Cantor Confusion
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From: MoeBlee on 3 Nov 2006 13:27 mueckenh(a)rz.fh-augsburg.de wrote: > The only thing to be shown is that omega is the limit of the > natural numbers. This is the fundamental assumption of set theory. What set theory? There's no axiom of ZFC that omega is the limit of the natural numbers. We prove, but do not assume, that omega is a limit ordinal. And 'limit' in 'limit ordinal' is not the same sense of 'limit' as in the limit of a function. MoeBlee
From: MoeBlee on 3 Nov 2006 14:59 mueckenh(a)rz.fh-augsburg.de wrote: > MoeBlee schrieb: > > > > > > > > > > > > > 0 > > > > > > 1 2 > > > > > > 3 4 5 > > > > > > 6 7 8 9 > > > > > > > > > > > > The entries surpass every finite entry. Nevertheless you call all of > > > > > them finite. > > > > > > > > I don't know what you're trying to say. > > > > > > Because you did not read what I wrote. I defined it above: "better say > > > finite sequences or numbers or entries" > > > > No, I read it over a few times. When I say I don't understand > > something, you can take me at my word that I mean just that - I read it > > a few times, thought about it, and don't understand it. Thus, you can > > save yourself the wasted typing of saying false things such as that I > > didn't read what you wrote. > > I don't know what you mean by entries SURPASSING every finite entries. > > What entries surpass which other entries? What does 'surpass' mean? If > > you give me ordinary discourse, then I'll have a better chance of > > understanding you, just as I defined each of my terms, 'sequence', > > 'entry', etc. in my own remarks. > > The digits of the numbers written down in your list (above) are not > bounded by a finite number. S is a denumerable sequence of finite sequences. But, yes, of course, every natural number is a member of the range of exactly one of the finite sequences that are in the range of S. So, yes, of course, the union of the ranges of the finite sequences is unbounded. So lets call that union, 'E' (which is, as I understand, what you would call the set of entries). > > > Maybe, if you say so. But omega is not the maximum of all finite > > > sequences. > > Yes, since omega is not a sequence at all, let alone being a finite > > sequence, let alone being the maximum of all finite sequences. > > > > > Therefore the width of the list is less than omega. > > > > My argument does not mention 'width of the list'. If YOU want to refer > > to 'width of the list', then YOU need to define it. And that means > > first proving that there exists a unique object that meets the > > description. > > The width of the list is the number of digits of the number with most > digits. As such a number does not exist, he width is the supremum, > namely omega. First, my proof says nothing about digits or numbers of digits. My proof doesn't even require that we have proven the basis representation theorem. Then, to make your definition precise, this is how I take it: The width of S is the number of digits in the element of E that has more digits than any other element of E, if such an element of E exists, and otherwise the width of S is the supremum (by ordinal ordering) of the set of numbers of digits of elements of E. And, so, yes, under that definition, the width of S is omega. > > > > > > The diagonal of the list is infinite. > > > > > > > > > > That is your assertion. But obviously the diagonal elements are > > > > > simultaneously elements of the entries. > > > > > > > > No, we trivially PROVE the diagonal sequence is infinite. > > > > > > You may also prove that the maximum of numbers less than 5 is 5. > > > Nevertheless it is false. > > > > No, I can't prove that. > > > > > The diagonal of a list of sequences with less than 5 terms is less than > > > 5. > > > The diagonal of a list of sequences with less than omega terms is less > > > than omega. > > > > > > This simple truth should convince you that ZFC is not acceptable. > > > > You claim it is a simple truth without proving it. And your claim is > > not even compatible with the simple intuitive picture that uses > > ellipses. So not only do you not have a mathematical proof of your > > claim, you don't have an intuitive explanation, except an argument by > > ANALOGY in which you analogize between the finite and infinite, only > > assuming, as a form of question begging, that what holds for a finite > > sequence must hold for an infinite sequence. > > I did not introduce a number omega which is larger than every natural > number. > But IF such a number is introduced, THEN we should be allowed to use > the inequality omega > n for every natural number n, i.e. for the n > digits of the n-th list entry. If '>' stands for the standard ordinal ordering, then, yes, of course, An(n in omega -> n < omega). But that doesn't entail that your argument by analogy has any merit whatsoever. > > > > > 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. > > > > > > > > In my example, I said nothing about summing up. And I said nothing > > > > about anything in S being larger than any natural number. > > > > > > You said the domain is omega. You said "we trivially PROVE the diagonal > > > sequence is infinite". omega is larger than any natural number. > > > "Infinite" means "larger than any natural number". > > > > The common definition of 'is infinite' I use is: > > > > x is infinite <-> ~En(n is a natural number & x is equinumerous with n) > > > > which in turn reduces to: > > > > x is infinite <-> ~Enf(n is a natural number & f is a bijection between > > x and n) > > > > No mention there of "larger". > > So you do not man that omega > n holds fo every n e N? Then we have no > dissent. No, of course, if '>' stands for the standard ordinal ordering, then omega > n holds for every n in omega. But all I said was that my proof does not mention that. My proof does not require mentioning that. Just because my proof does not mention it does not mean that my proof contradicts it. > > > > > Or put it so: Every segment of the diagonal is covered by an entry. > > > > > > > > Which 'entries'? > > > > > > > > > There is no segment which is not covered. > > > > > > > > What is the initial segment {<0 2>}, of the diagonal, covered by? And > > > > what does it matter? > > > > > > > > > If all entries are finite, > > > > > > > > Yes, all entries of S are finite sequences. > > > > > > Without a maximum. > > > > Yes, if you mean that there is no entry has a greater length (notice, > > by the way, that 'greater' here is just the usual 'greater than' > > relation among natural numbers; i.e., finite) than all other
From: MoeBlee on 3 Nov 2006 15:22 MoeBlee wrote: > S(n+1) = the unique finite sequence f such that length(f) = > length(S(n))+1 and such that f(0)= S(n)(max(dom(S(n)))) + 1 Oops, that should be: S(n+1) = the unique finite sequence f such that: length(f) = length(S(n))+1 and f(0)= S(n)(max(dom(S(n)))) + 1 and f is monotone per the standard ordering of omega In other words, I forget to mention that f is monotone. But that was obvious from my illustration anyway. So, here's the proof again for reference: A sequence is a function such that the domain of the function is an ordinal. A finite sequence has a natural number as its domain. A denumerable (countably infinite) sequence has omega as its domain. An uncountable sequence has an uncountable ordinal as its domain. The entries in a sequence are members of the range of the sequence. Each entry is indexed by a member of the domain. The elements of the sequence are ordered pairs of the form <x y> where x is a member of the domain (which is an ordinal) and y is a member of the range of the sequence (so the y's are the entries). dom(S) = the domain of S. length(S) = dom(S). In my example S is the denumerable sequence recursively defined as follows: S(0) = {<0 0>} S(n+1) = the unique finite sequence f such that: length(f) = length(S(n))+1 and f(0)= S(n)(max(dom(S(n)))) + 1 and f is monotone per the standard ordering of omega So S is a denumerable sequence such that each entry of S is a finite sequence. The diagonal of S = the unique denumerable sequence D such that, for all n in omega, D(n) = S(n)(n). The diagonal of S is a denumerable sequence. Therefore, there exists a denumerable sequence S of finite sequences such that the diagonal of S is denumerable. MoeBlee
From: David Marcus on 3 Nov 2006 15:36 MoeBlee wrote: > mueckenh(a)rz.fh-augsburg.de wrote: > > MoeBlee schrieb: > > > For reference, here is my proof: > > > > > > A sequence is a function such that the domain of the function is an > > > ordinal. A finite sequence has a natural number as its domain. A > > > denumerable (countably infinite) sequence has omega as its domain. An > > > uncountable sequence has an uncountable ordinal as its domain. > > > > > > The entries in a sequence are members of the range of the sequence. > > > Each entry is indexed by a member of the domain. The elements of the > > > sequence are ordered pairs of the form <x y> where x is a member of the > > > > > > domain (which is an ordinal) and y is a member of the range of the > > > sequence (so the y's are the entries). > > > > > > dom(S) = the domain of S. > > > > > > length(S) = dom(S). > > > > > > In my example S is the denumerable sequence recursively defined as > > > follows: > > > > > > S(0) = {<0 0>} > > > > > > S(n+1) = the unique finite sequence f such that length(f) = > > > length(S(n))+1 and such that f(0)= S(n)(max(dom(S(n)))) + 1 > > > > > > So S is a denumerable sequence such that each entry of S is a finite > > > sequence. > > > > > > The diagonal of S = the unique denumerable sequence D such that, for > > > all n in omega, D(n) = S(n)(n). > > > > > > The diagonal of S is a denumerable sequence. > > > > > > Therefore, there exists a denumerable sequence S of finite sequences > > > such that the diagonal of S is denumerable. > > > > Now map it on a line. It is longer than any line entry. > > The real line? 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. -- David Marcus
From: MoeBlee on 3 Nov 2006 16:26
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. MoeBlee |