Cantor Confusion
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From: Virgil on 6 Nov 2006 17:09 In article <1162826490.608758.156730(a)h48g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > David Marcus schrieb: > > > > > > "Whole number" is Cantor's name for his creation. > > > > > > > > > > My question is : Do you maintain omega > n for all n e N? > > > > > > > > Before I can answer the question, I need to know what you mean by the > > > > words/terms. So, please define "omega", "N", and ">". Also, by > > > > "maintain" do you mean that ZFC proves it? > > > > > > Do you keep on thinking that omega > n for all N, applying all > > > words/terms as Kunen would understand them. > > > > Using Kunen's definitions, then ZFC proves that if n is a natural > > number, then n < omega. Do you disagree that ZFC proves this? > > No. Therefore omega is larger than any finite natural number. > Every natural number is finite. > omega is not finite. > > Omega is the number of digits of the diagonal of > > 1 > 12 > 123 > ... > > The number of digits of the first and any other column is omega. The > number of digits of any line is less than omega. > > The length of the matrix is omega, as a maximum taken. Does meuckenh claim a line numbered omega? If so then what is the line number of the immediately preceding line? If not then it is only a supremum on the line numbers and not an actual maximum. > The width of the matrix is omega, as a supremum not taken. > > This small problem contradicts ZFC. No, it contradicts WM. Unless he can find a finite natural which is the immediate predecessor of omega.
From: Virgil on 6 Nov 2006 17:18 In article <1162827731.738172.234870(a)h54g2000cwb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > David Marcus schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > Virgil schrieb: > > > > > > > In article <1162470874.593282.36250(a)b28g2000cwb.googlegroups.com>, > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > > > WM merely repeated his automatic error several more times here. > > > > > > > > WM claims that a list in which the nth listed element is a string of > > > > length at least n characters cannot produce a diagonal of length > > > > greater that any finite number of characters. > > > > > > > > > > The diagonal needs an element from every line, the n-th element from > > > the n-th line. Therefore it cannot be longer than every line. > > > > By "it cannot be longer than every line", do you mean its length can't > > be greater than the sup of the lengths of the lines? > > Its length can't be the length of a column, i.e. omega, if the width of > the matrix has only the supremum omega. If every line has as many digits as its line number then the diagonal length MUST have a digit for every line number. > > There is a difference between maximum and supremum which cannot be > bridged. And to those who can see beyond the end of their nose, there is no need to bridge it, since it does not obtrude in this situation. There is no more a line numbered omega than there is a digit positions in any line numbered omega, but there is one digit position in the diagonal for every line.
From: Virgil on 6 Nov 2006 17:26 In article <1162828228.538226.143300(a)b28g2000cwb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > There is a mapping of the diagonal on a (each) column. > There is no mapping of the diagonal on any line. > The diagonal cannot have more elements than the width of the matrix is. For every n in N, there is a line longer than n, For every n in N, there is a line number larger than n. There is no last line. here is no last position in the diagonal. > > The number of elements of the diagonal is assumed to be omega. > That is wrong, because only the supremum is omega. WM assumes that omega must be a member of omega, which is false. To say that the number of elements in a well- ordered set is omega prohibits any of those members, taken in order, from corresponding to omega. To insist otherwise is to require omega to be a member of itself.
From: Virgil on 6 Nov 2006 17:32 In article <1162828907.384237.5300(a)h48g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1162563567.735020.246810(a)b28g2000cwb.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Virgil schrieb: > > > > > > > In article <1162470874.593282.36250(a)b28g2000cwb.googlegroups.com>, > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > > > WM merely repeated his automatic error several more times here. > > > > > > > > WM claims that a list in which the nth listed element is a string of > > > > length at least n characters cannot produce a diagonal of length > > > > greater that any finite number of characters. > > > > > > > > > > The diagonal needs an element from every line, the n-th element from > > > the n-th line. Therefore it cannot be longer than every line. > > > > The "diagonal"must be longer than every finite "line". > > But it cant, because each of its elements stems from a line. Which line(s) is the diagonal NOT longer than when line n is at least of length n? > > > The only way it > > will ever fail to be longer than some "line" is if that line is itself > > infiitely long. > > Which is impossible, because each of its elements stems from a line. When for all n in N, line n is of length n, which is quite possible, then for all n in N, the diagonal is of length >= n. That looks like an infinite diagonal to me. WM is assuming that since every natural is finite, there must be a largest natural, but in ZF and NBG, that is specifically false. And no assumptions, however hidden, by WM can make it otherwise.
From: MoeBlee on 6 Nov 2006 17:38
mueckenh(a)rz.fh-augsburg.de wrote: > MoeBlee schrieb: > > > David Marcus wrote: > > > Obviously, I could be wrong, but I think WM means map it on a line of > > > the list. He seems to think that because we construct the diagonal from > > > the list, the diagonal must be one of the lines in the list. Why he > > > thinks this, I have no clue. > > > > You mean map the diagonal (or the range of the diagonal, or whatever) > > onto one of the finite sequences that is in the range of the infinite > > sequence of those finite sequences? I.e., map the diagonal onto a > > member of the range of S? A 1-1 map? If so, yes, I would share your > > bafflement as to why we should think there is such a mapping or what > > contradiction there is in there not being such a mapping. > > There is a mapping of the diagonal on a (each) column. Okay, if you want to put it that way. > There is no mapping of the diagonal on any line. Okay. > The diagonal cannot have more elements than the width of the matrix is. I answered that already. If you define 'the width', then it turns out to be equal to omega, which is just what the length of the diagonal is. If you don't define 'the width', then 'the width' is empty language and is irrelevent. > The number of elements of the diagonal is assumed to be omega. > That is wrong, because only the supremum is omega. I answered that already. And the cardinality of the diagonal is not assumed, but is proven, to be omega. And the fact that omega is a limit ordinal greater than any finite segment of the diagonal is not in contradiction with anything in the proof nor does it entail that the cardinality of the diagonal cannot be omega. So, at this point, you have just returned with replies that I've already answered, and you haven't shown any step in my proof tha is not justified by first order logic applied to the Z axioms nor have you shown a sentence P and ~P of the language of set theory such that both are theorems of set theory. MoeBlee |