Cantor Confusion
in [Math]
|
Prev: Pi berechnen: Ramanujan oder BBP
Next: Group Theory
From: Andy Smith on 5 Jan 2007 02:04 On Fri, 05 Jan 2007 10:37:01 EST, Andy Smith wrote: >>>>> But anyway, what integer does your mapping map 1/3 (a real in [0,1]) >>>>> to? >>>> It would map to ..0101010 ! >>>That's not an integer >> It is the natural successor to ..0101001 ! >That's not an integer either. > But I agree it is hard to see how one can count up to > ..0101010 from 0. As I said in my earlier post, I am > suspicious of anything to do with infinite sets. But what >is wrong with my systematic technique for counting the >reals, and how does it differ substantively from mapping >the rationals onto the natural numbers? >I thought that was already answered. If your method >actually works, then >it should map some integer to 1/3. Which one is it? >> I should say that I am not sufficently barking to think >>that Cantor is wrong, just trying to get my head >> straight, and informed advice is welcome ... >My advice is, once an error in your reasoning has been pointed out, don't >simply ignore it and continue asking where the error is. Thank you for your (disdainful) advice. However, you didn't point out where the error was, you just baldly stated that ..0101010 was not an integer. You can count up to ..0101010 in a (countably) infinite number of steps M by the iteration: N0 = 2 { Nm = N(m-1)+ N(m-1)+2 } Do the first iteration in 1 second, the second in 1/2 second, third in 1/4 second etc. and in precisely 2 seconds you will have generated ..0101010 Like I said, I am suspicious of arguments involving infinite sets.
From: Dave Seaman on 5 Jan 2007 12:16 On Fri, 05 Jan 2007 12:04:03 EST, Andy Smith wrote: > On Fri, 05 Jan 2007 10:37:01 EST, Andy Smith wrote: >>>>>> But anyway, what integer does your mapping map 1/3 (a real in [0,1]) >>>>>> to? >>>>> It would map to ..0101010 ! >>>>That's not an integer >>> It is the natural successor to ..0101001 ! >>That's not an integer either. >> But I agree it is hard to see how one can count up to >> ..0101010 from 0. As I said in my earlier post, I am >> suspicious of anything to do with infinite sets. But what >>is wrong with my systematic technique for counting the >>reals, and how does it differ substantively from mapping >>the rationals onto the natural numbers? >>I thought that was already answered. If your method >actually works, then >>it should map some integer to 1/3. Which one is it? >>> I should say that I am not sufficently barking to think >>>that Cantor is wrong, just trying to get my head >>> straight, and informed advice is welcome ... >>My advice is, once an error in your reasoning has been pointed out, don't >>simply ignore it and continue asking where the error is. > Thank you for your (disdainful) advice. However, you didn't point out > where the error was, you just baldly stated that ..0101010 was > not an integer. You can count up to ..0101010 in a (countably) infinite number of steps M by the iteration: Are you familiar with the Peano axioms? Do you know what mathematical induction is? It's as easy proof by induction that every natural number can be represented by a finite number of digits. > N0 = 2 > { > Nm = N(m-1)+ N(m-1)+2 > } Each number that you reach by that iteration has a finite number of digits. > Do the first iteration in 1 second, the second in 1/2 second, > third in 1/4 second etc. and in precisely 2 seconds you will have generated > .0101010 On which step does it appear? > Like I said, I am suspicious of arguments involving infinite sets. That's why we have axioms. -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. <http://www.mumia2000.org/>
From: mueckenh on 5 Jan 2007 13:12 cbrown(a)cbrownsystems.com schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > cbrown(a)cbrownsystems.com schrieb: > > > > > You say that 2 > 3/2. What is an example of a surjection from the > > > rational number 2 onto the rational number 3/2? > > > > There is an injection from 3 into 4 which is not a surjection (namely > > 3/2 and 4/2). > > > > Truly amazing. > > I can only assume from your response that you have forgotten (or > perhaps just don't care) what a surjection is. Read carefully. I wrote "injection". > > A surjection is a function which maps the /members/ of one /set/ onto > the /members/ of another /set/; so that for every /member/ of the > latter /set/, there is a /member/ of the former /set/ which maps to > that /member/. > > What are the /members/ of the set 2 that you mean to act as the domain > of the surjection? Or is the set 2 the empty set? Let m, n be natural numbers. If n =/= m, then there is no one-to-one mapping of n onto m. If If n > m, then there is no injective mapping of n onto m. If If n < m, then there is no surjective mapping of n onto m. (In future, please look up these things in a book on set theory.) > I don't mean some finite collection of /subsets/ of the set 2, but the > /members/ of the set 2. You agree that there is in general a difference > between "x is a subset of Y" and "x is a member of Y", yes? n = {0,1,2,...,n-1} The members of the set 2 = {0, 1} are 0 and 1. > > You seem to claim that 1(/2) is a /member/ of the set 2 - at least when > you claim a surjection onto the set 3/2. 1 is a member of 2, yes. Each natural number is the set of smaller natural numbers, i.e., n = {m eps N | m < n} > But how then can 1(/2) cease to be a member of the set 2 when you claim > a surjection f from the set 7/3 onto the set 2? 1 does not cease to be a member of 2. The fact that 7/3 > 2 is proved by a surjection from 7 onto 6, because 2 = 6/3. > > > > Why is there no > > > surjection from the rational number 3/2 onto the rational number 2? > > > > There is no surjection from 3(/2) into 4(/2). > > But there is a surjection of 6(/4) onto 4(/2). If you compare halves with quarters, you will not get correct results. > 6 is greater than 4, is it not? Of course, take for instance f(1) = 1, f(2) = 2, f(3) = 3, f(4) = 4 = f(5) = f(6). > > > What is the basis for your claim that there is a surjection from the > > > rational 6/4 onto the rational 3/2, but no surjection from the rational > > > 4/6 onto the rational 3/2? > > > > There is a bijection from 6 onto 6 (namely 6/4 and 3*2/2*2 = 6/4). > > Therefore, you agree that 1(/4) is a member of 3/2. So as noted above, > 6(/4) > 4(/2), because there is a surjection from 6 onto 4. See above. > > > > > > > You of course understand what is meant by a surjection from a set A > > > onto a set B, yes? > > > > Of course. > > It doesn't appear so to me. I think perhaps at some point you did, but > perhaps not. Perhaps there is a gap in your understanding of numbers and their representation in set theory? Regards, WM
From: Andy Smith on 5 Jan 2007 05:04 >Are you familiar with the Peano axioms? Do you know what mathematical >induction is? Yes, and yes, even though my training is in physics. >It's as easy proof by induction that every natural number can be represented by >a finite number of digits. Yes, but I am suspicious of rules of inference and axioms when applied to infinite sets. Many things don't work or give indeterminate answers when applied to infinite series. >On which step does it appear? The last one! (NB, iteration formula was wrong, should have had Nm = N(m-1)+ N(m-1) + N(m-1)+ N(m-1)+ 2) >That's why we have axioms. Yes, but I am not wholly convinced about their validity to infinite sets or to the application of otherwise irrefutable rules of inference in these circumstances. To answer my own question, the difference between enumerating the rationals and the reals is one of process. The standard scheme for mapping the rationals means that one can pick any rational whatever and easily compute its corresponding mapping integer. In contrast, with my suggested scheme for mapping the reals, it is easy to pick any irrational number and demand its indes - and the answer is only available when all the counting is complete. So, fair enough. But I would turn around the question. Suppose we have the set of all integers {N} as a completed thing. It is easy enough to visualise that - e.g. label all the terms in the series 1 + 1/2 + 1/4 + 1/8 ..., and consider the interval [0,2] which then contains, in a bag as it were, all of them. View them as binary numbers, and reflect {N} about the binary point 010 becomes .010... etc. What is the largest fraction present in the reflected set? Is it .111111... ? Presumabaly you would say not. You would say, it can be as close to 111111... as you like, but is always epilson away from it. But, this is a complete set - there is no process. So, if 111111... is not the largest member, what is it? 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}. But I'm a physicist.
From: Dave Seaman on 5 Jan 2007 15:58
On Fri, 05 Jan 2007 15:04:01 EST, Andy Smith wrote: >>Are you familiar with the Peano axioms? Do you know what mathematical >>induction is? > Yes, and yes, even though my training is in physics. >>It's as easy proof by induction that every natural number can be represented by >>a finite number of digits. > Yes, but I am suspicious of rules of inference and axioms > when applied to infinite sets. Many things don't work or > give indeterminate answers when applied to infinite > series. That's an excellent point. Let's keep that in mind as this discussion proceeds and see which of us is prone to claiming without justification that properties of finite sets must also hold for infinite sets. The induction proof in question does not even refer to infinite sets, at least not directly. The conclusion of the proof is that each natural number can be represented by a finite number of digits. Where do you see any reference to an infinite set in that conclusion? Proposition. Each positive integer can be represented by a finite number of digits. Proof. We show by induction that each n > 0 can be represented by at most n digits. Step 1. The number 1 can be represented by one digit. Step 2 (the induction step). Suppose a number n can be represented by at most n digits. Then n+1 can likewise be represented by n digits unless a carry occurs from the highest-order digit, in which case n+1 is a power of 2 and requires exactly one more digit than n. In either case, n+1 digits are certainly sufficient. Conclusion. By induction, the proposition holds for each positive integer. Where have I mentioned an infinite set in this proof? >>On which step does it appear? > The last one! (NB, iteration formula was wrong, should have had Nm = N(m-1)+ N(m-1) + N(m-1)+ N(m-1)+ 2) Oh, I see. Each finite set of integers must have a last member, and therefore the set of all integers must likewise have a last member. Is that your reasoning here? What was that fallacy that you warned me about earlier? >>That's why we have axioms. > Yes, but I am not wholly convinced about their validity to infinite sets > or to the application of otherwise irrefutable rules of > inference in these circumstances. 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. If 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. > To answer my own question, the difference between enumerating the > rationals and the reals is one of process. The standard > scheme for mapping the rationals means that one can pick > any rational whatever and easily compute its corresponding > mapping integer. In contrast, with my suggested scheme for > mapping the reals, it is easy to pick any irrational > number and demand its indes - and the answer is only > available when all the counting is complete. So, fair > enough. 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. > But I would turn around the question. Suppose we have > the set of all integers {N} as a completed thing. It is easy > enough to visualise that - e.g. label all the terms > in the series 1 + 1/2 + 1/4 + 1/8 ..., and consider the > interval [0,2] which then contains, in a bag as it were, all > of them. > View them as binary numbers, and reflect {N} about the > binary point 010 becomes .010... etc. > What is the largest fraction present in the reflected > set? Is it .111111... ? That number is not present. Which place in the list would it occupy? In fact, there is no largest number in the list. > Presumabaly you would say not. You would say, it can be as close to > 111111... as you like, but is always epilson away from it. > But, this is a complete set - there is no process. So, if > 111111... is not the largest member, what is it? Why do you assume that every set has to have a largest member? Are you assuming here that since finite sets of rationals must have a largest member, then the same must surely hold for infinite sets of rationals? 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. > 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 I'm a physicist. -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. <http://www.mumia2000.org/> |