Cantor Confusion
in [Math]
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From: WM on 10 May 2007 07:18 On 10 Mai, 03:09, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > So in your opinion there are only finitely many paths in the infinite tree. > > > > I did not say that. > > Quote: "Only from those few you can ask for". I can ask only for finitely > many paths. That is correct. But that does not mean that I said that there are only finitely many paths in the tree, as you implied. > Not at all. I ask the single question: "are all other paths different from > a particular path?" That is circular, because by "other" you imply that path p can be distinguished by a node from all the others. But we know, that this not possible. Hence there is another path with p for all the time. If you state 1) p is not alone at any node. 2) There is no single path p' with p at all nodes. Then you should show at least two (or more) paths p' which are with p such that one of them is always with p but not always the same. Prove that more than one further path p' is required. You always state that this was true, but you cannot prove it. Is that mathematics? > In the same way: "what is sum{i = 1..n} i" a single > question. You want to split the first question into infinitely many > questions about all paths. In that case you should also split the second > question into infinitely many questions about all n. Using your logic > the split questions can only be answered for finitely many cases, so the > formula "sum{i = 1..n} i = (n + 1) * n / 2" is only valid for those cases > that you did ask. It is valid for those cases I did prove. I did not prove it (and one cannot prove it) for "all natural numbers which obey this formula", because all natural numbers which obey this formula do not obey this formula. > > > > > > There is a subtle difference. With the diagonal proof we always > > > > > remain in the finite. > > > > > > > > That's why it fails for numbers with infinitely many digits. > > > > > > No. > > > > Assertion, no argument. > > Yours was also an assertion without argument. My argument is: Why do you have to exclude the case 0.999... = 1.000... if you remain always in the finite? For every finite index both numbers are different. > > > > > No. Only for those p' you can ask for. There must be others, because > > > > there is no node which belongs only to p alone. > > > > > > I do not ask for each individual path. I show it for all. Or do you > > > reject the formula sum{i = 1..n} = (n + 1) * n / 2? You can not ask > > > the validity for all n, but only for finitely many n, according to your > > > thinking. So for each and every n you have to prove it again. > > > > This formula is obviously only correct for finite sets of numbers, > > Yes, so what? > > > because for the whole set the result aleph_0. > > Pray give a proof of this statement. Pray look it up yourself. Hrbacek and Jech, p. 188, for instance. > > > Do you think it would be > > possible to sum an infinite set by (n + 1) * n / 2? > > No, as such has not been defined in mathematics. But you reject the > formula above for any 'n' for which you did not ask. So each time you > use the formula with a different 'n' you should prove it for that particular > 'n'. It is obviously wrong for all n for which it holds, if such set of all n exists. > > > > > > > "completed" means nothing remains. > > > > > > > > > > Perhaps. > > > > > > > > Sure. > > > > > > Oh. A mathematical proof, please. > > > > The set of natural numbers is completed such that there is no natural > > number outside of that set. Do you disagree? > > Ah, now you define completed, I think. That meaning should be obvious without definition. > But it is not true that nothing > remains. We have still the rational numbers, the real numbers, and > whatever. Read again what I wrote, please: The set of natural numbers is completed such that there is no natural number outside of that set. > > > > > But again, in mathematics it is *not* necessary to test > > > > > each and every individual. Otherwise you could not even prove that > > > > > sum{i = 1..n} i = (n + 1) * n / 2 > > > > > for all n, but only for finitely many n. > > > > > > > > Of course this is only true for finitely many n. For infinitely many > > > > natural numbers, we get aleph_0. > > > > > > You are wrong. We do not get that. If so, pray show a proof. > > > > See, for instance, Hrbacek and Jech, p. 188, if you do not believe me. > > Again you misread what I state. I do not state "for finitely many i", I > state "for finitely many n". The formula is a statement about a single n. Only on the right-hand side. The left-hand side formula is a statement about a set. And here we can see again, that "infinite set of finite natural numbers" is a selfconradiction. > > > (There are more books which treat that topic, but this in my > > possession: > > It will not before Friday that I have again access to that book, but > context is *very* important. > > > "It is not very difficult to evaluate infinite sums. For example, > > consider > > 1 + 2 + 3 + ... + n + ... + ... (n in N). > > It is easy to see that this sum is equal to aleph_0." > > Off-hand I would state that they are wrong, because infinite sums are not > even defined in mathematics. To wit: > sum{i = 1..oo} > is shorthand for: > lim{n -> oo} sum{i = 1..n} > so there is a limit lurking behind. > > > What else should this sum be? It cannot be finite and it cannot be > > uncountable. > > In analysis it is undefined. As it is in algebra. And also in many > instances of set theory. But as I have stated already before, Hrbacek > and Jech give their particular definition of limits in ordinal arithmetic, > and with those definitions it makes sense. I think the statement is in > the context of that definition. Here is a part of the introduction, but I have not bothered to translate the Greek and other symbols: 10 ARITHMETIC OF CARDINAL NUMBERS 1. INFINITE SUMS AND PRODUCTS OF CARDINAL NUMBERS As we introduced arithmetic operations on infinite numbers, it is reasonable to generalize these operations and introduce sums and products of infinitely many numbers. For instance, it is natural to expect that 1 + 1 + ... = ï0 \____V____/ ï0 times or, more generally, ï« + ï« + ... = ï« ï ï¬. \____V____/ ï¬ times The sum of two cardinal numbers ï«1 and ï«2 was defined as the cardinality of A1 ï A2, where A1 and A2 are disjoint sets such that | A1| = ï«1 and |A2| = ï«2. Thus we generalize he notion of sum as follows. 1.1 Definition Let {Ai | i ï I} be a system of mutually disjoint sets, and let |Ai| = ï«i for all i ï I. We define the sum of {ï«i | i ï I} by Cardinality of Union The definition of ïi ïIï«i uses particular sets {Ai | i ï I}. In the finite case when I = {1, 2} and ï«1 + ï«2 = |A1 ï A2|, we have shown that the choice of A1 and A2 is irrelevant. We have proved that if A1', A2' is another pair of disjoint sets such that |A1'| = ï«1, |A2'| = ï«2, then |A1' ï A2'| = |A1 ï A2|. and so on. > > > > What is the relevance? But try to find the first occurrence where pi(x) > > > and Li(x) switch place. Nevertheless, it *is* a natural number > > > > Are you sure? > > Well, it should be a natural number, otherwise we could not even talk about > pi(x). We can talk about pink elephants. William enjoys it frequently. > > > What evidence is there that 2^65536 is a natural number? > > Why not? > > >http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&ha... > > Not mathematics, but logic. No mathematics without logic. Regards, WM
From: WM on 10 May 2007 07:22 On 10 Mai, 03:43, Virgil <vir...(a)comcast.net> wrote: > In article <1178737478.076726.140...(a)n59g2000hsh.googlegroups.com>, > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > I did not say that I could name a path which does never branch off > > from p. But there must be such a path. Why not call it p''. > > Why must there be any different path that is not also a separate path? > > > > > > <snip> > > > > >The proof in the tree is, that for > > > > every path p, and level M, there is another path *existing in the > > > tree* > > > which is not different from p at a level less than or equal > > > to M. > > > Yes. But why only see the one side of the medal? For every path p > > there is another path *existing in the tree* which is not different > > from p at any node, because p is not single at any node. > > False!!! > > Given any path p and any path p', if there is no node in one but not > the other, then p = p'. That is completely irrelevant! Give me an example of two nodes, such that p' is not with p at the first but the second, and p'' is not with p at the second but the first. Then I will believe you. Otherwise you are wrong. Regards, WM
From: William Hughes on 10 May 2007 07:42 On May 10, 6:24 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > On 9 Mai, 22:07, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > On May 9, 2:49 pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > On 9 Mai, 17:39, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > On May 9, 11:23 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > > On 9 Mai, 16:10, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > > > On May 9, 9:39 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > > > > On 9 Mai, 01:10, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > > > > > > > > But why do you write down these paths: > > > > > > > > > > > > Let p'(1) be the path 0111... > > > > > > > > > > > Let p'(2) be the path 00111... > > > > > > > > > > > Let p'(3) be the path 000111... > > > > > > > > > > > > ? > > > > > > > > > > > Any set p(1) up to p(n) can be substitued by a single path. > > > > > > > > > > In fact? Why then did you write down so many useless paths? > > > > > > > > > > > Look! Over there! A pink elephant! > > > > > > > > > > > The set P' can be substitued by a single path. > > > > > > > > > > Oh, does the set P' consist of any paths which are not paths p(n) with > > > > > > > > > a finite argument n? > > > > > > > > > i: Any set of paths {p(1) ... p(n)) can be replaced by a single > > > > > > > > path. > > > > > > > > Why then did you write the unnecessary paths? > > > > > > Why then did you write the unnecessary paths? > > > > Why then did you write the unnecessary paths? > > > > > > > > > ii: The set P' is the union of sets {p(1) ... p(n)} > > > > > > > > and as such a set of nodes. > > > > > > > > In order to avoid the clumsy expression of path bunches, I call a path > > > > > > > bunch now a finite path. A finite path ends at some node. > > > > > > > > The infinite path {0.000...} is the union of all finite paths with > > > > > > > only 0's, namely {0.} U { 0.0} U {0.00} U ... > > > > > > > The union is one infinite path and, therefore, has not more paths than > > > > > > > were unified. > > > > > > > There are only a countable number of finite paths in an infinite path > > > > > > The infinite path {0.000...} is the union of all finite paths with > > > > > only 0's, namely {0.} U { 0.0} U {0.00} U ... > > > > > accepted? > > > > > Yes, but don't let the pink elephants distract you. > > > > The infinite path {0.000...} is a union of finite paths, > > > > or a *set* of nodes. > > > > The number of these sets is 1, i.e., less than the number of sets > > > unioned. > > > The number of infinite paths equal to {0.0000....} is 1. > > > Therefore, the number of infinite paths equal to {0.000...} > > is less than the number of sets unioned. > > > Look! Over there! A pink elephant. > > > The number of infinite paths is [more] than the number of sets unioned. > > Did you want to write that? You unveil your secret! > Let N be the set of nodes. Let P_I be the set of infinite paths. We now have N is countably infinite. Each element of P_I is a countably infinite subset of N. Note: There are more countably infinite subsets of N than elements of N (no contradiction, each countably infinite subset uses more nodes, an infinite number, than it produces subsets, 1, but each node is used more than once). There are an uncountable number of countably infinite subsets of N. The number of infinite paths is the number of countably infinite subsets of N which are infinite paths. - William Hughes
From: William Hughes on 10 May 2007 07:52 On May 10, 6:32 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: <snip> > Give me two levels for > which two different paths are required. There do not exist two levels for which two different paths are required. Therefore Any finite set of paths can be replaced by a single path. Look! Over There! A pink elephant! Any infinite set of paths can be replaced by a single path. - William Hughes
From: WM on 10 May 2007 10:02
On 10 Mai, 03:40, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1178716585.114658.276...(a)h2g2000hsg.googlegroups.com> WM <mueck...(a)rz.fh-augsburg.de> writes: > ... > > > > > There are subsets of a finitely definable set which are not > > > > > finitely definitable. > ... > > There does not exist anything in mathematics, unless it is finitely > > definable. In particular, any infinite definition is not a definition, > > because there is a definition of "definition" which says: Definition: > > A definition has an end, i.e., a last word and a point. > > Try to find a finite definition of the set of Ulam's lucky numbers. Here it is. Use the prime sieve of Eratosthenes but don't count the numbers already eliminated. What is the problem? The set is finitely defined. Not all lucky numbers can get finitely defined. Regards, WM |