "establish". I thought we had > > > a common definition, but that is apparently not the case. > > > > It is. p(oo) = p(0) U p(1) U p(2) U ... > > The finite paths are elements of the union of sets of paths P(n). The > > tree unites some of these elements which fit together to form an > > infinite path. > >In that case, yes. > > > >But anyway, > > > yes, P(oo) contains infinite paths, and so can *not* be a subset of > > > U P(i), as that one contains (as you note) only finite paths. > > > > It need not be a subset of U P(i) in order to be in the tree. > >That is what I am arguing all the time, but you did maintain that it >*is* a subset of that union, and need that in your proof that P(oo) >is countable. sticking my nose in again, smack it if you wish. As I understand it : P(1) is the set of all numbers of 1 bit length - 2 elements (0.0 & 0.1) P(2) is the set of all numbers of 2 bits length - 4 elements (0.00, 0.01, 0.10, 0.11) P(n) is the set of all numbers of n bits length - with 2^n elements. The set P(n) includes all P(m) for m<n, if a right fill of 0 for the bits is inferred (i.e. 0.0 is the same as 0.00 is the same as 0.00..) For any bit-length (path-length) n, P(n) is countable, so for any finite n, P(n) is countable. But, real numbers require an infinite number of bits to represent them (or, if you prefer, we can define the set of reals as the set of numbers with an infinite random binary expansion). But if the number of bits is not finite, neither is P(oo), and the reals are uncountable. -- Andy Smith
From: Virgil on 16 Feb 2007 12:00 In article <1171615110.930410.270960(a)s48g2000cws.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 15 Feb., 14:05, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1171469941.993167.166...(a)h3g2000cwc.googlegroups.com> > > mueck...(a)rz.fh-augsburg.de writes: > > > > > On 14 Feb., 02:22, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > ... > > > > > > > > Numbers can express properties? You have lost me here. > > > > > > > > > > > > > > To have three elements is a property of a set. > > > > > > > > > > > > Oh. Your terminology is unfathomable. Indeed the number 3, when > > > > > > seen > > > > > > as a set, can have three elements. In the von Neumann model. > > > > > > However, > > > > > > I remember to also having seen another model, where the number > > > > > > three was > > > > > > {{{}}} > > > > > > > > > > It was page 93 of my book. > > > > > > > > I have seen it earlier than that. > > > > > > By the way, above is only number 2 given. > > > > Indeed, my error. So your comment: > > > When seen as a set of curly brackets it has 3 at the left sinde and 3 > > > at the right. > > was actually completely irrelevant. Let's get on with the actual > > representation of 3: {{{{}}}}. > 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. WM has not the power to command where the natural numbers shall start. If mathematics chooses to start them with with 0, then WM has no more power to change that than Canute to control the tides. > > Only set theory needs this absurd definition of nought to be the > "first" number Set theory may choos to start numbering wherever it choses to without let of hindrance from such impotent idiots as WM. > No, this is sham does not help: The number of > all natural numbers�, if existing, is a natural number, because the > natural numbers count themselves (so they were designed). Aping Kronecker's view that God made them? Does WM claim to have God's authority backing his ukases, or merely Kronecker's?
From: Virgil on 16 Feb 2007 12:25 In article <1171626693.139060.112450(a)v45g2000cwv.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 16 Feb., 02:06, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1171562525.973762.49...(a)v45g2000cwv.googlegroups.com> > > mueck...(a)rz.fh-augsburg.de writes: > > > > ... > > > One set of curly brackets would be enough to be number 3. But it is > > > correct that there are more obvious and less obvious numbers: > > > ... > > > > An ellipsis. > > > > > ||| > > > > A sequence of bars. > > > > > {{{ > > > > A sequence of braces. > > > > > abc > > > > The first letters in some of the alphabets that use the standard order. > > > > 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? That each, regarded as a set of objects, may be bijected with each of the others as sets. And that each may be bijected with the set { {}, {{}}, { {}, {{}} } }, which, in ZF and NBG, IS the number 3.
From: Lester Zick on 16 Feb 2007 12:49
On Fri, 16 Feb 2007 00:11:04 +0100, Carsten Schultz <carsten(a)codimi.de> wrote: >MoeBlee schrieb: >> On Feb 15, 1:42 pm, mueck...(a)rz.fh-augsburg.de wrote: >>> Every set of finite even numbers contains numbers which are larger >>> than the cardinal >>> number of the set, if the set has a cardinal number. >> >> Since even numbers are natural numbers, thus finite, I suppose that >> what you actually mean is: > >I think you are wrong. > >> Every finite set of even numbers has a >> member greater than the cardinality of the set. >> >> But that's not even true. >> >> {0 2} is a finite set of even numbers and no member is greater than >> the cardinality of the set. > >Guess why WM does not want 0 to be a natural number. Because he has fingers? >[...] > >>> Could you please look up the proof that the actually infnite set N >>> has a cardinal number larger than every natural? Usually it is done by >>> induction, and it starts: Certainly 0 < |N|. This is the fundamental >>> error. No, certainly 0 is not less than |N|. No axiom and no proof >>> lead to this statement. >> >> You're nuts. > >You are right on this one. Which is why your first statement was >probably wrong. > >Best, > >Carsten ~v~~ |