Cantor Confusion
in [Math]
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From: Dik T. Winter on 16 Feb 2007 09:58 In article <1171615110.930410.270960(a)s48g2000cws.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 15 Feb., 14:05, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > 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: {{{{}}}}. =20 > 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? > > > Only set theory needs this absurd definition of nought to be the > "first" number, because 3 counting the numbers up to 2 but not 3 > itself fits well with omega counting all the natural numbers but being > not a natural number. No, this is sham does not help: The number of > all natural numbers=B4, if existing, is a natural number, because the > natural numbers count themselves (so they were designed). So how many natural numbers precede the first natural numbers? We are counting natural numbers, so it should be a natural number? > |{1}| = 1 > |{1,2}| = 2 > |{1,2,3}| = 3 > ... > |{1,2,3,...}| = ... i.e. potentially infinite, not fixed, capable of > growing without bound, denoted by oo but not by a fixed number omega. And |{}| = ? -- 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 16 Feb 2007 09:50 In article <a0bat2ph8kof1gjb1qfrb06vjhmlmkpbke(a)4ax.com> G. Frege <nomail(a)invalid> writes: > On Fri, 16 Feb 2007 03:39:36 GMT, "Dik T. Winter" <Dik.Winter(a)cwi.nl> > wrote: > > > > Actually it has already been used in Alice in Wonderland, published when > > Frege was 17 years old ... > > > Actually I was referring to (1) the insight that there really a > problem is lurking and (2) the idea of using quotation marks to > prevent such confusion. :-) Actually I should have quoted the dialogue correctly. In the original there are quotation marks around the names. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: G. Frege on 16 Feb 2007 09:59 On Fri, 16 Feb 2007 14:50:48 GMT, "Dik T. Winter" <Dik.Winter(a)cwi.nl> wrote: > > Actually I should have quoted the dialogue correctly. In the original > there are quotation marks around the names. > Just what I've said: "Actually, this habit (NOT to use quotation marks) is still quite common among mathematicians (i.e. in mathematics in general). :-( :-) F. -- E-mail: info<at>simple-line<dot>de
From: G. Frege on 16 Feb 2007 10:13 On Fri, 16 Feb 2007 05:07:47 +0100, G. Frege <nomail(a)invalid> wrote: > > Actually I was referring to (1) the insight that there really a > problem is lurking and (2) the idea of using quotation marks to > prevent such confusion. :-) > > See: > http://plato.stanford.edu/entries/quotation/ > Would you PLEASE follow the link before talking even more nonsense? The following should be a correct quote from "Through the Looking Glass". ------------------------------------------- The name of the song is called "HADDOCKS' EYES."' 'Oh, that's the name of the song, is it?' Alice said, trying to feel interested. 'No, you don't understand,' the Knight said, looking a little vexed. 'That's what the name is CALLED. The name really IS "THE AGED AGED MAN."' 'Then I ought to have said "That's what the SONG is called"?' Alice corrected herself. 'No, you oughtn't: that's quite another thing! The SONG is called "WAYS AND MEANS": but that's only what it's CALLED, you know!' 'Well, what IS the song, then?' said Alice, who was by this time completely bewildered. 'I was coming to that,' the Knight said. 'The song really IS "A-SITTING ON A GATE": and the tune's my own invention.' ------------------------------------------- It's rather idiotic to compare that funny dialog with (very definite) things Frege formulated. Probably Lewis Carroll hat a different "problem" in mind here. See: http://www.ditext.com/carroll/tortoise.html F. -- E-mail: info<at>simple-line<dot>de
From: Dik T. Winter on 16 Feb 2007 10:34
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 not be 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. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |