Cantor Confusion
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From: Andy Smith on 5 Jan 2007 07:19 I forgot to say, regarding the existance of an actually infinite integer, if you consider the sequence 1 + 1/2 + 1/4 + 1/8 ... and label each element of the series 1,2,3, ... etc then when you look at the interval [0,2] representing the sum of the series I can point to the point at 2.00... and say with confidence that the label of the series member at that point is an actually infinite number?
From: Dik T. Winter on 5 Jan 2007 18:10 In article <7079661.1168035187072.JavaMail.jakarta(a)nitrogen.mathforum.org> Andy Smith <andy(a)phoenixsystems.co.uk> writes: .... > Yes, I was talking about axioms of arithmetic rather than Peano - > e.g. (according to Cantor) infinity + 1 is a new cardinal number but > 1 + infinity is infinity. This is for ordinal numbers, not for cardinal numbers. 1 + w = w != w + 1 and 1 + aleph_0 = aleph_0 = aleph_0 + 1. But what "axioms of arithmetic" are you talking about? > >Wrong. It's not a matter of whether the counting is complete. The > >problem is that when the counting is complete, there is no integer that > >maps to 1/3 in your scheme. By induction, each natural number maps to a > >rational that can be represented in a finite number of digits. > > Unless you concede that an actually infinite natural number exists. But (as Dave Seaman wrote), it can be proven that all natural numbers are finite. > Back to the flipped set of {N}, the resulting finished set > in [0,1] is bounded with one element at 0.000... > and one < 1.000... No. It is bounded by 0 and 1. And 0 is in the set but 1 is not in the set. > If we consider an element with N successive binary 1's > there will be an element with N+1 successive 1's, and this > is true for any N, so by induction there must be a number .1111.... ? You are again assuming that what goes on in the finite case also must go on in the infinite case. There is no such number in the set. > >I thought I did comment on that. The process produces an infinite digit > >string, which does not represent a member of N. > > But {N} represents the set of ALL natural numbers, > so there cannot be a member outside of it - so the > Cantor diagonalisation argument is fallacious (which was > rather the point) ? Yes, so the infinite digit string is not a natural number, and so the diagonal obtained does not represent a natural number. And so the Cantor argument for the natural numbers is fallacious. The same applies when you try to do it for the rational numbers in [0, 1]. The diagonalisation gives a digit string that is not present in the list, but you can not show that that digit string represents a rational number. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dave Seaman on 5 Jan 2007 19:04 On Fri, 05 Jan 2007 17:12:36 EST, Andy Smith wrote: > Thanks for the reponse. >>The Peano axioms are what define the natural numbers. If you don't >>accept those axioms, then for you the natural numbers do not exist. > I>f that's the case, then why are you arguing about the Cantor proof of >>the uncountability of the reals? After all, neither of the sets N or R >>exists for you and therefore the entire question is vacuous. > Yes, I was talking about axioms of arithmetic rather than Peano - e.g. (according to Cantor) infinity + 1 is a new cardinal number but 1 + infinity is infinity. You are mixing cardinals and ordinals here. If x is any infinite *ordinal*, then it is correct that 1+x = x < x+1, using ordinal addition. However, *cardinal* addition gives x = x+1 = 1+x whenever x is an infinite cardinal. And the reason this is true is not there are axioms that say so. There are no axioms in ZFC that actually mention ordinals, cardinals, or addition. Those are defined concepts, and the properties I mentioned are consequences of those definitions. A definition, unlike an axiom, does not entail an assumption. It's merely a convenience of language. >>Wrong. It's not a matter of whether the counting is complete. The >>problem is that when the counting is complete, there is no integer that >>maps to 1/3 in your scheme. By induction, each natural number maps to a >>rational that can be represented in a finite number of digits. > Unless you concede that an actually infinite natural number exists. By definition, a set is called "finite" if its cardinality is a natural number. Any set that is not finite is called "infinite". An "infinite natural number" is therefore a contradiction in terms. >>It seems that extrapolating without justification from the finite case to >>the infinite case is perfectly acceptable, as long as you are the one who >>is doing it. Extrapolating *with* justification (by appealing to the >>axioms) is *not* justified, in your view, if it is someone else's >>argument. > Fair enough, I'm not trying to get into a fight - just > trying to sort things out for myself. > Back to the flipped set of {N}, the resulting finished set > in [0,1] is bounded with one element at 0.000... > and one < 1.000... > If we consider an element with N successive binary 1's > there will be an element with N+1 successive 1's, and this is true for any N, so by induction there must be a number .1111.... ? By induction, your list contains a number with n successive 1's for each natural number n. However, the number .111... is not of that form, because the number of successive 1's is not a natural number. Therefore, you can't conclude by induction that .111... is in the list. >>> You didn't comment on the application of Cantor's diagonalisation argument >>> to {N} - again, with the inverted system above, that >>> would seem to suggest, if the argument is valid (and >>> in this case, the infinite matrix is definitely diagonal) >>> that there are integers not present in {N}. >>I thought I did comment on that. The process produces an infinite digit >>string, which does not represent a member of N. > But {N} represents the set of ALL natural numbers, No, N is the set of all natural numbers, and is therefore an infinite set; {N} is a finite set, having just one element, namely the set N. > so there cannot be a member outside of it - so the > Cantor diagonalisation argument is fallacious (which was > rather the point) ? No, the point is that when you attempt to apply the diagonal argument to the identity map on N, you get something that is not a natural number. The mere fact that some infinite digit string does not appear in the list is not a contradiction, because technically the list wasn't claimed to contain any digit strings. It contains only natural numbers, and the thing you produced isn't one. Contrast that with the situation where the argument is applied to a list of real numbers. In that case, the diagonal argument produces a real number that is not in the list. That's the difference; in one case you get something of the required kind, and in the other you don't. -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. <http://www.mumia2000.org/>
From: Dave Seaman on 5 Jan 2007 19:18 On Fri, 05 Jan 2007 17:19:56 EST, Andy Smith wrote: > I forgot to say, regarding the existance of an > actually infinite integer, if you consider the > sequence > 1 + 1/2 + 1/4 + 1/8 ... > and label each element of the series 1,2,3, ... etc > then when you look at the interval [0,2] representing > the sum of the series I can point to the point at 2.00... > and say with confidence that the label of the series > member at that point is an actually infinite number? The n-th term of the series is a_n = 1/2^n and the n-th partial sum is S_n = sum_{k=1}^n a_k = 2 - 1/2^n for each natural number n. The sum of the series is S = lim_{n->oo} S_n = 2. Although the symbol "oo" appears in that last statement, it goes away completely if we replace the statement by its definition, which is: For each epsilon > 0, there exists N > 0 such that | S_n - 2 | < epsilon for every n > N. There are no infinite integers here. The numbers N and n are finite, and epsilon is a real number, not an integer. -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. <http://www.mumia2000.org/>
From: Andy Smith on 5 Jan 2007 09:28
OK, thanks. I will go back to physics. I do now see what you mean, but it still strikes me as a bit Humpty Dumpty - " aword means what I see it means, nothing more and nothing less". So you can't have an actually infinite natural number, because by definition all natural numbers are finite, although the set of all natural numbers is infinite. So numbers such as ..1111 are not realisable, and the countability of the reals is fundamentally different from that of the integers. I still wonder what the label attached to the point exactly at 2.000... on the sum 1 + 1/2 + 1/4 ... is going to read though. Regards & thanks |