Cantor Confusion
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From: Virgil on 1 Nov 2006 15:14 In article <1162382668.224018.252560(a)k70g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > MoeBlee schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > All entries of the list have a finite number of letters. An infinite > > > sequence is larger than any finite sequence. The diagonal of a list > > > cannot have more letters than the lines. > > > > > > According to your logic the list can have infinitely many lines. But > > > even if that was correct it would not facilitate an infinte diagonal. > > > > > > The number of diagonal elements is the minimum of columns and lines. > > > > 0 > > 1 2 > > 3 4 5 > > 6 7 8 9 > > ............... > > ................. > > ................... > > > > infinitely downward for an infinite list of finite lists. > > > > The diagonal is 0 2 5 9 14 ... infinitely across. > > > > All entries in the infinite list are finite lists. > > better say finite sequences or numbers or entries > > > The infinite list is > > longer than any finite list. > > The entries surpass every finite entry. Nevertheless you call all of > them finite. > > > The diagonal of the list is infinite. > > That is your assertion. But obviously the diagonal elements are > simultaneously elements of the entries. Each position in the diagonal is determined by a position in one of the listed elements, but if that position number is an unbounded function of the list number, with each listed entry at least that long, then there is no problem with having an infinite diagonal. > > > > And that be formalized easily in set theory. > > That may be, therefore it is no wonder that set theory yields > selfcontradictions. That set theory yields results that WM does not like, and which cotradict things that WM chooses to assume, but nothing which contradicts any of its own axioms, which is the only relevant criterion for being self-contradictory. So that WM makes false claims. > > The diagonal elements are simultaneously elements of the entries. > Therefore the diagonal elements cannot sum up to a number which is > larger than any natural number unless also the elements of list entries > sum up to a number which is larger than any natural. WM deliberately blinds himself by assuming that if every string of characters in an infinite list of strings is finite that there must be a UNIFORM limit on their lengths. But a trivial example proves him wrong: For each n in the infinite completed set of naturals Net f(n) be a string of length at least n, then there is no maximally long string in the image of f, as for each m in N there is n in N with n > Length(f(m)). > > Or put it so: Every segment of the diagonal is covered by an entry. > There is no segment which is not covered. If all entries are finite, > then the diagonal cannot be infinite (if infinite omega is larger than > any finite n). Note that the "diagonal" in my example above MUST be greater than any finite n, so that either there is a number greater than every finite n and simultaneously less than omega or WM is wrong again.
From: Virgil on 1 Nov 2006 15:26 In article <1162383001.838477.151460(a)m7g2000cwm.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > All entries of the list have a finite number of letters. An infinite > > > sequence is larger than any finite sequence. The diagonal of a list > > > cannot have more letters than the lines. > > > > It can be, and must be, longer than any single line, > > It cannot be longer than any single line because it consists of > elements of the single lines. WM is under the mistaken impression that there must be a longest line in any list of lines. Let N be the infinite set of all finite naturals in ZF or NBG or any other set theory in which there is an infinite set of finite naturals. Let f be a function with domain N, and have as value f(n) a string of characters of length 2*n. In ZF or NBG, this is a perfectly well-definable sort of function. Any "diagonal" for this function will have to be of length greater than every n in N. > Every diagonal element is a line element. > Segment of the diagonal has less elements than some list element. If > the whole diagonal has more elements than every list element, then > there must be some point of crossing or overtaking. Not for my example above. > > > > > > > The number of diagonal elements is the minimum of columns and lines. > > > > Actually, it would be at least the maximum of both, and if neither is > > finitely bounded, then the diagonal must be infinite. > > The lenght of the diagonal of a matrix is not the maximum of both but > the minimum. And if one is finitely bounded, then the diagonal is also > finitely bounded. But in my example above, neither is bounded, and neither is the diagonal. > > Now, we know that the width is bounded by the condition that every line > is finite. It is not what WM doesn't know that hurts him, its what he knows that ain't so. In my example above every line is bounded but the set of lines is not. WM keeps trying to argue that since every finite set of naturals is finitely bounded the set of all naturals is also finitely bounded, but in ZF on NBG or most other set theories, it ain't so. > So the diagonal is finite. Only in your dreams. > > Regards, WM
From: Virgil on 1 Nov 2006 15:35 In article <1162383101.839370.87880(a)m73g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > > > Correct. And this sequence has to be defined somehow. But there are > > > > > only countably many definitions. > > > > > > > > That is true, if by "definition" you mean a finitary algorithm or > > > > recipe. If this constraint does not hold then there are aleph-1 > > > > "definitions" > > > > > > There are only countably many finite strings over a finite alphabet, > > > whichever may be their definition. > > > > But without both finitenesses, countability is not guaranteeable. > > Why do you emphasize this point? Every alphabet is finite. Every > definition is finite. Only to those who include both in their axiom systems. Which I do not. In ZF or NBG or similar, the set of all functions from the infinite set of all finite naturals, N, to any set of more than one member is uncountable. > > > WM > > All entries of the list have a finite number of letters. An infinite > > sequence is larger than any finite sequence. The diagonal of a list > > cannot have more letters than the lines. Since the number of "lines" need not be finite, neither need the diagonal be finite. > > > Virgil: > But there can be lines, and a diagonal, larger than any finite natural, > > if there is no finite natural upper bound on line length. > > WM: > Give an example of a finite natural which does not belong to a finite > sequence. What is your point? Would that prevent that natural from belonging to an infinite sequence of naturals? Or even infinitely many such infinite sequences? Wm seems to have a collection of short circuits in his thinker which lead him to believe all sorts of things follow from nonsense.
From: Virgil on 1 Nov 2006 15:40 In article <335f8$45489012$82a1e228$29179(a)news2.tudelft.nl>, Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> wrote: > Randy Poe wrote: > > > Han de Bruijn wrote: > >>A large mass (as with the balls in a vase close to noon) surely _will_ > >>halt time, according to the General Theory of Relativity. > > > > You are no doubt remembering something you heard about > > time inside the Schwarzchild radius of a black hole. > > > > Point to ponder: Black holes exist in our galaxy. Yet here we > > are, with time ticking on regardless. Is it possible you remember > > something wrong? > > That's besides the point. First of all, it has NOT been established that > black holes indeed exist in our galaxy. It has been established beyond reasonable doubt that huge black holes exist at the center of a great many galaxies, including our own. If HdB is to pose as a physicist, he should keep up on his physics.
From: Virgil on 1 Nov 2006 15:55
In article <1162404540.740063.144740(a)m73g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > > Correct. And given any set of lines we can find a finite number which > > > is larger. Otherwise, at least one of the lines would contain a number > > > of letters which was larger than any natural number, i.e. which was > > > omega. > > > > But what is important is not the number of letters in any given > > line but the number of lines. These are two different things. > > You say so. I strongly disagree, because the number of every segment of > lines is the same as the maximum of letters in the lines of the > segment. But luckily this dissent is not relevant, because the diagonal > is not only limited by the length but also by the width of a matrix. Let us consider the "matrix" in which there is one line for each n in the infinite set of all finite naturals, N, and the nth line is of width n. Then neither the "length" nor "width" of this "matrix" has any finite bound. Therefore neither does the "diagonal". > > The maximum of a set of finite numbers which has no maximum is simply > not present. It is *not* an infinite number which is larger than any > finite number, because a maximum must belong to the set. And a supremum > not belonging to the elements of the set does not yield a diagonal > digit. Does WM choose to claim that a set of naturals which has no maximum must still have a finite upper bound? Or is it true or such a set S that for every n in S there is an m in S with m > n. > > > > Why do you say incorrect? Given any integer N, we certainly have more > > letters > > than N. > > But the smallest number which is larger than any integer is omega. And > we have excluded a line with omega letters. But we also exclude any lesser bound. > > If it does not exist you cannot use it to limit the > > diagonal. > > If it does not exist, then you cannot build a diagonal of that length. > > > > > > So the greatest number of letters the diagonal can have is > > > > greater than any integer. > > > > > > Incorrect. See above. > > > > Your claim is that the number of letters the diagonal can have > > is limited above by the number of letters in the list. > > It is limited by the number of letters in *one line* of the list. Which line? > > The maximum of a set of finite numbers which has no maximum is simply > not present. That's the point, stupid. You keep requiring a finite maximum when there is none. The length of the diagonal, like the set of line lengths and the set of line numbers, need not have any finite maximum. But WM in his self-imposed blindness will not allow himself to see that. |