An uncountable countable set
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From: mueckenh on 26 Jul 2006 17:14 Dik T. Winter schrieb: > > > > > > > The proof is given by the list: > > > > > > > > > > > > 1 0.1 > > > > > > 2 0.11 > > > > > > 3 0.111 > > > > > > ... ... > > > > > > w 0.111... > > > > > > > > > > That is neither a proof, nor a list. > > > > > > > > The list is given in the upper part. Call the whole an extended list. > > > > It shows that w is not the number of the naturals, because w appears in > > > > the sequence of ordinal numbers only once. > > > > > > That is only a statement, *not* a proof. > > > > Why is the ordinality of the left column of the extended list another > > than the ordinality of the right column? > > I do not state so. The ordinality of both "lists" is w+1. Why do you think > they are different? Look at the list. The ordinality of the number 1 2 3 .... n is the same as that of the unary number in the corresponding line 0.111...1 (with indices 1,2,3,...,n) as long as n is a natural number. But the ordinality of 1 2 3 .... w differs from the ordinality of the unary of the corresponding line 0.111... This proves that there is no possibility to interpret w as the number which counts the naturals. And this interruption is independent of the first ordinal, whether you may choose 0 or 411 as the first one. If you arrange the list such that the unary numbers are the same as the numbers in the left column, then it is impossible to continue into the infinite. Here is an example starting with 3: 3 0.111 4 0.1111 5 0.11111 .... w 0.111... > > In general, the ordinal number of any initial segment of ordinal numbers > is the smallest ordinal number that is larger than all ordinal numbers in > that segment. (This can be done because the ordinal numbers form a > well-ordered set.) And note that 0 is also an ordinal number (of the empty > set). So the ordinal number of {} is 0, of {0, 1, 2} is 3, etc. > Finally, the ordinal number of {0, 1, 2, 3, 4, ..., w} is w+1. And so > is the ordinal number of {1, 2, 3, 4, 5, ..., w}, because there is an > order preserving bijection between the two. > > You apparently did not appreciate that 0 is an ordinal number. (And that > is independent of Bourbaki.) > -- That makes no difference. 0 0. 1 0.1 2 0.11 3 0.111 .... w 0.111... The interruption between finite and infinite remains. Regards, WM
From: mueckenh on 26 Jul 2006 17:18 Dik T. Winter schrieb: > > > An arbitrary union of finite sequences is not necessarily a finite > > > sequence. This is similar to: the union of finite sets is a finite set. > > > But this argues that the union of {1}, {1, 2}, {1, 2, 3}, ... is a > > > finite set. Consider the elements of those sets as indices to the > > > elements of the sequences. This is only true when N is finite, but that > > > is the negation of the axiom of infinity. > > > > True is that the elements of n are finite. Hence there union cannot be > > infinite, because every union of natural numbers is a natural number. > > Now you are talking about unions of natural numbers. Makes no sense to > me in this context. Sorry, my fault. I meant unions of finite segments like {1}, {1, 2}, {1, 2, 3}, ... remain a finite segment. Because we can identify tem with natural numbers which always remain finite. Regards, WM
From: Franziska Neugebauer on 26 Jul 2006 17:20 mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > >> > T. Jech is one of the leading set theorist. >> > He says up to the advent of set theory there was no use for actual >> > infinity. >> >> 1. This utterance is not set theroretic use. > > It explains and distinguishes what infinity means in set theory in > comparison with analysis. You are obviously the victim of a homonym. > Aren't you even able to understand such > simple texts? It has *nothing* to do with his wife. > > >> 2. There is also no use of "actual" vs. "potential" infinity after >> the advent of set theory either. > > You continuously contradict yourself. Compare Hilbert's statement, > which he uttered *after* the advent of set theory. I neither will compare Hilbert's statements nor will I discuss the timeline of set theory. > You had the chance to read it most recently and to learn something > about that topic. No thank you. There's no demand for antiquarian books. >> If you can't take criticism you'd better not post. > > Critizism is welcome, senseless gossip of ignorants is not. May I call your attention to the correct spelling of criticism? F. N. -- xyz
From: Dik T. Winter on 26 Jul 2006 18:46 In article <1153897104.622376.322420(a)b28g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > In article <1153862905.806404.152570(a)i3g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > > > > Dik T. Winter schrieb: > > > > > > > > I think that my "list" has proved that (among other proofs). > > > > > > > > Your list proofs nothing. > > > > > > (every digit of n can be indexed) ==> (n is in the list) > > > > I ask for a proof, not for a restatement. Pray prove the above using > > common mathematical and logic methods. It is false. > > It is the *definition* of the list! Every digit which can be indexed > can be indexed by a natural number. Every natural number is in the > list. Yes, every digit of n can be indexed, so every index is in the list. How do you conclude that hence 'n' is in the list? By your logic we can state: (every element of a list can be indexed) ==> (the list is in the list). > I think it is in vain to discuss with you if you do not accept the > basics of logic. I completely reciprocate the complement. > > > I do not mention these details because for doing practical mathematics > > > they are not important. > > > > Oh, so you are distinguishing practical mathematics and mathematics. And > > you are teaching pupils things that (according to you) things that make no > > sense. > > Of course they make sense. I teach Newtonian mechanics too though I > know the relativity is the better theory. And you also have stated that they do not make sense? What is it? > > > So the triangle gets infinitely long but not infinitely broad. This > > > means the Diagonal must bend. > > > > Eh? Where do you conclude *that* from? The triangle gets infinitely long > > and infinitely broad. > > Wrong. Infinitely broad would mean infinite numbers. Why don't you see > such things? If you allow the triangle to complete, as you wish, that would be true. But I use "becomes infinitely" in a loose sense. I properly should have said: "The triangle grows without bound, bot in length as in breadth". > > It remains a straight line, see above. > > > Infinitely broad would mean infinite numbers. Infinitely long would also mean infinite numbers. But the common terminology is the loose sense I mention above. You use the term as if it were actually possible for one of them to actually become infinite. > > > In any case the number of nodes or edges is larger than that of paths. > > > > Yup. > > > > > On the other hand, if you start with a > > > > path: > > > > o > > > > | > > > > and use the basic figure: > > > > | > > > > o > > > > / \ > > > > you get other information. > > > > > > Not at all. The number of infinite paths is always less than the number > > > of edges. > > > > But in this case neither the paths, nor the edges are terminated. > > So what? In an infinite tree there is no terminaton. But according to > the axiom of infinity "all levels of the tree exist". Right, what is the problem? All levels do exist. With the first picture that means that all terminating paths do exist in that picture, and not the non-terminating paths. With the second picture that means that also all non-terminating paths do exist. You are still strugling with the concept that something that grows without bound does not terminate. > > > So while > > in all cases the number of edges is less > > it is larger by a factor of 2. I have some problems with the first picture when we let it grow without bound, I have to think about that (but it fits in a way in Conway's book). With the second picture there are no such problems. We start with a single node and edge. After n additions there are: n+1 nodes, 2n+1 edges and n+1 paths. (I thought you also meant intermediate paths that stop at any node, apparently you do not). When we allow the final picture, we get aleph-0 nodes, but 'c' edges and paths. You can assign the symbols 'l' and 'r' to the paths, edges and nodes at each split. When you properly do that you will find that the nodes represent all *terminating* sequences of those symbols and the paths represent all *non-terminating* sequences of those symbols. (On purpose I do not assign numerical values here, because in that case we have to bother about multiple representations.) The edges represent a combination of all *terminating* sequences and all *non-terminating* sequences. The diagonal argument does not apply to the set of all *terminating* sequences, because the diagonal is (of necessity) non-terminating. And indeed, we can assign a natural number to each terminating sequence. Count the number of sequences of smaller size (that can easily be done). Order the sequences of the given size lexicographically (that also can easily be done), and count along that list until you reach the sequence you want. You might even derive a formula that given a sequence gives you the natural number assigned to it. The diagonal argument *does* apply to the set of non-terminating sequences. There is no way to assign a natural number to each of those sequences. So the cardinality of them is 'c' (or at least larger than aleph-0, but introducting numerical representations, allowing for multiple representations, etc., we can show that it actually *is* c). The cardinality of the set of edges is aleph-0 + c = c. Apparently you are concerned that n+1 can lead to either aleph-0 or to c, when we complete the picture. Well, yes, that is the way it is. That is not in itself an inconsistency. > > I need to do nothing in this sense at all. You apparently think that I > > have to, in fact, change each a_nn. But there is no need for that. I > > just state that if a_nn is 4, than I can use 5, if it is not, I use 4. > > It does not bother me whether it is 4 of not, what is important to me > > is that what I find is definitely not equal to an. > > You cannot find it and cannot know the number found without having it > determined. The list given together with my rule define the diagonal. My rule is independent of the list. But it is easily shown that given a list and the rule, the diagonal is not in that list, because it is different from each element, and so the list is
From: Dik T. Winter on 26 Jul 2006 19:10
In article <1153947128.476764.274120(a)b28g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > Let's state that more proper. The index of every digit that can be > > indexed is in the true list (of natural numbers). > > Because this is the list of all indices. > > > But later you use > > the word "index" to mean something completely different. You are > > stating there that if every digit of a number can be indexed the number > > *itself* is in the list. But that does not follow. You need a proof > > for that. > > It does not *follow* but it is he definition of the true list. This > list contains all naturals, hence all indices. It is the complete set > N. Therefore there is no proof required (and possible). This is illogic. What you are indexing is the list of digits of K. You are claiming that K is in the list of indices. That makes no sense at all. > > And there is no such proof. > > In order to save set theory with its "infinite set of finite numbers" > you are forced to assert that the list of all natural numbers does not > contain the list of all natural numbers. Yes, but that is obvious. As the list of all natural numbers is not a natural number, it is not in the list of all natural numbers. > > As I stated earlier, every > > *terminating* number of which all digits can be indexed through numbers > > in the list is itself in the list. That is trivial. The last digit > > index position is the number itself. But that is *not* true for a > > non-terminating list, as it does not have a last index position. > > But every index number has a last position. Yes, so what? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |