Cantor Confusion
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From: mueckenh on 20 Dec 2006 07:04 cbrown(a)cbrownsystems.com schrieb: > > No. But I need no proof in order to see that > > 1 + 1/2 + 1/4 + ...+ 1/2^n + ... >= 2 - 1/2^n > 3/2 for n > 2. > > > > Yes, it appears that you need no proof to make any of your assertions. > You simply "see" them; or "feel" them. That's not what I would call > "really correct mathematics". Call it whatever you want. I can calculate 1 + 1/2. But considering the fact that you would call the following result "correct mathematics", I prefer to stay with my version. Down to level n (for n = 1, 2, 3, ...) we can distinguish P(n) = 2^n different paths and E(n) = 2*2^n - 2 different edges. So we find lim{n-->oo} E(n)/P(n) = 2, or, if we are unable to determine limits, E(n)/P(n) >= 1. Nevertheless "E(n)/P(n) < 1 in the actual infinity". Regards, WM
From: mueckenh on 20 Dec 2006 07:14 Dik T. Winter schrieb: > Right. That is the difference between scientific press and vanity press. > In the former there is review and editing assistence, in the latter you > can publish anything you want to publish. In the former you do not pay > anything, Never heard of "page charge"? I remember that was usual with Physical Review already 30 years ago. > I think there will not be much sale, except amongst your students for > which it will be a required read. (Is that the case?) No. Much too expensive for students. But I will put some copies in the library. > But, indeed, if you pay for it, anything can be published. But > that was already true in Galilei's time. You are in error. > > > > Oh, and as a nicety, the author will receive parts of the > > > sale cost for each book outside the initial print run. It is 10%, but > > > I do not know whether that is before taxes or after. > > > > I would be glad to give these 10 % to any customer who really > > understands my ideas. > > Let me do some calculations. You are required to take 30 books of the > initial run for half the price of EUR 28.50. That means 30 times 14.25 > or EUR 427.50. My estimate is that the initial run is exactly 30 copies > and further printing is only on demand. Wrong. > > But whatever, send me a copy of the book if you want a review in this > newsgroup. I promise to donate it to our library when finished. I'll do it. > > But it still makes me wonder why you are referring to your book in a > newsgroup where only a small portion of the readers is able to read > German. Better a small portion than an empty set. Regards, WM
From: mueckenh on 20 Dec 2006 07:23 Dik T. Winter schrieb: > > For the evaluation of this *sum* we need *no limit* but only the > > computation > > 1 + 1/2 + 1/4 + ...+ 1/2^n + ... >= 2 - 1/2^n > 3/2 for n > 2. > > We were talking about *shares*. In how many shares is the edge going left > from the root divided, and how do you do that division? Pray keep to the > subject. We are talking about shares, but we do not need more than 3/2 per path.> > I cannot do it (and it has been proved that nobody can do it because > > there is no *definable* well-ordering). So I would prefer you > > demonstrate it. > > That is similar to asking me to prove the parallel axiom without the > parallel axiom as basis. No, Dik. That is an error, perhaps it is that one whole set theory rests on. I did not ask you to prove the well ordering of the reals. That has already been done by Zermelo. I asked for a well ordering, in form of an explicit formula, or a recursion or a list or what ever you will present to enable me to find out which real number follows on which real number. That cannot be done. Therefore I doubt that the proof has any relevance. > But I explicitly stated how you could construct > a well-ordering when AC is true. You can not do it in general, and > nobody can, because AC is not necessarily true. Are there axioms which are necessarily true? Not according to your point of view. > If somebody came up > with a construced well-ordering of the reals that would imply that AC > is true. In a similar way, if somebody could construct a line parallel > to another and show that there are no other parallel lines, that would > prove that the parallel axiom is true. The parallel axiom is true on a sheet of paper. Where is AC true? Regards, WM
From: William Hughes on 20 Dec 2006 07:26 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > > > > No. Only properties of the diagonal that you > > > > have accepted are used. > > > > The conclusion is that there is no line. > > > > > > Every element of the diagonal is n some line. > > > Every element of the diagonal is a finite natural number, by > > > definition. > > > > > > Up to the finite natural number n: Every element =< n of the diagonal > > > is contained in line n. Induction shows this is valid for any finite > > > natural number. > > > > > > There are only finite numbers in the diagonal. QED. > > > > Given that the diagonal is the potentially infinite > > set of natural numbers, the fact that there are > > only fiinite numbers in the diagonal is immediate. > > The only property of the diagonal that is used is > > the fact that it is not bounded. In particular > > > > -it is assumed that the diagonal contains > > only finite numbers > > > > -it is not assumed that all the elments of the > > diagonal exist > > > > The contradiction holds, using only properties > > of the potentially infinite set of natural numbers that > > you have stated. > > It does not as I have written in my last letter to you. But even if > your proof would hold, then we would have the fact that two > contradictory results can be derived. No Before you had A: There is a single line that contains every element of the diagonal B: If the diagonal is assumed infinite, then there is no single line that contains every element of the diagonal. Your conclusion was, assuming the diagonal infinite leads to a contradiction. Now you have A: There is a single line that contains every element of the diagonal B': If the diagonal is assumed unbounded, then there is no single line that contains every element of the diagonal. The contradiction now involves only assumptions you have made. Declaring that one cannot assume that the diagonal is infinite does not make this contradiction go away. Note, giving another putative proof of A will not change the fact that a contradiction exists. You need to deal with the proof of B'. - William Hughes
From: mueckenh on 20 Dec 2006 07:28
Virgil schrieb: > I said that there are infinitely many equations of form "1+1+1+...+1 = > n" with equal finite values on each side, but made no claim of any such > equation with any infinite umber o either side. If there are infinitely many different equations of form "1+1+1+...+1 = n, then there must be infinuitely many different sums of ones. Actual infinity of he umber of equations implies actual infi8nuity o at least one sum of ones. > > This statement is in complete compatibility with ZFC and NBG An exactly that is the reason why ZFC is wrong. You cannot have an actually infinite finite number. Regards, WM |