An uncountable countable set
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From: Dik T. Winter on 23 Jul 2006 21:15 In article <1153674914.649461.71180(a)m73g2000cwd.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > In article <1153478871.303029.5360(a)i3g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > But it is obvious for *all* *finite* *unary* number. And only finite > > > unary numbers are involved in the second list below: > > > > > > The *- sum of > > > > > > 0.1 > > > 0.11 > > > 0.111 > > > ... > > > 0.111... > > > > 0.111... is not a finite unary number. Rather, I would say it is not a > > unary number at all. > > 0.111.. is not a terminating rational number. Eh, can we please keep to the subject. You state you have a list that only finite numbers are involved in the "list" above. I state that 0.111... is not a finite unary number. Pray show that 0.111... is a finite unary number. (I neither have stated that it is a terminating rational number, because it is not.) -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 23 Jul 2006 21:49 In article <1153676088.467784.268720(a)b28g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > In article <1153482158.663876.226910(a)p79g2000cwp.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > Your hinting at any doubt about the clear meaning of Cantor's statement > > > is outrageous. > > > > Perhaps. In that case his statement was clearly wrong. > > But nobody seems to have observed that up to now. Yes, so what? Those papers are even no longer valid in current set theory. When developing a theory everybody is liable to take false turns. There is no need to note each and every false turn. > > > Wrong. My arguing is directed against the complete existence of the set > > > of all naturals, the set of all digits of a real number, the actual > > > infinity, the first transfinite number. That all is purest Cantor. > > > > As I said before. You are arguing agains the axiom of infinity. But your > > inconsitency proofs show nothing, because in part of your proofs you always > > use the negation of that axion. > > I do not use the negation of the axiom, but accept that every digit of > 0.111... can be indexed by a finite natural. Hence every sequence of You omit the "finite" between "every" and "sequence". > digits of 0.111... can be covered by a natural. (There is no digit by > which 0.111... could be distinguished from every natural.) But 0.111... > is the union of all of its finite sequences, i.e., of the naturals. Yes. And so? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 23 Jul 2006 21:42 In article <1153675669.168444.262950(a)h48g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > [reformatting to standard discussion style] > > > > In article <1153479321.840071.258130(a)b28g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > Dik T. Winter wrote: > > > > You again refrain from answering questions. The definition of 0.999... > > > > is as follows: > > > > 0.999... = lim{n -> oo} sum{k = 1 .. n} 10^(-k) > > > > You said that definition is silly. What is silly about it? > > > > > > Sorry, could you help me with "silly". My find function shows with > > > "silly" only the sentence: "It is only in modern set theory that such > > > silly things as the rejection of the binary tree occur." > > > > You actually said that it made no sense. But I will try to find the > > correct quote. > > Not necessary. I confirm: It makes no sense because omega makes no > sense. In what way makes the definition above no sense? I asked you before and now you finally reply that that is because omega makes no sense. While I do not understand that at all, this would mean that you think that all limits of the form lim{n -> oo} 1/n = 0 make no sense. > But may also be that I argued with respect to the finite number > of protons in the universe, which prohibits that natural numbers > indexes with 10^100 irregular digits can be written down or noted in > any form. I am not sure, because I could not find the original quote. No you did not state that at that time. It is nonsense. Mathematics is not concerned with the number of protons in the universe. > > You are confused. The count of paths have no limit. But if you wish > > to interprete the count of paths as a real number, be my guest, and > > tell me how to represent the count of paths as a real number. But > > whatever, if you were ever to complete your picture, both the number > > of edges and the number of paths would become uncountable (you did > > confuse me a bit by first talking about nodes and now about edges). > > It is the number of nodes that remains countable. The number of edges > > after adding the n-th splitting is 2n+1, the number of paths is 2n, and > > the number of nodes is n. > > ? In the tree the number of edges is the number of nodes (except for > the first one) because every node (except the first one) is the end of > an edge. In your drawing the edges are not terminated by nodes. At least, I would think that the symbol 'o' denotes a node. Apparently not. So let's see (I made an error in the number of paths, I think, but it is too hot here for too kong now to do any proper thinking). So you are comparing terminating paths with nodes. I have no problem there, the number of both is countable. But that way you will never get the representation of (say) 1/3, because that would involve a non-terminating path (that is, a non terminating sequence of binary digits). > > > > Again, you need something like a limit here, because adding nodes is a > > > > sequential process. > > > > > > Adding lines in Cantor's list is not a sequential process? > > > > Perhaps, but the list is given by some (unidentified process). Without > > the given list, there is even no start on the determination of the > > diagonal. The proof clearly states "given a list", so the assumption > > is there that there is a list. > > And the tree is given too by some unidentified process. Yup, so what? Given the tree it is easy to show that the number of paths is not countable but that the number of nodes is countable. Each node represents a finite path. Given the list, it is easy to show that there is some element that should be in the list but is not. What is the difference? > > Nope, the assumption is that you are able to give a list. And the proof > > shows that you are not able to. > > The proof accepts that I am able to give a *countable* list, i.e., a > list which is complete with repect to its enumeration by natural > numbers n e N. No. The proof *assumes* that the list you give is not only complete with respect to its enumeration by naturals, but also complete with respect to the numbers that have to be enumerated. > And my tree requires that I am able to give a tree with > *countably* many levels, complete with respect to N too. Your tree is. > > Yes. I state that adding levels (or individual nodes, but I grant that > > all nodes of the same level can be added simultaneously), is a > > sequential process. The definition of the list may or may not be a > > sequential process, but that is irrelevant. Given a list, the definition > > of the diagonal number is not sequential. > > 1) Every counting and every identification of the n-th digit is a > sequential process. Why? The list is as given (that is a mapping from N to the real numbers given in decimal notation). Getting the n-th member of the list is *not* a sequential process, so getting the n-th digit of the diagonal is *not* a sequential process. > 2) Given the tree, the calculation of edges per path however is *not* > sequential. As long as you are talking about finite paths, I agree. > > When > > something happens for natural n, and we can define a limit when n > > grows without bound, it is common to state that that limit is the infinite > > case in informal discourse. > > Just this is the mess in set theory. Infinity is not present in > magnitudes of numbers. It is most important to distinguish this. Eh? > > Yes, for each finite number n An covers all of A1 to An. No dispute about > > that (and I never disputed that). In shorthand: > > For each n, An = (*+){i = 1 .. n} Ai. > > True, without a problem. The problem is, what is: > > (*+){i = 1 .. oo} Ai > > you have *never* defined that one. You claim it is An for some n, but as > > you do not define it, and give no proof... oo is not a finite number, > > so we need a definition to reasonably argue about it. > > oo is not a finite number, but all numbers Ai of the sum (*+){i = 1 > .. oo} Ai are finite numbers. We have a definition for the case of > finite numbers. And only that counts. Wrong. In this case i ranges from 1 without bounds. Whe have a definition when i ranges to some fini
From: mueckenh on 24 Jul 2006 09:33 Virgil schrieb: > > Just that is in question. More precisely: Are there actually or only > > potentially infinitely many numbers. > > With respect to what axiom system are you asking this? > > Or do you maintain that there is some sort of absolute truth in > mathematics that does not require any assumptions at all? Of course. Real space is Euclidean or non-Euclidian irrespective of the axioms which mathematicians prefer. Aithmetics is finite or infinite irrespective of the axioms which mathematicians prefer. Do you really think reality would care if you set up axioms? > > > > The proof is given by the list: > > > > 1 0.1 > > 2 0.11 > > 3 0.111 > > ... ... > > w 0.111... > > That is not a list as presented. The upper part is a list. The last line is the LUB. All together ths may be called an "extended list" > > > > The set of unary numbers cannot result in the *+sum 0.111... . (I > > defined: the *+ sum is 1 if there is at least one 1 in it.) With > > respect to the left column this means the cardinality of 1,2,3,... > > cannot be w without w in it. > > Since "mueckenh"'s definition of "*+"is in all cases cited equivalent to > taking the LUB of the corresponding set of ordinals, omega (w) as an ordinal number appears only one n the sequence of ordinals. It does not appear in the list. 1/9 0 0.111... is the LUB of the list, if considered as a list of rationals, but w does not give the number of natural numbers. Because obviously their number is less than the number of numbers including w. To spell it out clearly: 1/9 is the supremum of many sets, including the list above, the set {1/9} and the set {1/10, 1/9}. But it is not expected to count the elements of the sets to which it is supremum. omega, however, is, according to Cantor, the number of elements of N. It can be put in order with other ordinals like the naturals and omega + n, to name only few. This is contradicted because we see from the above extended list, that the number of elements of N is *not* yet omega. It is less than omega unless omega appeared twice in the order of ordinals. > > Again you forgot the special from of unary representations. Is the > > special property of unary representation so difficult to see, so hard > > to remember? > > Properties of sets of numbers are not dependent on any representational > scheme. But their recognition depends on the representation. When I realized that the set of naturals cannot be counted by omega, I did not yet have this lucid and plausible scheme. Regards, WM
From: mueckenh on 24 Jul 2006 10:09
Dik T. Winter schrieb: > With the axiom of infinity there are infinitely many finite numbers. > And I still do not know the exact difference between actual infinite > and potential infinite. But that is (in my opinion) something that is > best left to philosphy. In set theory "infinity" means "actual infinity". Potential infinity can neither be counted nor be surpassed. > > 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. Here we see, that it appears only after all naturals and measures the cardinality of the extended list, not of the list of the naturals alone. > > > The set of unary numbers cannot result in the *+sum 0.111... . (I > > defined: the *+ sum is 1 if there is at least one 1 in it.) With > > respect to the left column this means the cardinality of 1,2,3,... > > cannot be w without w in it. > > You again state such without a proof, prove it, assuming that there is > no last element. Let's start with the list: > A1 = 0.1 > A2 = 0.11 > A3 = 0.111 > assuming decimal notation (because in that way we can define everything > we need in the easiest possible way). Now it is easy to define *+ on > two operands, Ap *+ Aq = Ap when p > q else it is Aq. And now it is > easy to do a finite set of operations, and we can proof that: > (*+){i = 1 .. n} Ai = An > no problem so far. We get a problem when we try to define the operation > for *all* Ai. As there are not finitely many Ai, we can not use the > finite case to extend to all elements of the list. The finite case applies as long a only finite numbers are concerned. How should a rule which is proven for any finite number become invalid for any other finite number? > In my opinion the > only sane way to define the *+ sum of all elements is to define it as > lim{n -> oo} An > but you apparently disagree. That is only a convention. We do not know if it holds. In order to find out, my list was designed. Therefore we cannot use it as input. > Whatever, with this definition we get that > (*+){i = 1 .. oo} Ai = 1/9 > and 1/9 is not equal to any of the Ai. We cannot arbitrarily forget that for all finite numbers (*+){i = 1 .. n} Ai = An. Because this is not an arbitrary definition but a result of the precise defintion of the *+sum of a column, namely to be 1 if at least one 1 is present. Therefore your arbitrary definition(*+){i = 1 ... oo} Ai = 1/9 cannot be correct, because 1/9 is not equal to any of the Ai. > > > So, why do you think that it is in the list, when the list is a list of > > > natural numbers in unary notation? > > > > It is not *in* the list, but has been added at that position which is > > due to its 1's, namely omega. This is justified because all numbers are > > listed at that position which is due to the 1's in their unary > > representations and because omega is a number (though not a natural) > > which covers only *one* place in the sequence of ordinals. > > Except that there is no digit position with ordinal number w in the > standard decimal notation. Yes, just that is the problem. All numbers of the true list have a finite number of digits. If it is claimed that all digits of 0.111... can be indexed, then 0.111... has omega 1's and then it belongs to the position omega and then omega is necessary to index it. > > > In both cases, when you insert the *+ sum early in the list you get either > > > (case 1) a non-natural number in the list, so it makes no sense, or > > > (case 2) a different rational in the list, and in this case the *+ sum > > > is that rational. But in the latter case I see no problem. > > > > In the list there are only terminating rationals. 0.111... has omega > > digits. Therefore we see perfect parallels. > > Again, you make no sense here. I think you are arguing (case 2). If > you add the *+ sum in the list the list no longer contains only terminating > rationals. Forall terminating rationals we have a terminating *+sum. > Apparently you are thinking that 0.111... is a terminating > rational. It has two different definitions, which in general are intermingled. The *+sum of the list is one of its members ( = natural numbers or terminating rationals). All members are terminating. Although we cannot determine their numbers of digits, we know: There are not infinitely many - in no number of the list. Transfer this knowledge to the *+sum. The second definition of 0.111... is the set theoretic actual infinite sequence of aleph_0 or omega 1's. You know and accept the difference between the list numbers and this latter definition. Why don't you accept the fact that a *+sum of finite list numbers cannot lead to this latter definition? Take the case of infinitely many numbers 0.1 0.1 0.1 .... Will you claim that the *+sum were something other than 0.1? It would in principle not be different from your claim that the rules change if there are infinitely many finite numbers as in my list. > What is the ordinal number of its last digit (and please mind > that in decimal notation the ordinal numbers of the digits are natural > numbers)? The last digit of 0.111... has no ordinal number. Therefore the set of its 1's and the set N has no ordinal number. Therefore it is wrong to define omega as the LUB of the naturals *and simultaneously as their cardinality*. omega is not a number which could be put in trichotomy with n e N and other ordinals like omega + n. > > > > > This is only possible, if it was in the list before already. > > > > > > Wrong. It is because all elements are idempotent under your *+ operation. > > > Union {a} with {b} is the same as union {a} with {b} and {a, b}. > > > > Again you forgot the special from of unary representations. Is the > > special property of unary representation so difficult to see, so hard > > to remember? > > And again switching back to unary representation. In unary representation > 0.111... is *not* a natural number. As the list contains natural numbers > only, that "number" is not in the list. However, what I said still holds. > Seeing the things 0.1, etc. as strin |