Cantor Confusion
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From: mueckenh on 1 Nov 2006 13:09 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. 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. > 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. > > > > > > Greater than any integer is no integer but only omega (and > > its successors). > > > > Call the number of letters anything you want. If it exists it must be > greater > than any integer. Correct. If it exists, it must be infinite (or omega). Only then the number of lines can be omega. But we have excluded that the number of letters is omega. > 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. 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. Regards, WM
From: mueckenh on 1 Nov 2006 13:12 David Marcus schrieb: > > I know of a really good logician who told me that he is unable to > > translate my proof into ZFC. > > Hardly surprising. If he is truly a good logician, then he didn't want > to tell you that you were talking gibberish. No, he (and he is not the only one to do so) does understand the arguing but does not know how to translate it into ZF language. That is all. > > Fine. Then stop claiming that you can prove ZFC is inconsistent. If you > don't like ZFC, that's your concern. But, that is completely different > from saying that you have a proof within ZFC of a contradiction. I have a proof which shows that the real numbers have less elements than a countable set. That is all. ZFC and what there can be formalized does not bother me at all. > > > > OK, let's use our logic (or standard logic or ZFC). The list has > > > infinitely many lines and columns, so the number of diagonal elements is > > > infinite. Your statement that the number of diagonal elements is finite > > > is wrong. You've jumped from "each entry is finite" to "the number of > > > columns is finite". > > > > Please give one letter which requires an actually infinite number omega > > of columns. > > If you can't, please stop the nonsense talk about an infinite number of > > finite numbers. > > You seem to have skipped right over my statement that I was using > standard logic (or ZFC). In standard logic (or ZFC), "the number of > columns is finite" is not the same as "each entry in the list has a > finite number of columns". It seems simply bizarre that you would mix > these up. It seems that you do not understand what logic is. That branch which you call "standard logic" is some kind of theology which is not useful to promote clear thinking. 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. > > > > If the list consists of finite sequences, then the diaogonal is a > > > > finite sequence too. Because it cannot be broader than the list > > > > > In ZFC or in some other system? > > > > Everywhere. > > How would you know since you admitted above that your proofs don't work > in ZFC? Why make claims that you then immediately contradict? I know that ZFC claims there are more reals than any countable set has elements. Hence my proof contradicts ZFC. Regards, WM
From: mueckenh on 1 Nov 2006 13:19 Virgil schrieb: > So that they can waste it listening to your foolishness instead? Sorry, Virgil, I can well understand what you must think now, after having studied set theory for many years probably. You cannot agree that it is nonsense. Even if you wanted. On the other hand the proof that I am right is given by the finity of the universe and the mathematics done inside this universe. So every infinity is nonsense, with or without axiom. There can be no doubt. Therefore you cannot expect that I hold back my finding and help many young students to avoid this trap. Regards, WM > A list with endlessly many strings of endlessly increasing lengths > trivially and obviously does not contradict anything except WMspeak. Do you believe that A n: n < omega holds? Take: n as the columns. 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. > Give an example of a digit that does not belong to an infinite > sequence, WM. If you cannot do so, then every digit belongs > to am infinite sequence. Here are many digits which do not belong to infinite sequences: 1,2,3,... > In the infinite binary tree, uncountably many paths will be made out of > countably many edges. So that there are some paths without any edge? > No number can be "put in trichotomy" with all other numbers. What is X - Y, if X = [Pi*10^10^100] and Y is X with the last decimal digit exchanged by 7? > In an infinite binary tree, there are infinitely many paths through each > edge and through each node of that tree. So how does WM isolate one of > those paths from all the others to be related to only that one edge and > no other edges? In mathematics, there are infinitely many irrational numbers beginning with 2,718281828... How do you isolate one of them? Regards, WM
From: mueckenh on 1 Nov 2006 13:22 Dik T. Winter schrieb: > In article <1162299756.756951.78990(a)b28g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > > Dik T. Winter schrieb: > > > > > > For the diagonal number of Cantor's list it is nnot sufficient to come > > > > arbitrarily close to a number which is different from any list number. > > > > > > Sorry, but numbers are fixed and not variable. The sentence "number ... > > > to come arbitrarily close" is nonsense. Numbers do not come arbitrarily > > > close to each other, it is sequences that can come arbitrarily close to > > > each other. > > > > 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. . > > Yes. So what? The sequence comes as close as one wishes to some other > sequence. What is the problem with that? > > > > > > Representatives of the equivalence classes that actually are the real > > > > > numbers. > > > > > > > > How many different representatives can be chosen in Cantor's list for > > > > one equivalence class? > > > > > > As many as you want. > > > > 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. Regards, WM
From: mueckenh on 1 Nov 2006 13:25
Dik T. Winter schrieb: > In article <1162299524.423928.41670(a)k70g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > > > > The definition is sufficiently special (=3D non-general) to determine > > > > that lim [n-->oo] {1,2,3,...,n} =3D N is correct. > > > > > > Yes, of course, because that was the definition you provided. So because > > > it is so defined, the definition is correct. > > > > > > > > > > > The operator "lim [n-->oo]" defines N. In your example lim {n -> oo} > > > > {-1, 0, 1, ..., n} we have N too but in addition the numbers -1 and 0. > > > > > > "lim {n -> oo}" is an operator that works on sequences, apparently. And > > > you have defined that operator for precisely one sequence of sets. > > > > It is defined for infinitely many sets of integers. > > (Infinitely many sequences of sets.) > > > lim [n-->oo] {-k,-k+1,..., 0, 1,2,3,...,n} = {-k,-k+1, ...,0} u N for > > every k e N. > > lim [n-->oo] {k, k+1, k+2,...,n} = N \ {1,2,3,...,k-1} for every k e N. > > Finally you give a definition. Why did it take so long? I thought that this was so clear that no explanation was required. > > > > > > > case, we have 1/n < epsilon for every positive epsilon and we may > > > > > > *define or put* > > > > > > lim [n-->oo] 1/n = 0. > > > > > > In the second case we have without further ado > > > > > > lim [n-->oo] 1/n = 0. > > > > > > That is the difference between potential and actual infinity. > > > > > > > > > > Well, in mathematics the first form is valid, the second is not > > > > > valid > > > > > > > > If actual infinity is assumed to exist, then the second case is valid. > > > > > > No. The second is never valid, because the limit notation is defined > > > in such a way that the limit point is *never* reached (all definitions > > > of limits in mathematics are formulated in such a way that the limit > > > point itself, or possible function values at the limit point, are not > > > used). 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? > > > Die Anzahl einer unendlichen Menge [ist] eine durch das Gesetz der > > Zählung mitbestimmte unendliche ganze Zahl. (G. Cantor, Collected > > Works p. 174) > > ... kann also omega sowohl als eine gerade, wie als eine ungerade Zahl > > aufgefaßt werden. (G. Cantor, Collected Works p. 178) > > 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. > > > > > > But now you are talking in analysis where limits are properly defined > > > > > (because there is a topology and a metric). > > > > > > > > Set theory was discovered and defined as being based on analysis. Cp. > > > > Cantor's first proof. > > > > > > Yes, so what? Set theory is more basic than analysis. In set theory > > > limits are in general not defined. > > > > 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. > Especially limits of > sequences of sets are in general not defined. > > > > > So apply this knowledge to the case of the balls and the vase. > > > > > > it can not be applied to it. > > > > What a lucky accident! > > Don't you see that the whole aim of Newspeak is to narrow the range of > > thought? In the end we shall make thoughtcrime literally impossible, > > because there will be no words in which to express it. (George Orwell > > in "1984") > > 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. Regards, WM |