Cantor Confusion
in [Math]
|
From: WM on 21 May 2007 08:30 On 21 Mai, 04:01, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > > No. Then sum{i = 1 to oo} i could also yield another result. But > > > > that is certaily false. It can neither be finite nor can it be > > > > uncountable. The same holds for sum{i = 1 to oo} 1 = aleph_0. > > > > > > Why? That depends entirely on how you define it. I have looked, what you > > > are missing is that in the book cardinal arithmetic is defined, and that is > > > not the same as standard arithmetic, and with cardinal arithmetic the use > > > of 'oo' is extremely misleading, and actually also your notation. Better > > > is (and I think the writers of the book use that notation): > > > sum{i in N} i = aleph_0, > > > which is quite different because it does not suggest sequencing. > > > > The writers sum the *sequence* 1, 2, 3, ... with a definite and unique > > result (no definition could lead to another result. Your claim that > > this was not done or was impossible is wrong. > > No. They do not state it is a sequence. They write 1 + 2 + 3 + .... That is the sum of a sequence, i.e., a series. > But definitions *can* lead to > another result. For instance? > It depends on the definitions. The sum of the set of > all natural numbers is defined *by them*. Other definitions can give > other results, because the result is not defined without their definitions. Which result do you think could be possible? Note, 2*aleph_0 = aleph_0. Of course you are wrong. The only result in set theory is aleph_0. The only result in mathematics is infinite. > > > > And indeed > > > also sum{i in S} i is defined when S is an arbitrary set of cardinal > > > numbers. But indeed, the book *gives* some additional definitions to > > > get such results. In addition, it is also stated that those additional > > > definitions require the axiom of choice. Did you really read it? In > > > standard arithmetic sum{i = 1..oo} i is not defined. > > > > Here the sum *is* defined. You said it was not, but it is. > > It is not defined in standard arithmetic. It is defined in that book using > some set theoretical definitions, that are not common. You must not think that only such things are common which you happen to know. > > > > In cardinal arithmetic division (and subtraction, I think) are not defined. > > > And I clearly stated that the formula was about natural numbers, so I think > > > I implied sufficiently that I meant ordinary arithmetic. > > > > I said that the sequence of natural numbers can be summed up. This is > > true. > > Depends. Did you miss the point that the axiom of choice was required? It is required for many tasks. > But > *their* sum is not a sum of natural numbers, but of cardinal numbers, and > it is not a sequence. Their notation shows the sum of the sequence of natural numbers 1 + 2 + 3 + ... + n + ... (n in N) They write, it is easy to see that this sum is equal to aleph_0. > > > > But whatever the > > > case when I use their definition of the limit of sequences of ordinal > > > numbers, I can define: > > > sum{i = 1..omega} i = lim{n -> omega} sum{1 = 1..n} i = > > > lim{n -> omega} (n + 1) * n / 2 = omega. > > > > omega and aleph are not distinct in this case. Ordinal and cardinal > > summation are not different in this case. Hrbacek and Jech show that 1 > > + 2 + 3 + ... = aleph_0 (you can also call it omega). > > They do only show it when you use *their* definitions (and when you > consider cardinal arithmetic). It is a common and natural definition. That is proven by the fact that there is no other result possible. > > > And in fact it > > cannot be else. But the formula which is true for all natural numbers > > and for the sum of every initial segment of natural numbers is not > > true for every initial segment of natural numbers. > > Oh. I can define division on cardinals such that > (aleph_0 + 1) * aleph_0 / 2 = aleph_0, > which would perfectly fit. Or do you have some other ideas why such a > definition would not be possible? The only problem is that such definitions > are not common. The second problem is that according to he proof f the formula by complete induction, the number appearing on the right-hand side of the formula is in N. Your definition would prove aleph_0 in N. And n fact, if there were all the natural numbers actually existing, then an infinite one must be among them. Regards, WM
From: WM on 21 May 2007 08:36 On 21 Mai, 04:10, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1179652465.304840.3...(a)e65g2000hsc.googlegroups.com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > On 19 Mai, 04:50, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > > > > The set is finitely defined. Not all lucky numbers can get > > > > > > > > finitely defined. > > > > > > > > > > > > > > In general a recursive definition is not considered a finite > > > > > > > definition. > > > > > > > > > > > > Every definition which ends after finitely many words is a finite > > > > > > definition. > > > > > > > > > > Ah, so you disagree with common mathematical terminology. > > > > > > > > No. A finite definition means a definition by a finite number of > > > > words. Every other definition is nonsense. I agree with the common > > > > mathematical definition which implies that there are only finitely > > > > many definitions. > > > > > > Which implication? Again contradicting the axiom of infinity? > > > > Pardon, I meant "countably many definitions". This is implied by the > > finity of every definition. If there were infinite definitions, then > > there were uncountably many definitions. > > Yup. So the question remains: "you disagree with common mathematical > terminology?" No, it is obvious is that a finite definition means a definition which is given by a finite number of words. Infinite definitions are nonsense, according to Cantor. > But whatever, you can not apply the diagonal argument > to "finitely defined numbers". You can supply a list of finite definitions, > but not all of them define a number. And for the diagonal argument a list > of numbers is needed. Correct. Therefore it is clear that the finitely defined numbers cannot be put in a list, i.e., they cannot be in bijection wit N. My argument should only show that there are countable sets (like the paths in the tree) which cannot be put in a bijection with N. Reards, WM
From: WM on 21 May 2007 08:42 On 21 Mai, 05:00, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1179663816.282116.232...(a)h2g2000hsg.googlegroups.com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > On 19 Mai, 04:20, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > ... > > > But, again, that is *not* the diagonal proof of Cantor. > > > > The following wm-proof certainly even in your opinion belongs to the > > diagonal proofs considered by Cantor: > > > > 0) mmm... > > 1) wmmm... > > 2) wwmmm... > > 3) wwwmmm... > > 4) wwwwmmm... > > ... .......... > > > > And if the list can be considered as a completed entity, then there > > must be all natural numbers in the first column. And there must be a > > line with all natural indexes mapped on w's, i.e., no w must be > > missing (as would be the case if one m was present). > > Why? Show a proof. You are again assuming that there is a last natural > number. If the diagonal is complete, then the list must also be complete, because the diagonal cannot be longer or broader than the list. If the list is complete, then there must be a line with only w's, because otherwise the list is not complete. It is easy to see that for every number of w's in the diagonal, there is a line with as much (well one less) w's. This is also shown by considering the limit for n --> oo (line with n w's - diagonal) = 0 Regards, WM
From: WM on 21 May 2007 10:11 On 21 Mai, 13:47, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > WM says... > > > > > > >On 20 Mai, 17:57, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > >> WM says... > > >> >> > It is. Set theory is simply biased. Consider the list > > >> >> > 0.666... > >> >> > 0.3666... > >> >> > 0.33666... > >> >> > 0.333666... > >> >> > ... > > >> >> > If the diagonal number is defined by "replace 6 by 3", then we have > >> >> > two answers none of which can be preferred by logic, but the second of > >> >> > which is suppressed by convention. > > >> The diagonal number is 0.333... which is not on the list. > > >The diagonal number cannot have more 3's than every list number has > >3's. > > Of course it can. Consider the following infinite matrix M: xooo... xxooo... xxxooo... .... You claim the (unchanged) diagonal can have more x's than any line? Than is wrong because the diagonal consists of line elements. > With your rule for the diagonal: replace 3 by 6, then > if we start with the following list: > > 0.6000 > 0.06000 > 0.006000 > 0.0006000 > 0.00006000 > ... > > then the diagonal will be the same: 0.333... > There is one "3" for each number in the list. What do you think why I chose the special list? In that list the diagonal number cannot have more 3's than at least one line of the list (well, one more). > > >If the diagonal number is complete in the sense that at every > >position indexed by a natural number there is a 3, then there must be > >a complete sequence of 3's among the list entries too. > > No, that's false. It's *provably* false. It is also provably right. See matrix M above. > For your particular > example, what is true is this (letting r_j be the jth number > in the list) > > 1. For each natural number n > 0, there exists an index j > such that r_j agrees with the diagonal number in the first > n decimal places. > > 2. For each index j, there exists a natural number n > 0 > such that r_j does *not* agree with the diagonal number in > the first n decimal places. > > Statement 1. says that the diagonal number can be approximated > arbitrarily well. Statement 2. says that the diagonal number does > not occur on the list. Both are true. But they are contradictory, because more than "each natural" number is not available. Forall n in the first n decimal places means for *all* natural indexes. > > These two statements have the logical forms: > > 1. forall n, exists j, Phi(r_j, d, n) > 2. forall j, exists n, ~Phi(r_j, d, n) > > where Phi(r_j, d, n) is the statement "r_j agrees with the diagonal > number d in the first n decimal places". > > >Otherwise the list is not complete, i.e., there is always a 6 at > >some natural index. > > The definition of the diagonal number d is the following: (letting > r[n] mean the nth decimal place of the real r) > > d[n] = 9 - r_n[n] > > The definition of the numbers r_j is this: > > r_j[n] = 6, if j=n > = 3, otherwise > > It follows that > > r_n[n] = 6 > > Putting those together, we have > > d[n] = 9 - 6 = 3 > > So every digit of the diagonal number is 3. If there are infinitely many 3's in the diagonal, then there are infinitely (- 1) many 3's in an r_n. > > >If the list is not complete however, the diagonal number cannot be > >complete either. > > The list is "complete" in the sense that for every natural > number j, there is a corresponding real number r_j. Similarly, > the diagonal number is complete in the sense that for every > natural number j, there is a corresponding digit d[j]. > > >And the saying that at *every* position there is a > >digit b_n =/= a_n,n is nonsense. > > It's provably true. (I'm using d, rather than b, and I'm > using r_n instead a_n). > > d[n] = 3 > r_n[n] = 6 > > Clearly d[n] is never equal to r_n[n]. That is only one side of the medal which does not take into account that hat each real number r_n is not better determined than by |r_n - r_n(k)| < epsilon(k), or to express it opposite: An irrational real number r_n cannot be determined better than by a sequence r_n(k) with |r_n - r_n(k)| > anti-epsilon(k). In other words: An irrational real number *is not* defined better than up to an anti-epsilon > 0 because for every approximation r_n(k) we can find such anti-epsilon > 0 which is less than the difference |r_n - r_n(k)|. > > >> >For the entries E(n) of the list we find > >> >lim[n-->oo] (E(n) - 0.333...) = 0. > >> >It is the same case as lim[n-->oo] (1 - 0.999...9 with n 9's) = 0. > > >> That's true. In this particular case, the limit of the sequence > >> is equal to the diagonal of the sequence. So what? > > >Why only in this particular case? Is lim[i --> oo] (b_i - a_i) * 10^-i > >= 0 correct only in one special case? > > I say in this case because not every sequence has a limit. For example, > the sequence > > 0.600... > 0.330... > 0.6660... > 0.33300... > 0.666600... > 0.3333000... > > has no limit. But it still has a diagonal, namely > 0.363636... But if this diagonal is to be considered a real number, then we need the factors 10^-i. The same is true for lim[i --> oo] (b_i - a_i) * 10^-i > > >> >If every initial segment of the diagonal number is represented by the > >> >initial segment of an entry of the list, then the full diagonal number > >> >is represented by an entry of the list. > > >> That's false. > > >No, that's correct. > > It's provably false. The diagonal number in your case satisfies > > d[n] = 3 > > Every element has r_n[n] = 6. Clearly they are not the same, so > d is not equal to r_n. > > >It is the only valid interpretation of the notion infinity. > > I don't know what that's supposed to mean, but you seem to be > confusing three different concepts: > > 1. The number d appears on the list r_j. For this to be true, > it must be the case that Exists j, Forall n, d[n] = r_j[n]. > That is *false*. Provably false. No. It may be provably false for the diagonal sequence, but not for the diagonal number. > > 2. The number d is the limit of the numbers r_j. For that to be > true, it must be the case that Forall n > 0, Exists j, Forall m < n, > d[m] = r_j[m]. That's true in your case. And that is *all* we can consider in a list f n-ary expansions. Any infinite n-ary expansion of a real number (not ending by zeros) *is not* defined better than up to an anti-epsilon > 0 because for every approximation r_n(k) we can find an anti-epsilon > 0 which is less than the difference |r_n - r_n(k)|. > > >> To say that the number r appears on the list r_0, r_1, ... > >> is to say that there is some natural number j such that r = r_j. If > >> we let D_j = |r - r_j|, then the criterion for r appearing on the list > >> is that > > >> exists j such that D_j = 0 > > >This criterion is false. > > If D_j is not zero, then that means that the diagonal number > is not equal to r_j. > > >In case lim [i --> oo] (10 - 9) * 10^-i = 0 we see it clearly. > > What do limits have to do with anything? There are two different > questions: > > 1. Does the diagonal number d appear on the list? The answer is no. This answer is given at most for the diagonal sequence, not for the diagonal number. > 2. Is the diagonal number the limit of the numbers on the list? The > answer for your case is yes. And that is *all* we can consider in a list f n-ary expansions. Any infinite n-ary expansion of a real number (not ending by zeros) *is not* defined better than up to an anti-epsilon > 0 because for every approximation r_n(k) we can find an anti-epsilon > 0 which is less than the difference |r_n - r_n(k)|. > > >Forall n e N we have 0 < 1 - 0.999...9 (with n 9's). > > Yes, that's true. 1 does not appear on the list > > 0.9 > 0.99 > 0.999 > 0.9999 > 0.99999 > etc. > > and 0.33333 does not appear on the list > > 0.6666... > 0.36666 > 0.336666 > 0.3336666 > etc. > > >> So r does not appear on the list. > > >So the problem of replacing 0 by 9 is not existing. > > What are you talking about? Of course the number 0.9999... > (all 9s) exists. And it is equal to 1. But it is not on the > list > > 0.9 > 0.99 > 0.999 > 0.9999 > 0.99999 > etc. > > >But we have in the limit j --> oo: D_j = 0. > > Yes. That's true, but irrelevant. For the diagonal to be > on the list, it must be that D_j = 0 for some finite j. That is only required for a finite list. > It's not good enough that limit of D_j = 0. Consider Dave L. Renfro's example: 0.49999999... 0.49111111... 0.44911111... 0.44491111... 0.44449111... .. . . . . . . Replacing 4 by 5 and 9 by 0 gives the diagonal number 0.5000...0999... =/= 0.5 for any finite j. Only in the limit j --> oo the diagonal number is 0.5000... = 0.499..., the first entry. Here it is good enough to consider the limit to make the proof fail? Regards, WM
From: Daryl McCullough on 21 May 2007 11:44
WM says... >On 21 Mai, 13:47, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: >> WM says... >You claim the (unchanged) diagonal can have more x's than any line? >Than is wrong because the diagonal consists of line elements. > >> With your rule for the diagonal: replace 3 by 6, then >> if we start with the following list: >> >> 0.6000 >> 0.06000 >> 0.006000 >> 0.0006000 >> 0.00006000 >> ... >> >> then the diagonal will be the same: 0.333... >> There is one "3" for each number in the list. > >What do you think why I chose the special list? Because with a different starting list, it is more immediately obvious that your claims are false. >In that list the diagonal number cannot have more 3's than at least >one line of the list (well, one more). The point about mathematics is that you aren't allowed to just say something because you feel it ought to be true. You have to show that your claim *follows* from your assumptions. >> >If the diagonal number is complete in the sense that at every >> >position indexed by a natural number there is a 3, then there must be >> >a complete sequence of 3's among the list entries too. >> >> No, that's false. It's *provably* false. > >It is also provably right. See matrix M above. A proof consists of a sequence of statements such that each statement is either an assumption, or follows from previous statements from the rules of inference. There is such a proof of the claim "For all j, the diagonal is unequal to r_j" where r_j is the jth real in the list. You are claiming, what, exactly? That the diagonal *is* on the list? In other words, you are claiming "There exists a j such that the diagonal is equal to r_j" What's the proof of that? Give a sequence of statements, such that each statement is either an assumption, or follows from previous statements by the rules of inference. >> For your particular >> example, what is true is this (letting r_j be the jth number >> in the list) >> >> 1. For each natural number n > 0, there exists an index j >> such that r_j agrees with the diagonal number in the first >> n decimal places. >> >> 2. For each index j, there exists a natural number n > 0 >> such that r_j does *not* agree with the diagonal number in >> the first n decimal places. >> >> Statement 1. says that the diagonal number can be approximated >> arbitrarily well. Statement 2. says that the diagonal number does >> not occur on the list. Both are true. > >But they are contradictory They are both *true*. They can't be contradictory. If you claim they are contradictory, then it's easy enough to prove such a claim. If A and B are contradictory, that means that A implies the negation of B. Okay, so you need to show that 1. above implies the negation of 2. Statement 1 is S1: Forall n > 0, exists j r_j agrees with the diagonal number in the first n decimal places. The negation of statement 2 is NS2: Exists j, forall n > 0, r_j agrees with the diagonal in the first n decimal places. So you need to prove that S1 implies NS2. Go ahead. Notice that S1 starts with "forall", while NS2 starts with "Exists". There is a difference between "Forall x, Exists y, Phi(x,y)" and "Exists x, Forall y, Phi(x,y)". For example, let Phi(x,y) be "y is the mother of x". Then Forall x, Exists y, y is the mother of x is true, because every person has a mother. But Exists y, Forall x, y is the mother of x is false, because not every person has the *same* mother. >> These two statements have the logical forms: >> >> 1. forall n, exists j, Phi(r_j, d, n) >> 2. forall j, exists n, ~Phi(r_j, d, n) >> >> where Phi(r_j, d, n) is the statement "r_j agrees with the diagonal >> number d in the first n decimal places". >> >> >Otherwise the list is not complete, i.e., there is always a 6 at >> >some natural index. >> >> The definition of the diagonal number d is the following: (letting >> r[n] mean the nth decimal place of the real r) >> >> d[n] = 9 - r_n[n] >> >> The definition of the numbers r_j is this: >> >> r_j[n] = 6, if j=n >> = 3, otherwise >> >> It follows that >> >> r_n[n] = 6 >> >> Putting those together, we have >> >> d[n] = 9 - 6 = 3 >> >> So every digit of the diagonal number is 3. > >If there are infinitely many 3's in the diagonal, then there are >infinitely (- 1) many 3's in an r_n. No. By the definition of the diagonal, d, if there are infinitely many 3s in d, then that means that there are infinitely many values of n such that r_n[n] = 6. And there are. >> It's provably true. (I'm using d, rather than b, and I'm >> using r_n instead a_n). >> >> d[n] = 3 >> r_n[n] = 6 >> >> Clearly d[n] is never equal to r_n[n]. > >That is only one side There is no "other side". d is not equal to r_0. d is not equal to r_1. d is not equal to r_2. For each natural number n, d is not equal to r_n. >of the medal which does not take into account >that hat each real number r_n is not better determined than by |r_n - >r_n(k)| < epsilon(k), or to express it opposite: An irrational real >number r_n cannot be determined better than by a sequence r_n(k) with >|r_n - r_n(k)| > anti-epsilon(k). To the extent that I understand what you're trying to say, yes. Every irrational number can be written as a limit of a sequence of rational numbers. However, if r is irrational, and r_0, r_1, ... is a sequence of rationals converging to r, then it follows that: 1. Forall n, r is not equal to r_n. 2. r = Limit as n --> infinity of r_n Both statements are true. But only the first is relevant to the question: Does r appear on the list r_0, r_1, ...? >> I say in this case because not every sequence has a limit. For example, >> the sequence >> >> 0.600... >> 0.330... >> 0.6660... >> 0.33300... >> 0.666600... >> 0.3333000... >> >> has no limit. But it still has a diagonal, namely >> 0.363636... > >But if this diagonal is to be considered a real number, then we need >the factors 10^-i. The same is true for > lim[i --> oo] (b_i - a_i) * 10^-i That limit is zero. But so what? To say that the limit of a sequence is zero is *different* from saying that any element in the sequence is equal to zero. For example, the sequence 1, 1/2, 1/3, 1/4, ... has the limit 0. But no number in the sequence is equal to zero. >> 1. The number d appears on the list r_j. For this to be true, >> it must be the case that Exists j, Forall n, d[n] = r_j[n]. >> That is *false*. Provably false. > >No. It may be provably false for the diagonal sequence, but not for >the diagonal number. What are you talking about? If you have a sequence of reals r_0, r_1, r_2, ..., then to say that a number d appears on the list means, by definition, that d is equal to r_n for some number n. To say that d = r_n means that every decimal place of d is equal to the corresponding decimal place of r_n. In other words d is on the list <-> exists n, d = r_n <-> exists n, forall j, d[j] = r_n[j] And that's false. For every n, d[n] is never equal to r_n[n]. So for every n, d is not equal to r_n. So d is not on the list. >> 2. The number d is the limit of the numbers r_j. For that to be >> true, it must be the case that Forall n > 0, Exists j, Forall m < n, >> d[m] = r_j[m]. That's true in your case. > >And that is *all* we can consider in a list f n-ary expansions. No, in the case r_0 = 0.666..., r_1 = 0.3666..., r_2 = 0.63666..., etc., and the case d = 0.3333..., we can prove d is not equal to r_0 d is not equal to r_1 d is not equal to r_2 etc. >Any infinite n-ary expansion of a real number (not ending by zeros) *is >not* defined better than up to an anti-epsilon > 0 because for every >approximation r_n(k) we can find an anti-epsilon > 0 which is less >than the difference |r_n - r_n(k)|. To know that d is not equal to r_0, we only need to look in the first decimal place. To know that d is not equal to r_1, we only need to look in two decimal places. To know that d is not equal to r_2, we only need to look at three decimal places. For every n, we only need to look at the first n decimal places to see that d is not equal to r_n. >> What do limits have to do with anything? There are two different >> questions: >> >> 1. Does the diagonal number d appear on the list? The answer is no. > >This answer is given at most for the diagonal sequence, not for the >diagonal number. I repeat: Does the diagonal number d appear on the list? The answer is no. There is no number n such that r_n = 0.3333.... >> Yes, that's true. 1 does not appear on the list >> >> 0.9 >> 0.99 >> 0.999 >> 0.9999 >> 0.99999 >> etc. >> >> and 0.33333 does not appear on the list >> >> 0.6666... >> 0.36666 >> 0.336666 >> 0.3336666 >> etc. >> >> >> So r does not appear on the list. >> >> >So the problem of replacing 0 by 9 is not existing. >> >> What are you talking about? Of course the number 0.9999... >> (all 9s) exists. And it is equal to 1. But it is not on the >> list >> >> 0.9 >> 0.99 >> 0.999 >> 0.9999 >> 0.99999 >> etc. >> >> >But we have in the limit j --> oo: D_j = 0. >> >> Yes. That's true, but irrelevant. For the diagonal to be >> on the list, it must be that D_j = 0 for some finite j. > >That is only required for a finite list. The words "d appears on the list" means the SAME THING as "there exists a j such that d = r_j", which means the SAME THING as "there exists a j such that D_j = 0". >> It's not good enough that limit of D_j = 0. > >Consider Dave L. Renfro's example: > >0.49999999... >0.49111111... >0.44911111... >0.44491111... >0.44449111... >. . . . . . . > >Replacing 4 by 5 and 9 by 0 gives the diagonal number >0.5000...0999... =/= 0.5 for any finite j. Only in the limit j --> oo >the diagonal number is 0.5000... = 0.499..., the first entry. Here it >is good enough to consider the limit to make the proof fail? That's right. To make sure that the diagonal number d does not appear on the list, you need to prove that for each n, d is unequal to r_n. If r_n ends with all 0s or all 9s, then that means that r_n has two decimal expansions, r1_n and r2_n. You need to show that d[n] is unequal to r1_n[n] *and* d[n] is unequal to r2_n[n]. The easiest way is to make sure that d[n] is never 0 and is never 9. A diagonal procedure that does this is the following: replace 9 by 1 replace 8 by 1 replace 7 by 1 replace 6 by 1 replace 5 by 1 replace 4 by 1 replace 3 by 1 replace 2 by 1 replace 1 by 2 replace 0 by 1 -- Daryl McCullough Ithaca, NY |