Cantor Confusion
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From: mueckenh on 27 Dec 2006 16:20 Virgil schrieb: > In article <1166952847.829226.324240(a)79g2000cws.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Virgil schrieb: > > > > > > > > As not all natural numbers do exist, the set is potentially infinite, > > > > i.e., it is finite. It has a maximum. L_D. Taking this maximum and > > > > adding 1 or takin L_D ^ L_D or so yields another maximum. In any cae > > > > there is a maximum. > > > > > > Which is always exceeded by a larger maximum. > > > > Which then is the maximum. > > As the creation of your "maximum" simultaneously creates its successor, > the time span for any natural to be maximal is zero. There is no creation in zero time, because then, at the same time, something would exist and would not exist (Aristoteles). ==================================================== >> Nobody can name it, because the set is not fixed. > In what set theory are sets not fixed. Wm's theories of volatile > sets do not work in mathematics. It is the only theory about sets which works in reality and does not require cranks and their beliefs on the infinite, but is based upon clear facts like: The infinite sequence of natural numbers is the union of all finite initial segments of natural numbers. An infinite tree is a tree which covers every finite level. ================================ >> The absence of more than 10^100 bits limits the existence of objects >> which require more than 10^100 bits to exist > In physical existence one may need physixcal bits, but not otherwise. Existence = physical existence, unless you have a strong religious intuition. (This intuition and its results, however, can be traced back on physical circumstances like drugs or hormones or electrical currents.) ==================================================== > The Cantor construction rule produces for any countably infinite list > a number which necessarily differs from every number in the list. The argument produces it only for finite segments of the list. How can you conclude from the finite domain on the infinite domain where completely other rules govern the numbers? ======================================= >> You cannot find such a number a_n without having pi already. But this >> number shall be used to establish the existence of pi. This is >> impossible. pi does not exist *as a number* (it exists as an idea and >> in form of close approximations). > Which idea, in a non-physical world populated only by ideas, suffices. But not as a number in Cantor's list, a digit of which has to be exchanged. > On the contrary, it is WM's claim that the limit of a quotient must > always equal the quotient of limits that is the problem. > WM claims lim n -> oo E(n)/P(n) = (lim n -> oo E(n))/(lim n -> oo P(n)), > but it does not hold for any form of limit definition in standard > mathematics. The truth is easily stated without quotients: lim n --> oo E(n) is not less than lim n --> oo P(n)). ==================== > The same " limit" argument will conclude the in a tree in which no path > has a terminal node, every path has a terminal node. > For every finite tree, the number of terminal nodes (leaf nosed) is > exaclty equal to the number of paths, so that if the limit argument is > valid for the edge to path ratio, it must equally be valid for the terminal node to path ratio. No, limit arguments do not hold in the infinite case (but only in the finite case. Therefore they are called "limit", i.e., something bounded and bounding.) You can see it best by the following example: In every finite initial segment of natural numbers, there are as many even numbers as odd numbers (plus or minus one number). But in the infinite set N there are 250 times more even numbers than odd numbers. Regards, WM
From: mueckenh on 27 Dec 2006 16:29 Dik T. Winter schrieb: > In article <1166962708.594095.140560(a)f1g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > In article <1166907306.745536.315650(a)h40g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > ... > > > > It was Dik who insisted that a well ordering of the reals can be done > > > > (not only be proved), if AC is true. Here is the discssion: > > > > > > Indeed, it can be done *when AC is true*. But perhaps we have a different > > > view on the meaning of the sentence "can be done"? > > > > It has been proved that there is *no definable well-ordering* of the > > reals. So it cannot be *done*. The only thing that can be done is to > > prove that a well-ordering exists. Zermelo did it. > > Oh well, I was slightly incorrect. You need also V=L. You were incorrect - but not slightly. > "...; it is also possible to show that the ZFC axioms are not sufficient to > prove the existence of a definiable (by a formula) well-order of the > reals. However it is consistent with ZFC that a definable well- > ordering of the reals exists -- I have not read the proof (because I don't like to spend my time with useless topics), but a real expert of set theory told me that it has been proved that there is no definable well ordering in ZFC > it follows from ZFC+V=L that a particular formula > well-orders the reals, or indeed any set." Instead of casting assertions: Why don't you simply give the formula for the definable well ordering? The formla could be evaluated, and, step by step, we would get a list of all reals. Regards, WM
From: mueckenh on 27 Dec 2006 16:34 cbrown(a)cbrownsystems.com schrieb: > > This bijection between sets (initial segments {1,2,3,...n} and > > {2,4,6,...2n}) is only valid for finite sets. > > Suppose I instead counter-claim that this bijection is obviously "not > valid" for finite sets; only for infinite sets. > > Is that what you call a "correct mathematical argument"? Simply > claiming something is valid or not valid? How is it different from your > own argument? It would be as impossible to contradict as your other claims or as the following: In every finite initial segment of natural numbers, there are as many even numbers as odd numbers (plus or minus one number). But in the infinite set N there are 250 times more even numbers than odd numbers. > > > Thus "|N|/|R| < 1" is nonsense; as I previously stated > quite clearly. I did not calculate |N|/|R| but edges per path. This calculation yields a real number for any finite path - and the infinite path is nothing else but the union of all finite paths. > > > The number of edges accumulated by one path up to level n is a function > > like that above: > > f(n) = 2 - 1/2^n. Very easy. > > Yes; when you write it that way, it "looks" just like a valid statement > about the real numbers. So, you "feel" that it is a both a sensible and > a valid statement about things which are not real numbers; when instead > it is nonsense. What is your definition of an infinite path? You feel that it is not the union of all finite paths but something much much greater and longer, more exciting and most mysterious? Do you have a definition how or any reason why the infinite path is not the union of all finite paths? Regards, WM
From: Virgil on 27 Dec 2006 16:45 In article <1167254457.021768.100380(a)i12g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1166952847.829226.324240(a)79g2000cws.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Virgil schrieb: > > > > > > > > > > > As not all natural numbers do exist, the set is potentially infinite, > > > > > i.e., it is finite. It has a maximum. L_D. Taking this maximum and > > > > > adding 1 or takin L_D ^ L_D or so yields another maximum. In any cae > > > > > there is a maximum. > > > > > > > > Which is always exceeded by a larger maximum. > > > > > > Which then is the maximum. > > > > As the creation of your "maximum" simultaneously creates its successor, > > the time span for any natural to be maximal is zero. > > There is no creation in zero time, because then, at the same time, > something would exist and would not exist (Aristoteles). But creation in the immaterial world of ideas does not require physical time, so we can have infinite amounts of immaterial time in finite real time intervals. > > For every finite tree, the number of terminal nodes (leaf nosed) is > > exaclty equal to the number of paths, so that if the limit argument is > > valid for the edge to path ratio, it must equally be valid for the > terminal node to path ratio. > > > No, limit arguments do not hold in the infinite case (but only in the > finite case. Therefore they are called "limit", i.e., something bounded > and bounding.) Then why do you argue that the limit of nodes/edges holds in the infinite case? > > You can see it best by the following example: > > In every finite initial segment of natural numbers, there are as many > even numbers as odd numbers (plus or minus one number). But in the > infinite set N there are 250 times more even numbers than odd numbers. What sort of limit does WM invoke to justify that result? In the Cantor sense there are the same number of odds as evens. If WM is arguing for some other measure, he should let us in on what he means.
From: mueckenh on 27 Dec 2006 16:47
Dik T. Winter schrieb: > > The diagonal digits a_nn of Cantor's argument have to be > > multiplierd by 10^-n in order to yield the diagonal number SUM a_nn * > > 10^-n. But there is no problem with yielding zero for n --> oo? > > --- Remarkable. > > Apparently you do not understand the working of limits. Apparently your limits work only for real numbers in lists but no for real numbers in trees. > It has been > proven that *every* Cauchy sequence yields a real number. And it is > easy to see that *every* decimal expansion is in fact a Cauchy sequence. > 10^-n is *never* zero. You need limits. Yes. And they say: As long as 10^-k is not zero, the sum {n=1 to k} a_n + 10^-n is not an irrational number. > > > There is no edge in the *finite* domain which is "passed in full". In > > the infinite series there is an edge passed in full. > > Which edge is passed in full to the path that goes at each step to > the left? I don't claim the existence of this representation of 0. > > > Otherwise the path > > does not exist at all. > > That is what you are trying to prove. But you can not just state it. You > hav to *prove* it. I need not prove that two entities which have no different property, cannot be distinguished. > > > The infinite sum 1 + 1/2 + 1/4 + ... yields 2, although there is no > > smallest term 2^-n. > > Indeed. > > > It can be applied to calculate the edges per path. You should know that > > absolutely converging series can be reordered. > > You should know that you can *not* reverse absolutely converging series. Proof? > And that is what you are doing. Because that would result in a series > without start. No reordering of an absolutely converging series changes its sum. > > > =============================== > > > I now understand what you were trying to do. > > > > It lasted rather a while. > > Consider it a complaint about your clarity and lack of definitions. Here are some readers who understood it quite easily. > > > > But in the complete tree > > >the parts of edges that are assigned to an infinite path are not 1, > > >1/2, 1/4 etc. There is *no* edge that is completely assigned to a > > > path, > > > > Then the tree is incomplete. You fail to understand infiniteness > > already on this low level? > > Circular reasoning abounds here. Care to give a *proof* that the tree > is not complete in that case? This statement is completely similar to > the statement that the completed set of natural numbers should have a > last element. There is no largest element, but the infinite path is nothing than the union of all finite paths. Therefore we can calculate the union of all finite sequences 1 + 1/2 + 1/4 + ...+1/2^n. > > Please state when you intend to no longer reason within ZF (especially > the axiom of infinity). As long as you are reasoningin in ZF, there > is *no* largest element of the (completed) set of natural numbers. > > > > Note that in each finite tree, the complete edge assigned to a path is > > > the last edge on the path. As there is no last edge in an infinite > > > path, this does not hold in the infinite tree. > > > > So there are different paths without different edges? > > No. I never did state that. So at least one path has its own edge? Regards, WM |