Cantor Confusion
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From: mueckenh on 1 Nov 2006 06:56 Virgil schrieb: > In article <1162219026.697699.10850(a)m73g2000cwd.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > It is unbounded but always finite. > > Unbounded implies not finite, as finite implies bounded. That is the definition of potential infinity. But for this definition we never have a completet set and never can be sure whether a set is Dedeking infinite. Sorry, to say, but you always mix up these two very different things. It should be comprehensible that potential infinity is possible without the axiom of infinity. But that is not what set theory requires. Therefore it is correct to say that in set theory theory there is no infinity present or detectable without the axiom of infinity. WM: > Sorry, I can't see that conclusion. You can always define a new and > larger base., a new and more efficient way to use the bits. Virgil; Unless this each time erases from existence some previously defined number(s), you are admitting that the number of numbers it is possible in infinite time to define is infinite. You cannot have it both ways. WM If a definition of a number requires all he bits you have, then it must erase everything you defined before. That is unavoidable. You cannot define more than 100 numbers with 100 bits, but maybe you decide to define only one number with your 100 bits. Do you think these considerations are comprehensible (although perhaps not describable in ZFC)? WM: > As long as your strings have a finite number of characters, you have a > finite number of strings. Virgil: That may be how things work in in WM's world, but not in mine. Since I start, in ZF or equivalent, with at least one actually infinite set, |N. I can get as many actually infinite other sets as I wish. If WM claims that his world does not work that way, he is only making equally unprovable assumptions himself. But WM does not have the power to determine reality, only the power to assume that he knows it. WM: The power to recognize it, I hope. WM: > > > Try to draw the graph of the function f(n) = n. The points lay on the > > > diagonal of the first quadrant. As long as n is finite, f(n) is finite > > > too and vice versa. Virgil: > > So what? WM: > This shows that both are finite or both are not. Virgil: So what? The set of even naturals and the set of odd naturals are both finite or both infinite, too. But that does not mean that either is finite. WM But you assert that the number f(n) of numbers n is infinite while the magnitude n of numbers n remains finite. Regards, WM
From: mueckenh on 1 Nov 2006 06:59 Virgil schrieb: > In article <1162220492.974057.66580(a)b28g2000cwb.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Virgil schrieb: > > > > > > > > What kind of well ordering of the reals do you claim to exist? > > > > > > A well ordering in which each non-empty subset has a smallest member. > > > > That is the definition. > > Exactly so! > > > > > > > Defined? > > > > Catalogue? > > > > List? > > > > Else? (Please specify) > > > > > > A well ordering in which each non-empty subset has a smallest member. > > > > > > That's what well orderings are all about. > > > > I know. But the definition does not guarantee existence. In particular > > it does not say how the order in R differs from the order in N or Q. > > > > > > If you are asking for a rule for determining which objects come before > > > others, you should know that no such explicit rule is possible, but that > > > does not make their existence imposible. > > > > What kind of existence do you have in mind? > > In a system that has assumed the axiom of choice, the kind that that > axiom insists on. > > These things do not exist Why then do you assert the existence of things which do not exist? Regards, WM
From: mueckenh on 1 Nov 2006 07:04 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. > > And that be formalized easily in set theory. That may be, therefore it is no wonder that set theory yields selfcontradictions. 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. 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). Regards, WM
From: mueckenh on 1 Nov 2006 07:10 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. 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. > > > > 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. Now, we know that the width is bounded by the condition that every line is finite. So the diagonal is finite. Regards, WM
From: mueckenh on 1 Nov 2006 07:11
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. 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. 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. Regards, WM |