An uncountable countable set
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From: Virgil on 21 Jul 2006 14:24 In article <1153482677.919976.211910(a)b28g2000cwb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > > The "first" successor. Tere is only one "first" successor! > > > > Conveniently snipping your own text again? You wrote in > > <"news:1153052131.852951.273540(a)b28g2000cwb.googlegroups.com">: > > > Hence, > > > 0.111... as the succesor of all naturals must consist of more 1's, than > > > any natural, if it is to be a number. "Mueckenh"'s "*+" operation is merely taking the SUP ( or, equivalently, LUB) of a bounded set of ordinals, so is a well defined ordinal. > You said "for all p". If p is a natural number, then the segment K[p] > is the unary representation of a natural number and, hence, is covered > by a natural number. Equally, for every bounded set of ordinals there is a unique SUP (or, equivalently, LUB) which is defined as the smallest ordinal greater than or equal to every member of the set. Thus the list of all finite ordinals has the same SUP as the list with its LUB appended as a member. Finding the SUP is really what "mueckenh"'s "*+" operation is really all about. And for that you need to consider the properties of ordinals.
From: Virgil on 21 Jul 2006 14:44 In article <1153483340.080263.55660(a)m79g2000cwm.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Again: For any *finite* natural n we have a logical equivalence, even > better an inclusion: indexing the n-th digit includes covering at least > all digits from 1 to n. > If each digit of 0.111... is indexed by one natural, then each segment > of 0.111... is covered by at least one natural too. > As long as only natural numbers are concerned there is no outcome. "Mueckenh"'s so called "*+" operation on the finite ordinals, with or without inclusion of their limit ordinal, is logically identical to the LUB or SUP operation on bounded sets of ordinals. And the LUB of the set of finite ordinals is the same as the LUB including the limit ordinal. > > > > > > The reason is that > > > > > > 0.1 > > > 0.11 > > > 0.111 > > > ... > > > gives allegedly the same sum as > > > > > > 0.111... (as representation of aleph 0) > > > 0.1 > > > 0.11 > > > 0.111 > > > ... > > If aleph 0 is a number larger than any natural, then it cannot be the > *+ sum of a list containing only naturals excluding itself and > simultaneously the sum of a list including itself. "Mueckenh" is misnaming his operation as a "sum" when it is in reality a least upper bound (LUB) or supremum (SUP) of a bounded set of ordinals. And then his 0.111..., is merely the LUB of {0.1,0.11,0.111,...} and is equally the LUB of ({0.1,0.11,0.111,...} union {0.111...}). > > > You apparently have another definition in mind. Pray give *your* > > definition. > > There are two definitions conceivable. Both lead to a contradiction. > 1) If "..." means "forall" n, e N then 0.111... is not different from > all n. (Because for naturals indexing includes covering.) But we know > that 0.111... is not n the list of naturals. Contradiction. > 2) If "..." means --> oo including aleph 0 as an index, then the digit > with this index is not defined. The contradiction arises only by assuming that the "*+" has properties other than that or the SUP or LUB operator on bounded sets of ordinals. Clearly for every finite subset of {0.1, 0.11, 0.111, ...} the "*+" operator and the SUP (or LUB) operator produces identical results, and for any infinite subset, both produce 0.111... . When viewed correctly, "mueckenh"'s claim of contradiction vanishes. > I ask questions too which you do not answer. The most important one: > How can a finite natural number index digit number n but not cover all > digits from 1 to n? Because a finite ordinal has only one ordinal value. > > > > 0.1 > > > 0.11 > > > 0.111 > > > ... > > > > As my interpretation of your "..." was apparently wrong above, pray, finally > > supply a definition of that notation. > > See above. 1) and 2). > > > > How many digits has 1/9 in decimal representation? > > > > That is not a natural number. > > No, but the set of digits has a cardinality, namely aleph 0. > > > There is no such column. And 0.111... is not a natural. > > That is a contradiction. And I told you why: > > You said: all digits of 0.111... could be indexed by natural numbers, > but not all could be covered by natural numbers. This is wrong as one > can easily prove: Every natural which indexes a digit covers all digits > up to that one. If all digits of 0.111... can be indexed than all > digits up to every (and that is all digits) can be covered. > > What is in *your* > > opinion the result of *+-ing all natural numbers togther? > > As we have seen it is simply impossible to *+ sum all naturals. Every > result yields a contradiction. The set of all naturals does not exist. It is quite possible, and even trivial, using von Neumann ordinals, to do a SUP or LUB of any infinite set of finite ordinals, and that SUP or LUB is trivially another ordinal greater than any finite ordinal.
From: Virgil on 21 Jul 2006 14:56 In article <1153499310.820621.218440(a)m73g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > > Now you should be able to see the following about *+ sums: Mueckenh"'s "*+" sums are, in fact, merely LUBs of bounded set of ordinals in drag. There is no instance in which "mueckenh" can show that his "*+" operation gives a different result than the LUB operation on the set of finite ordinal numbers. NOTE: every bounded set of ordinals has a unique LUB. Some such sets contain their LUB and others do not. E.G. : S = {0.1, 0.11, 0.111, ...} has LUB 0.111... not in S the set itself. T =({0.1, 0.11, 0.111, ...} union {0.111...} has the same LUB which IS in T. LUB(S) = LUB(T) but that LUB is NOT a member of S and IS a member of T. Perhaps if "mueckenh" cold ever get his thoughts in order, he would be less disordered by ordinal numbers.
From: Dik T. Winter on 21 Jul 2006 20:31 In article <1153478099.434931.273560(a)m79g2000cwm.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > Where in the guidelines, and where in the > > consitution, is it layed down that zero is considered a natural number? > > I have access to both, so pray give pointers. > > I do not remember the source, but it was a trustworthy one. Yes, so what am I to say about it? I simply do not believe it. And do *not* come up with the objection that I should show that it is not there, unless you want me to post all the guidelines plus the constition to let you show it is not there. > > When you can derive, > > within that system, contradicting conclusions, your axioms are inconsistent. > > In principle, no other knowledge than the axioms and the definitions is > > needed. > > Within this system I defined a *+ sum and showed that > > The *- sum of > > 0.1 > 0.11 > 0.111 > ... > 0.111... Nope. You did *not* define how to do a sum of order w, nor did you define how to do a sum of order w+1. And when I tried to interprete it you said I was wrong. So *define* the meaning of "...", and not through examples. > This is a contradiction, because by tertium non datur only one case can > apply. One of these results is wrong if 0.111... represents a number. You are again making no sense. Using your *+ notation: 0.10 *+ 0.01 = 0.11 (sum not in the list) 0.10 *+ 0.01 *+ 0.11 = 0.11 (sum in the list) And here I use how you defined *+ addition, and used finite addition only (the only thing you have ever defined). And do not come with the objection that I use 0.01. Because all of 0.10, 0.01 and 0.11 can represent a number (1/5, 1/25 and 6/25 to give an example). -- 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 21 Jul 2006 20:59
In article <1153478388.042474.295770(a)m79g2000cwm.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > In article <1153301881.393720.57330(a)75g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > Dik T. Winter schrieb: > > > > > If the *+ sum of the list is a unary representation of some 1's, > > > > > then, > by the construction of the unary numbers of the list, > > > > > this sum must be a unary numer of the list. > > > > > > > > As long as your list is finite. > > > > > > As long as it contains *numbers* which can be distinguised from each > > > other and which obey trichotomy. > > > > No, as long as your list is finite. > > The numbers count the lines. As long as there are only finite numbers, > the list is finite. Why? There are infinitely many finite numbers. You are trying to establish that the Axiom of Infinity leads to a contradiction. But your very statement "As long as there are only finite numbers, the list is finite" is in itself in contradiction to the axiom. The statement is just opinion, without proof. So there is not yet a contradiction. Please show a proof of your statement. > But as you have not the understanding, consider the related case: > A finite number covers what it indexes. This does not depend on how > much numbers are in the list. > From this discovery you may obtain the result for what you call an > infinity list. Pray, can you not write in an understandable manner? I have really *no* idea what you are writing here. > > > If we assume that non-terminating fractions have aleph_0 digits, then > > > this case is realized by > > > > > > 0.1 > > > 0.11 > > > 0.111 > > > ... > > > 0.111... > > > > > > Here the last number of an infinite list is in this list. > > > > I thought we were talking about natural numbers. I have not yet seen a > > definition that calls the last one a natural number. > > I never said so. So, why do you think that it is in the list, when the list is a list of natural numbers in unary notation? > > You should *not* > > switch between representations during the process. > > It is important for the discovery. But they give different information. When the "notations" are unary representations of natural numbers, their *+ sum is not the representation of natural numbers, and so is not in the list. When the "notations" are decimal (or ternary, or whatever) representations of rational numbers their *+ sum is a representation of a rational number that is not in the list. 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. > > Either you have a > > list of natural numbers (in that case 0.111... does not belong to it) > > or you have a list of rational numbers in decimal notation. The latter > > is easier to reason about because it can be tackled easier. In that > > case your list consists of: > > An = sum{i = 1..n} 10^(-n) = (1 - 10^(-n))/9. > > Also we can easily show what repeated *+ is (notated here as SUM): > > SUM{i = 1 .. k} Ai = Ak. > > Now we can also get a proper definition about what your notation ... means > > in the list: > > lim{k -> oo} SUM{i = 1 .. k} Ak = 1/9 = K. > > And K is (by convention) decimally notated as 0.111..., which makes sense, > > in my opinion. K clearly is not equal to any of the An, unless there is > > a natural number for which 10^(-n) = 0. > > Nevertheless we have the *+ sum 0.111... from > > 0.1 > 0.11 > 0.111 > ... > > Now we can insert this in the list and we find that there is no change > in the *+ sum. Yes, so what? There are enough, even finite, cases where the *+ sum does not change when you include the *+ sum in the list. The reason is simple, All elements you use are idempotent under *+. > 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}. > Then call > > 0.1 > 0.11 > 0.111 > ... > 0.111... > > an extended list or an ordered set or what you like. The proof should > not fail by lack of names. Which proof? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |