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From: WM on 12 Jun 2010 08:51 On 12 Jun., 01:35, Owen Jacobson <angrybald...(a)gmail.com> wrote: > >>>>> 0.1 > >>>>> 0.11 > >>>>> 0.111 > >>>>> ... > > >> For all elements n of N, let F_n be a function from N to {0, 1} as follows: > > >> F_n(x) = 1 if x <= n > >> F_n(x) = 0 if x > n > > >> Let G be a function from N to {0, 1} as follows: > > >> G(x) = 1 > > >> Which of the following statements accurately characterizes your argument? > > >> 1. If N exists, then there exists some natural number z in N such that > >> L(z) = F_z = G. > > > Yes. But z does not and cannot exist. Therefore N cannot exists as a > > completed entity. > > Can you frame an existence argument for a number z in N with the > properties above, either by deriving it directly, Proof by complete induction. If the existence of all nodes 1 in different lines of the list is claimed, then at least two lines must be shown which, with respect to their contents of nodes 1, cannot be replaced by one line. This is obviously impossible. On the contrary, it can be proved by complete induction, hence for every line of the list, that all nodes which are in that line and in all previous lines, also are in the next line. (Complete induction holds for all natural numbers, though not for the set of them. But that ist not required here.) > or by showing that it > must exist, in terms of ZFC? ZFC is rubbish. I do not use it. > > Try to avoid analogies: Try to use complete induction. > you don't actually need binary trees or decimal > expansions to frame your argument, and your argument would be much > clearer without them. At present we are not in the binary tree. We have only a very simple list. See above. > Framing your argument in terms of functions would > help me, obviously, but framing it in terms of sets and subsets would > work as well. Your approach stands above. I did not delete it. As far as I can see we discuss it. > It's not at all obvious from the arguments you've > presented that z must exist. Why do you resist complete induction? Please understand: Complete induction works for every line of the list above. Therefore, if all 1's are ther, then they all must be in one and the same line together. > It's very obvious that you're convinced it > must exist, but your argument is so buried in your "tree" analogy and > in invented terms like "cover" that it's very hard to follow. Please forget the tree. Consider this list 0.1 0.11 0.111 .... and consider this line 0.111... If the latter exists, then it must exist also in the list. That has been shown by induction And it can be shown by super task. (Some people deny super-task, but then they should also deny finished infinity.) Proof by construction of an infinite set. Construct the above list, but remove always line number n after having constructed the next line number n + 1. Then the list shrinks to a single line but this single line contains the same as the list because never anything has been removed that had not been added before to an existing line. Therefore this single line contains all nodes 1. It is the limit p = 0.111... If, however, line number n is not removed after line number n + 1 has been constructed, there cannot be an effect of this marginalia on line n + 1 and later lines. > > > But there are only countably many finite definitons. > > That was never in dispute. Look up "computable numbers," for example. But it is frequently disputed that all real numbers form a countable set. Therefore besides finite definition, there must be infinite sequences "standing alone". But there is no infinite sequence without finite definition. A finite definition implies an infinite sequence. You can obtain every term of the sequence from the definition. But you cannot reverse that direction. You can never obtain a definition or a number from an infinite sequence (which is not given by a finite definition). Regards, WM
From: Gus Gassmann on 12 Jun 2010 09:53 On Jun 11, 12:01 pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: > On 11 Jun., 14:34, Gus Gassmann <horand.gassm...(a)googlemail.com> > wrote: > > > > But I can estimate that all together form a set that is not > > > larger than a countable set. > > > By making a false continuity assumption. Big deal. > > What kind of discontinuity do you expect??? > Infinity means going on and on and on. Since you do not provide the exact proof you have in mind, I can't comment specifically (not that I am particularly interested in doing so here, given that you did not understand previous attempts at explaining this to you, your status as professor notwithstanding). However the general gist of these "proofs" is this: 1. For every integer n, card (T_n) < b_n for some structure T_n and real number b_n 2. T_n converges to some T in some intuitive sense that might even be made rigorous. The fallacy (or circular argument, to be precise) is then to infer from this that card(T) <= lim (b_n). > Do you believe that there is an abyss "after all"? Do you think along > the lines of the medieval sophists who expected to fall from the earth > after passing its edge? > > Well, your reasoning sounds like that. Don't worry. There is no abyss > and there is no dragon and there is no discontinuity . As a mathematician, I demand proof of this "fact", which you are of course impossible to provide (see above). That your mental defects prevent you from understanding this is not really my concern.
From: WM on 12 Jun 2010 13:18 On 12 Jun., 15:53, Gus Gassmann <horand.gassm...(a)googlemail.com> wrote: > On Jun 11, 12:01 pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > On 11 Jun., 14:34, Gus Gassmann <horand.gassm...(a)googlemail.com> > > wrote: > > > > > But I can estimate that all together form a set that is not > > > > larger than a countable set. > > > > By making a false continuity assumption. Big deal. > > > What kind of discontinuity do you expect??? > > Infinity means going on and on and on. > > Since you do not provide the exact proof you have in mind, I can't > comment specifically But usually you comment without knowingthe exact topic you comment on. > (not that I am particularly interested in doing > so here, given that you did not understand previous attempts at > explaining this to you, You did not express yourself in clear words but always only quoted beliefs. > However the general gist of these "proofs" is this: > > 1. For every integer n, card (T_n) < b_n for some structure T_n and > real number b_n > 2. T_n converges to some T in some intuitive sense that might even be > made rigorous. > > The fallacy (or circular argument, to be precise) is then to infer > from this that > > card(T) <= lim (b_n). For some reciprocal values of integer sums, i.e. convergent series, we conclude: If for all n in N: SUM(k= 1...n) a_k < SUM(k= 1...n) b_k, then SUM(k= 1...oo) a_k =< SUM(k= 1...oo) b_k. Why is that assumed? Nobody really knows what happens in the limit. Nobody ever could try. Why is this definition assumed to be true? > > > Do you believe that there is an abyss "after all"? Do you think along > > the lines of the medieval sophists who expected to fall from the earth > > after passing its edge? > > > Well, your reasoning sounds like that. Don't worry. There is no abyss > > and there is no dragon and there is no discontinuity . > > As a mathematician, I demand proof of this "fact", But as a "matheologician" you believe that Tristram Shandy can finish his diary, writing every day of his life, although he needs a full year for each day. The "argument" is: For each day of his life we know the year when it will be written. The other side is: For every year of his life we can calculate that the number of unwritten days has grown by 364 and never decreases. But according to your mathematical spirit this continuous increase does not count because "there is no continuity". You can be happy, that such "mathematics" fortunately has no influence on real mathematics. Regards, WM
From: Owen Jacobson on 12 Jun 2010 14:17 On 2010-06-12 08:51:48 -0400, WM said: > On 12 Jun., 01:35, Owen Jacobson <angrybald...(a)gmail.com> wrote: >> >>>> For all elements n of N, let F_n be a function from N to {0, 1} as follows: >> >>>> F_n(x) = 1 if x <= n >>>> F_n(x) = 0 if x > n >> >>>> Let G be a function from N to {0, 1} as follows: >> >>>> G(x) = 1 >> >>>> Which of the following statements accurately characterizes your argument? >> >>>> 1. If N exists, then there exists some natural number z in N such that >>>> L(z) = F_z = G. >> >>> Yes. But z does not and cannot exist. Therefore N cannot exists as a >>> completed entity. >> >> Can you frame an existence argument for a number z in N with the >> properties above, either by deriving it directly, > > Proof by complete induction. If the existence of all nodes 1 in > different lines of the list is claimed, This is exactly the sort of analogy-based presentation I'd like to avoid. I *think* what you mean is something similar to 1. ∀x ∃n F_n(x) = 1 where x, n ∈ N which is fairly easy to prove given how F_n is defined. > then at least two lines must be shown which, with respect to their > contents of nodes 1, cannot be replaced by one line. And (this phrase is particularly unclear; bear with me if I misrepresent it) 2. ∃x ∀n F_n(x) = 0 where x, n ∈ N which is fairly easy to disprove given how F_n is defined. It's not obvious to me that statement 1 implies statement 2. Assuming for the moment that I've correctly represented your argument, would you care to draw out the proof symbolically? If I've misrepresented your statement, could you offer a better version, as well as drawing out the rest of the proof? Note also that even if you can prove statement 2 or something similar to it, that doesn't imply anything about the relationship between the functions F_n and the function G. > On the contrary, it can be proved by complete induction What are the axioms related to "complete induction"? You've introduced a new term; define it. > hence for every line of the list, that all nodes which are in that line and in > all previous lines, also are in the next line. I think you mean 3. ∀x ∀y ∀n (F_x(n) = 1) ∧ (y > x) → (F_y(n) = 1) where x, y, n ∈ N which is easily proven as well, but implies nothing about the relationships between F_n and G: in particular, it does not imply ∃z ∀n (F_z(n) = G(n)). >> or by showing that it >> must exist, in terms of ZFC? > > ZFC is rubbish. I do not use it. Without a set theory, there is no set N. What set theory *are* you using? What are its axioms and primitives? >> Try to avoid analogies: > > Try to use complete induction. Define "complete induction". In particular, how is it different from the following: 4. φ(0) ∧ (φ(n) → φ(S(n))) → ∀x φ(x) where 0, n, x ∈ N and S(n) and N are defined in the usual way Note that this induction axiom schema does not allow you to prove, from φ(0), that φ(N). In ZFC, no axiom allowing such a conclusion exists, so if your "complete induction" allows this conclusion, you'll need to show the additional axioms (or the axioms you've changed) that are necessary to reach that conclusion. -o
From: Owen Jacobson on 12 Jun 2010 14:24
On 2010-06-12 14:17:45 -0400, Owen Jacobson said: > > 4. 4. φ(0) ∧ (φ(n) → φ(S(n))) → ∀x φ(x) > where 0, n, x â N and S(n) and N are defined in the usual way That should read: 4. 4. φ(0) ∧ (∀n φ(n) → φ(S(n))) → ∀x φ(x) Sorry for the confusion. -o |