Cantor Confusion
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From: mueckenh on 2 Nov 2006 03:39 MoeBlee schrieb: > 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 > > 'sequence' and 'list' are synonymous here. a list is an injective sequence > > I'm pefectly happy to use just 'sequence'. Doing so does not at all > harm my argument. > > > > The infinite list is > > > longer than any finite list. > > > > 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" >Even using just the word > 'sequence', my point is correct. > > We have an infinite sequence S of finite sequences. Being an infinite > sequence, the length of S is longer than the length of any finite > sequence. Maybe, if you say so. But omega is not the maximum of all finite sequences. Therefore the width of the list is less than 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. 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. > > 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". > > 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. Without a sequence of infinite length. > > > then the diagonal cannot be infinite (if infinite omega is larger than > > any finite n). > > In this post I PROVED that the diagonal of S is infinite. The diagonal > of S is an infinite set. It is an infinite sequence. It has an infinite > domain. It is an infinite set of ordered paris. (And, by the way, it > has an infinite range.) And you've not shown that that contradicts any > theorem of any Z set theory. That it may contradict your own confused > word jumbles is not of concern to me. You have derived a nice contradiction. The diagonal is an infinite sequence. So the diagonal is longer than any of the finite sequences. But the diagonal consists of elements of the finite sequences. So it cannot be longer than the maximum of the finite sequences. If this maximum does not exist, you cannot take the supremum omega for it, because the supremum is not a member of the sequences and does not supply elements of the diagonal. Regards, WM
From: David Marcus on 2 Nov 2006 04:25 mueckenh(a)rz.fh-augsburg.de wrote: > The diagonal is an infinite sequence. So the diagonal is longer than > any of the finite sequences. But the diagonal consists of elements of > the finite sequences. So it cannot be longer than the maximum of the > finite sequences. If this maximum does not exist, you cannot take the > supremum omega for it, because the supremum is not a member of the > sequences and does not supply elements of the diagonal. Let's try a simpler problem. Consider the following list. 1 1 1 .... In other words, consider the sequence x where x(n) = 1 for n a natural number. How long is this sequence? -- David Marcus
From: Virgil on 2 Nov 2006 04:30 In article <1162456773.687420.34030(a)h48g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > > 'sequence' and 'list' are synonymous here. > > a list is an injective sequence I find my lists are far from injective. > > > > I'm pefectly happy to use just 'sequence'. Doing so does not at all > > harm my argument. > > > > > > The infinite list is > > > > longer than any finite list. > > > > > > 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" You only defined you meaning, but that is not the standard meaning. > > >Even using just the word > > 'sequence', my point is correct. > > > > We have an infinite sequence S of finite sequences. Being an infinite > > sequence, the length of S is longer than the length of any finite > > sequence. > > Maybe, if you say so. But omega is not the maximum of all finite > sequences. Therefore the width of the list is less than omega. You keep making that same mistake over and over again, as if you cannot learn. The width of a list can be equal to its length even when that length is unbounded, merely by having the "width" of each finite entry equal to its position in the list. Please do not make that same silly mistake again. > > You may also prove that the maximum of numbers less than 5 is 5. > Nevertheless it is false. It is only in such self-contradictory systems as WM moves in that one can "prove" falsehoods. In our systems that does not happen. > > 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. WM made that silly mistake again already. WM conflates the finiteness of each single line with the boundedness of an infinite set of finite lines. The former does not imply the latter as I have shown WM many times. > > This simple truth should convince you that ZFC is not acceptable. It is not acceptable to people whose logical skills are not up to its demands. But it works fine for those of is who have those skills. > > > > In this post I PROVED that the diagonal of S is infinite. The diagonal > > of S is an infinite set. It is an infinite sequence. It has an infinite > > domain. It is an infinite set of ordered paris. (And, by the way, it > > has an infinite range.) And you've not shown that that contradicts any > > theorem of any Z set theory. That it may contradict your own confused > > word jumbles is not of concern to me. > > You have derived a nice contradiction. All that it contradicts are WM's claims that he hs found any contradictions WITHIN ZF. Any contradictions are between ZF and WM's assumptions outside of ZF. > > The diagonal is an infinite sequence. So the diagonal is longer than > any of the finite sequences. But since those finite sequences can , for every n in N, includes a sequence longer than n characters, the diagonal can contain a position for every n in N. That makes the third time that WM has made that same silly mistake in just one post. > But the diagonal consists of elements of > the finite sequences. Let the nth sequence be of length > n and the diagonal is immediately unending. > So it cannot be longer than the maximum of the > finite sequences. Which need not exist. That's 4 times. >If this maximum does not exist, you cannot take the > supremum omega for it, But you can have a diagonal with an nth term for every n in N, provided the nth line contains an nth term for every n in N. Imagine an infinite triangle with the diagonal going from upper left to lower right as the lines go down and get longer and longer witthout end.
From: mueckenh on 2 Nov 2006 07:26 Dik T. Winter schrieb: > In article <1162405520.008395.100850(a)e64g2000cwd.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > ... > > > Finally you give a definition. Why did it take so long? > > > > I thought that this was so clear that no explanation was required. > > I asked you for a definition quite a few times. I would have thought > that that was enough indication that it was not clear at all. > > > > > > And in mathematics 1/oo is *not* defined. > > > > > > > > Not in mathematics. But in a theory which assumes omega to be a whole > > > > number. > > > > > > Which theory? > > > > Set theory. > > Cantor invented omega and defined omega as a whole number. > > Who changed this standard meaning? > > Why do you think this meaning was changed? > > When do you think the contrary meaning became standard? > > What is the contrary meaning? > > Do you agree that A n: n < omega is incorrect? > > If not, why do you complain about on-standard meaning on Cantor's > > definition? > > A nice rant. Why don't you answer? > Where did Cantor define 1/oo? A quote might be > appropriate. see below > > > Where in the above quote is 1/oo defined? > > > > It is defined that omega (which Cantor used later instead of oo) is a > > number larger than any natural number n. Omega is the limit ordinal > > number. Therefore 1/omega must be a number smaller than every fraction > > 1/n. > > Why? As long as 1/omega is not defined you can not talk about it. You > simply assume that 1/omega is a number. But that is not the case, it > is not defined. If omega is a number > n, then 1/omega is a number < 1/n. For all n e N. > > > > > Therefore we have there limit ordinal numbers? > > > > > > Yes. So what? Limits are in general not defined. > > > > In particular the limit of all segments of N is not defined. But set > > theory does it. > > No. And if so, show me where. Es ist sogar erlaubt, sich die neugeschaffene Zahl omega als Grenze zu denken, welcher die Zahlen n zustreben, wenn darunter nichts anderes verstanden wird, als daß omega die erste ganze Zahl sein soll, welche auf alle Zahlen n folgt, d. h. größer zu nennen ist als jede der Zahlen n. (p. 195) This definition has been conserved up to our days: Limesordinalzahl or limit ordinal number. > > > Nonsense. Suppose I define an ordering relation on sets. How can I apply > > > that knowledge to numbers? Defining something in one context does not > > > make it immediately applicable in another context. > > > > Everything is a set (in ZF). Numbers are sets too. > > In the Von Neumann model of ZF, I would think. In ZF. Karel Hrbacek and Thomas Jech: "Introduction to set theory" Marcel Dekker Inc., New York, 1984, 2nd edition, p. 2: "So the only objects with which we are concerned from now on are sets." Please revise: "Nonsense. Suppose I define an ordering relation on sets. How can I apply that knowledge to numbers?" Regards, WM
From: mueckenh on 2 Nov 2006 07:28
Dik T. Winter schrieb: > In article <1162405329.073198.286680(a)h48g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > ... > > > > Irrational numbers have no last digit. Therefore, with a sequence of > > > > digits like the diagonal number is, one can never have a completed > > > > number but only come as close as possible to any number --- or avoid to > > > > do so. > > > > The problem is that: > > For the diagonal number of Cantor's list it is not sufficient to come > > arbitrarily close to a number which is different from any list number > > --- or avoid to do so. . > > You lost me here. Numbers do not come arbitrarily close to each other. > Numbers are fixed entities. Sequences can come arbitrarily close to > each other. Irrational numbers are sequences. > > > > > Not so. Of course we talk about a fixed base like 10. > > > > > > Ah. In that case one or two, depending on the number involved. But > > > the diagonal obviously depends on the actual representative chosen. > > > > So we have no arbitrary choice but , in case of irrational numbers, > > exactly one representation. And this representation is *the limit* of > > all the sequences of the due equivalence class. > > You are pretty wrong here. There are a lot of rational numbers for which > there is only one representative in decimals. Did I dispute that? > And, representatives are > *not* limits. When considering the equivalence classes, most sequences > have one limit: the equivalence class it is sitting in. > > I think that you are still thinking that *some* represenation defines > a real number; but that is not the case. In Cantor's list there are those unique representations required. > That is especially not the > case when you consider only representations to some integral base. > There are other methods to define numbers. You do not like to call > them numbers, but ideas. But you can not prevent me to call something > like sqrt(2) a (real) number. And in common mathematics that is just > what it is. > > By definition, every sequence (use any definition, I know Cantor, > Dedekind, Baudet and Weierstrass, they all lead to the same): > {sum{k = 1...n} a_k/10^k} > is a representative of a "number". It is just a sequence of rationals. Therefore I do not understand why you say "Numbers are fixed entities". They are merely defined by sequences. Regards, WM |