constructible = unconstructible
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From: WM on 13 Jun 2010 05:35 On 13 Jun., 07:15, Owen Jacobson <angrybald...(a)gmail.com> wrote: > On 2010-06-12 16:58:50 -0400, WM said: > > > On 12 Jun., 20:17, Owen Jacobson <angrybald...(a)gmail.com> wrote: > > For every natural number n, > F_n : N â {0, 1} > > > F_n(x) = 1 if x <= n > > F_n(x) = 0 if x > n > > > n F_n(x) > > > 1 100000... > > 2 110000... > > 3 111000... > > ... ... > > That table is a handy illustration, but it is not the definition of the > functions F_n; the equations above are. There is no definition of definition. What you claim is that only definitions like those usual in ZFC + FOPL are definitions. I do not subscribe to that opinion. A definition is every abbreviation for a mathematical object, that can be understood. As Kronecker already said: There are no definitions. "Man hat häufig gesagt, die Mathematik müÃte mit Definitionen beginnen, und aus ihnen zusammen mit den postulierten Grundsätzen seien die mathematischen Sätze abzuleiten. Nun sind aber Definitionen an sich schon eine Unmöglichkeit, wie Kirchhoff zu sagen pflegte, denn jede Definition braucht ihre Begriffe, welche wieder zu definieren sind u.s.f." ["Sur le concept de nombre en mathematique" Cours inédit de Leopold Kronecker à Berlin (1891) Retranscrit et commenté par Jacqueline Boniface et Norbert Schappacher: Revue dâhistoire des mathématiques 7 (2001), p. 207â 275.] ÃBER DEN BEGRIFF DER ZAHL IN DER MATHEMATIK http://smf4.emath.fr/Publications/RevueHistoireMath/7/pdf/smf_rhm_7_207-275..pdf Therefore my definitions above are as good as yours, at least. > > And, for completeness: > > G : N â {0, 1} > G(x) = 1 > > L : N â {F_x | x in N} > L(x) = F_x > > >>> 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. âx ân F_n(x) = 1 > >> where x, n â N > > > No. This would yield only for every finite x a finite n. > > The claim of set theorists is that there is an infinite number of 1's > > in the whole list, but that there is no F_n with infinitely many 1's. > > My claim is that *if this was true* then there must be a function, > > call it F_oo, with F_oo(x) = 1 for all x. > > Yes, I'm aware that that's your claim, but I still see no proof. I repeat it below by induction. > The > function F_oo(x) = G(x) = 1 exists, but nothing about the set of > functions labelled F_n that implies that it exists, or that there is > some (natural) number x such that L(x) = F_oo = G. That is wrong. If all bits 1 of G(x) exist, then they must exist in the union of all F_n. But I proved that there are no two different F_n that contain more bits 1 than one of them. I am sorry, but I cannot see what is difficult to understand here. > > > Hence > > E oo A x F_oo(x) = 1. > > Of course this F_oo cannot be fixed. > > It certainly can be fixed, but it's not possible for your 'oo' to be a > natural number. However: It cannot be fixed as a natural number n. But all F_n of your (and of my definition) have only natural numbers n. > > For all _ordinal_ numbers n, there exists a function F_n(x) from the > natural numbers to the set {0, 1} such that: > > F_n(x) = 1 if x <= n > F_n(x) = 0 if x > n > > For all natural numbers x, x < Ï, therefore F_Ï(x) = G = 1 (where Ï is > the first limit ordinal greater than zero). > > However, Ï is provably not a natural number. So it is. Therefore the only possible outcome is: If all bits 1 exist in the list, then they exist not in one F_n alone. But that can be disproved. > > >>> Try to use complete induction. > > >> Define "complete induction". In particular, how is it different from > >> the following: > > >> where 0, n, x â N and S(n) and N are defined in the usual way > > > In exactly that way I prove that for all n F_n(x) can be removed from > > the list and nevertheless all 1's remain the list. > > A vague description of a mechanical process on a physical object is not > a mathematically-useful definition. You can certainly imagine a set and the subtraction of elements from that set? > Can you define "complete induction" > or at least distinguish it properly from "induction" as given in my > last post There is no difference. In German it is called complete induction because Euler and Gauus used the term "induction" for a less stringent arguing. > (or in any presentation of Peano arithmetic)? You appear to > be relying on "complete induction" being able to prove statements that > induction, as normally defined, cannot prove, so there must be some > distinction between the two. Sorry, my proof uses exactly your induction. > > >> Note that this induction axiom schema does not allow you to prove, from > >> Ï(0), that Ï(N). > > > I don't need that. > > I prove that A n Ï(x) where Ï(x) : F_n can be removed from the list > > without loss of any 1. > > This is incoherent. "A n Ï(x)" has an unbound variable (x) and does not > use the only bound variable (n) at all. Your description of the > predicate Ï has the same problem - 'n' simply appears out of nowhere in > it. It's possible to figure out what you meant, but as a professor you > should really be doing a better job with your notation. > > > Proof: > > F_1 can be removed, because its 1 appears in F_2. > > F_n can be removed, because its 1's appear in F_(n + 1). > > Therefore every finite function can be removed. > > What do "removed" and "appears" mean in the context of the three groups > of functions (L, G, and F_n) mentioned? Sorry, you should know subtraction of elements of sets. The set "List" of all F_n contains all bits 1, that are in G = 0.111..., as bits of the function F_n. From List I can subtract F_1 and the remaining F_n contain all bits 1 of G. From List I can subtract F_k and the remaining F_n contain all bits 1 of G. This means: I can subtract every F_k for k in N without loss of a single bit 1 of G. All remain in the set List, although this set is empty. This is a contradiction. > > I'm also still interested in an answer to this question from my last post: > > > 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? Where can I read more about > it by other mathematicians? For my proof above I use the axioms of ZFC. In particular I accept that all bits 1 of G exists. For proving I use what you call induction and what in German is called vollständige Induktion (translated as complete induction). This proof has been given above. More is not required. Regards, WM
From: WM on 13 Jun 2010 05:39 On 12 Jun., 23:50, Virgil <Vir...(a)home.esc> wrote: > In article > <adf3fd69-402e-4ad2-8cc8-aa543e1b2...(a)y4g2000yqy.googlegroups.com>, > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > F_1 can be removed, because its 1 appears in F_2. > > F_n can be removed, because its 1's appear in F_(n + 1). > > Therefore every finite function can be removed. > > You have only proved that an F_n can be removes PROVIDED some F_m, with > m > n remains unremoved, which is false when ALL F_k are removed. No. I have proved that every F_n can be removed, because for every natural number n there is per se a natural number n + 1. Therefore I need not explicitly state that. If all natural nunbers can be removed and if then nothing remains, then there must be something wrong with the notion "all natural numbers", not with my proof. Or when, do you think, should I start to pay attention that the n considered in fact has an n + 1 above itself? Regards, WM
From: WM on 13 Jun 2010 06:42 On 12 Jun., 19:57, Gus Gassmann <horand.gassm...(a)googlemail.com> wrote: > On Jun 12, 2:18 pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > > > 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? > > You are the biggest and you have no behaviour, no education, and no self-control. Or do you hope to defend your nonsensical opnion by technical knock out? > > Your claim that > > card(lim T_n) <= lim b_n can be proved IF card(lim T_n) = lim (card > T_n), which is true if card is a continuous function. Here are some examples where you abyss is missing for non continous functions. I can prove that the geometric series 1/2 + 1/4 + 1/8 + ... converges, such that the infinite sum is 1. I can prove that the harmonic series 1 + 1/2 + 1/3 + ... diverges, such that the infinite sum is infinite. I can prove that in Cantors list the first n lines do not contain the diagonal number. From that it is said to be possible to prove that the diagonal is not in the infinite list. And I can give infinitely many further proofs of that kind . But you will believe (or better you pretend to believe) that the implication from |{2, 4, 6, ..., 2n}| contains several elements larger than its cardinality n will fail for the complete set, such that we get |{2, 4, 6, ...}| is larger every element of the set. That is so wrong that I cannot understand how so many rather intelligent people have been believing it for about 100 years. Last regards, WM
From: Virgil on 13 Jun 2010 14:14 In article <1df9a467-33ff-437b-a198-ff07d8dc29fb(a)c10g2000yqi.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 13 Jun., 07:15, Owen Jacobson <angrybald...(a)gmail.com> wrote: > > On 2010-06-12 16:58:50 -0400, WM said: > > > > > On 12 Jun., 20:17, Owen Jacobson <angrybald...(a)gmail.com> wrote: > > > > For every natural number n, > > F_n : N → {0, 1} > > > > > F_n(x) = 1 if x <= n > > > F_n(x) = 0 if x > n > > > > > n F_n(x) > > > > > 1 100000... > > > 2 110000... > > > 3 111000... > > > ... ... > > > > That table is a handy illustration, but it is not the definition of the > > functions F_n; the equations above are. > > There is no definition of definition. > What you claim is that only definitions like those usual in ZFC + FOPL > are definitions. I do not subscribe to that opinion. A definition is > every abbreviation for a mathematical object, that can be understood. > As Kronecker already said: There are no definitions. One can quite easily define things which can be shown not to exist, which would be quite impossible if definitions were merely abbreviations of mathematical objects. Or does WM's world insist on having mathematical objects in it which do not exist? > > Therefore my definitions above are as good as yours, at least. Nonsense, as usual!
From: Virgil on 13 Jun 2010 14:22
In article <15b42a8a-c63f-4ada-a415-0bb6318420f2(a)c10g2000yqi.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 12 Jun., 23:50, Virgil <Vir...(a)home.esc> wrote: > > In article > > <adf3fd69-402e-4ad2-8cc8-aa543e1b2...(a)y4g2000yqy.googlegroups.com>, > > > > �WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > F_1 can be removed, because its 1 appears in F_2. > > > F_n can be removed, because its 1's appear in F_(n + 1). > > > Therefore every finite function can be removed. > > > > You have only proved that an F_n can be removes PROVIDED some F_m, with > > m > n remains unremoved, which is false when ALL F_k are removed. > > No. I have proved that every F_n can be removed, because for every > natural number n there is per se a natural number n + 1. Therefore I > need not explicitly state that. Because it is implicit, it must be considered, and you cannot remove any 1 and have any left unless there is at least one 1 left. To claim otherwise is to be a fool. > If all natural nunbers can be removed and if then nothing remains, > then there must be something wrong with the notion "all natural > numbers", not with my proof. There is, as usual, something wrong with your proof. > > Or when, do you think, should I start to pay attention that the n > considered in fact has an n + 1 above itself? As long as there are OTHER 1's you can remove any particular set of 1's and have some left, but you are claiming to be able to remove ALL of them and still have some left. There is an error there, but it is yours, not that of the set. |