constructible = unconstructible
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From: Virgil on 13 Jun 2010 14:27 In article <9716415a-dded-4076-9360-75bbfb33344c(a)d37g2000yqm.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > 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. Then WM has misread the proof or misunderstood it, because that is not the justification Cantor uses. The Cantor justification shows that for ANY ( and therefore for EVERY) n in N, the constructed sequence differs from the nth listed sequence. > > 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. Possibly because WM's alternatives are so much wronger.
From: WM on 13 Jun 2010 15:17 On 13 Jun., 20:22, Virgil <Vir...(a)home.esc> wrote: > In article > <15b42a8a-c63f-4ada-a415-0bb631842...(a)c10g2000yqi.googlegroups.com>, > > > > > > WM <mueck...(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, No. There is no other way for a natural number. There must be a larger one. If this can be wrong, then the theory which allows that is wrong. In particular set theory is wrong. > 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. Yes I agree. But unfortunately, all set theorists claim that. They claim that they can do this or that with all natural numbers. In particular they claim that they can write or remove all of them. That is nonsense. > > > 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. I claim to be able to remove all because set theory says so. I know that this saying in wrong. Regards, WM
From: WM on 13 Jun 2010 15:19 On 13 Jun., 20:27, Virgil <Vir...(a)home.esc> wrote: > > 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. > > Then WM has misread the proof or misunderstood it, because that is not > the justification Cantor uses. > > The Cantor justification shows that for ANY ( and therefore for EVERY) > n in N, the constructed sequence differs from the nth listed sequence. Every n is the last number of a finite sequence, isn't it? Do you know of an n that is not the last term of a finite sequence? Regards, WM
From: Virgil on 13 Jun 2010 19:14 In article <afd8fc58-0cf4-4219-bb26-26c8c17d60b7(a)c10g2000yqi.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 13 Jun., 20:22, Virgil <Vir...(a)home.esc> wrote: > > In article > > <15b42a8a-c63f-4ada-a415-0bb631842...(a)c10g2000yqi.googlegroups.com>, > > > > > > > > > > > > �WM <mueck...(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, > > No. Yes!!!! Things which are implicit in a situation cannot be treated as if they do not exist in that situation, at least not in honest mathematics. > There is no other way for a natural number. What does that mean in English? > There must be a larger > one. If this can be wrong, then the theory which allows that is wrong. While for each natural number in the set of natural numbers there is a larger natural number in the set of all natural numbers, this is not necessarily true for sets other than the set of natural numbers. > In particular set theory is wrong. WM's version of it is wrong, but only because WM is wrong. > > > 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. > > Yes I agree. But unfortunately, all set theorists claim that. No they don't. What they DO claim is that one cannot remove an element from a set and have a non-empty set left UNLESS there are other elements in that set. Which is quite different from what WM claims. They > claim that they can do this or that with all natural numbers. In > particular they claim that they can write or remove all of them. That > is nonsense. One can remove all the elements from any set, leaving the empty set. At least that is how it works in standard set theories, regardless of how WM claims it works in WM's Pseudo set theory. > > > > > 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. WM appears to be claiming that when all the elements of a set are removed there must be some elements still left in it. That would make for a very peculiar set theory. > > > > > > > > > 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. > > I claim to be able to remove all because set theory says so. I know > that this saying in wrong. What is wrong with removing all the members of a set from that set? In standard set theory, that is always possible and merely results in the standard empty set. For any set, S, and any subset of it, T, there is in standard set theories a set S\T = { x in S: not x in T }. In fact, T does not even have to be a subset of S for S\T to exist. In particular, for ANY set S, (including N) one has S\S = {}. This works for N as well as for any other set, despite what WM says..
From: Virgil on 13 Jun 2010 19:22
In article <49ddcc54-89ab-4f99-9170-c05b6fd5e644(a)d8g2000yqf.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 13 Jun., 20:27, Virgil <Vir...(a)home.esc> wrote: > > > > 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. > > > > Then WM has misread the proof or misunderstood it, because that is not > > the justification Cantor uses. > > > > The Cantor justification shows that for ANY ( and therefore for EVERY) > > �n in N, his constructed sequence differs from the nth listed sequence. > > Every n is the last number of a finite sequence, isn't it? Do you know > of an n that is not the last term of a finite sequence? Since whether it is is irrelevant here , it does not matter here. Given any sequence of binary sequences, one can define a binary sequence not found in that sequence of sequences. In fact, one can define infinitely many of them, one for each natural. |