An uncountable countable set
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From: Tony Orlow on 25 Aug 2006 09:26 Virgil wrote: > In article <44ed9670(a)news2.lightlink.com>, > Tony Orlow <tony(a)lightlink.com> wrote: > >> MoeBlee wrote: > >>> Please say what sentence and its negation you believe are both theorems >>> of set theory. >> I'm not sure how this would be proven in set theory (I don't think it >> is), but it appears to be a belief, anyway, that all sets can be >> classified through cardinality. > > In ZF it is NOT the case that every set must have a cardinality > comparible with that of other sets, but it is a thereom in both ZFC and > NBG. > Good, so there must be a cardinality for the set of bit positions in the naturals. > > >> However, the set of bit positions >> required to list the naturals in binary defies classification in this >> system. > > On the contrary, that set of bit positions has cardinality equal to that > of N. The fact that the same set of bit positions is capable of more > does not mean that there need be any smaller set which is sufficient. That depends on how many more it produces. An extra bit doubles the number of strings, so if you have gotten to the least bit string necessary for n naturals, you will have at most n-1 unused strings. Do you have at most aleph_0-1 unused strings, when using aleph_0 bit positions for the naturals? No, you have an uncountable number of unused strings, according to your theory. > > There are exact analogies in finite cases. E.g., the number of bits > required to ennumerate all the naturals up to, say, seventy is > sufficient to ennumerate considerably more than seventy. But less than twice seventy. Can you get within a factor of 2 for your set of naturals? Nope. > > In fact the only cases in which the same set of binary digits will NOT > enumerate more is when you want them to enumerate every natural up to > 2^n-1 for n=some natural n. Uh huh. So? > > As these cases become ever more rare as n increases, they are in effect > infinitely rare for infinite n, so the result that TO object to is quite > normal . and any other would be infinitely unusual. I am not sking for an exact number of bits, but just an acceptable cardinality for this set. There is none. >>>> I have presented a system >>> No you haven't. You've posted disconnected pieces of undefined >>> terminology. >> No, I've shown how IFR works with the notion of Big'un. > > You have made all sorts of unsupported claims, but one of them work in > any extant system. > If you say so. > >>>>>> Define "truth". >>>>> Formal definition is given in a formal meta-theory. Greatly simplified, >>>>> a sentence S is true in a model M iff the evaluation function per M >>>>> (definition of this function given courtesy of the defintion by >>>>> recursion theorem) with the sentence as argument yields the set of all >>>>> functions on the variables into the domain of the model. >>>> Maybe that's a little simplified. Not sure what you're saying. Sorry. >>> My statement is too compressed and not pinpoint accurate given the >>> limitations of a one paragraph answer. See Enderton's mathematical >>> logic textbook. The full formalization is culminated in the exercise in >>> which you are asked to make sets of functions from the variables into >>> the universe the values of a certain recursively defined function. >>> >>> MoeBlee >>> >> Okay. That's not a very fundamental definition, as far as I can tell. > > > Let's see if TO can do any better, then. > > So, TO, what is YOUR definition of TRUTH? Truth is the value put on a statement, from 0 to 1. Truth is consistency with reality. There are a few ways to talk about truth.
From: Dik T. Winter on 25 Aug 2006 10:32 In article <1156500654.035798.315570(a)m79g2000cwm.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > > > You apparently do not know what the word "computable" does mean. To > > > > be precise: all algebraic numbers are computable in the mathematical > > > > sense. > > > > > > To be precise: Each algebraic number is computable. But not all. > > > Because then the list of all algebraic numbers was computable. > > > > But it is. > > So you can compute all solutions of a polynomial equation even of > higher than fourth degree in finite time? I doubt that. I did not state that. I said that the numbers were computable, where I use the mathematical sense of computable. -- 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 25 Aug 2006 10:36 In article <1156501674.444594.242900(a)74g2000cwt.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > > It does to any practicable approximation (not by my logic but in > > > reality). > > > > Mathematics is not about reality... > > It is. You have not yet recognized it. it is not. But you have not yet recognised it. > > > > And why is the notation lim n --> oo not clear. You have defined it just > > > > a few days ago: > > > > > > > The same is meant as in sequences and series like: > > > > > > > lim [n --> oo] (1/n) = 0 > > > > > > > n becoming arbitrarily large, running through all natural numbers, but > > > > > > > always remaining a finite number < aleph_0. > > > > > > That is correct as long as n is a natural number. > > > > You said (see above) that the notation was not clear, while it was clearly > > a natural number. > > The notion n --> oo is not clear, often, because it is intermingled: > 1/n = 0 in the limit and n remaining a natural which leads to 1/n > 0 > in the limit. THe notion n --> oo is not clear, indeed. But you said above that the notion lim n --> oo is not clear, while you gave a definition (that is not clear in my opinion at all). However, in mathematics, the notion lim{n -> oo} 1/n is clear and well defined. > > How do you *define* remainings? And I see now that you switched from your > > original, which read: > > > lim [n --> oo] |{2, 4, 6, ..., 2n}| / |{1, 2, 3, ..., n}| = 1 (because > > > omega is a fixed, well defined and well deteremined quantum, according > > > to Cantor) > > What are the remainings in this case? > > There are none. I did not ask you whether there wer, I asked you how you define them. > > > Division was possible and was practised in fact long before rings and > > > fields were known. > > > > Yes? On cardinal numbers? > > Of course, on all numbers which were known at that time, on all numbers > which deserved the name number. The old Greek practiced it in geometric > form. > > > > For *finite* cardinals. Let's see: > > |{1, 2, 3, ...}| / |{2, 4, 6, ...}| > > I can do (among many others): > > (1) I start with a bijection between {1, 2, 3, ...} and {2, 4, 6, ...}; > > so the quotient is 1 without remainder. > > (2) I start with a bijection between {1, 3, 5, ...} and {2, 4, 6, ...}; > > so the quotient is 1 with a remaining that also yields a bijection, > > so the total quotient is 2 without remainder. > > (3) I start with a bijection between {2, 3, 4, ...} and {2, 4, 6, ...}; > > so the quotient is 1, with remainder 1. > > Depending on how I chose my bijections I can get every finite quotient > > with every finite (non-negative) remainder. So, which one should I chose? > > We need not get into these details. It is enough to know that the > result is > 1 or not. We need. If you do not properly define the notions you are using, it is not possible to get any sense out of a discussion about those notions. > But we can do without any cardinal division: > A n e |N: |{2, 4, 6, ..., 2n}| / 2n < 1 > but lim[n-->oo] |{2, 4, 6, ..., 2n}| / 2n > 10 (in order to avoid > infinity) > > > And, what is: > > |{2, 3, 5, 7, 11, 13, ...}| / |{1, 4, 6, 8, 9, 10, 12, ...}| ? > > Not necessary to define for my proof. No. But you said that division could be defined. -- 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 25 Aug 2006 10:40 In article <1156501872.777344.96270(a)m79g2000cwm.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > Some clarification. > > > > [ About Cantor's proofs:] > > I was not even completely correct in that statement. After completely > > reading his first proof and his second proof (the diagonal proof), I am > > quite sure that both his proofs were targeted at the following theorem: > > There are sets with a cardinality larger than that of the natural > > numbers. Note that this follows from the first part of the first paragraph on page 278 of his Werke. > > The first part of his first proof shows that a complete ordered field > > has cardinality larger than the natural numbers. In his proof he did > > not rely an any properties of the reals other than that they form a > > complete ordered field (he uses reals to exemplify). > > "Wenn eine nach irgendeinem Gesetze gegebene unendliche Reihe > voneinander verschiedener reeller Zahlgr=F6=DFen .... vorliegt, so l=E4=DFt > sich in jedem vorgegeben Intervalle ...eine Zahl ... (und folglich > unendlich viele solcher Zahlen) bestimmen, welche in der Reihe (4) > nicht vorkommt; dies soll nun bewiesen werden. > Can you read German? Reelle Zahl, Zahlgr=F6=DFe, Zahl, Zahl, Zahl. 10 > times "Zahl" appears in this paper. Note what I wrote: "in his proof he did not rely on any properties of the reals other than that they form a complete ordered field". What other properties of the reals did he use? > Cantor uses neither "field" nor "ordered" nor "complete". Yes. Pray read again what I wrote. > > Cantor's one and only purpose was a proof of the theorem: > > There are sets with cardinality larger than that of the natural numbers. > > He wrote that it was a simpler proof for his first theorem, i.e. the > theorem that the reals are uncountable. And he did *not* write that. -- 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 25 Aug 2006 11:10
In article <44eef4aa(a)news2.lightlink.com> Tony Orlow <tony(a)lightlink.com> writes: > Dik T. Winter wrote: .... > > > And, with the addition of which natural number to the set does the set > > > achieve this unnatural size? > > > > There is no such specific natural number. It is when we have them all, > > but as there is no largest number, this can not be achieved by taking > > them one by one. > > The set of all naturals numbers consists of only natural numbers. There > is NO natural number where the count becomes infinite. So there is no > point in the set, even if you COULD get to the "last" one, where any > infinite set has been achieved. And there is no point in the set where you have the complete set. Yes. Indeed. So what? > > > > Then show that the set of all natural numbers does not have > > > > cardinal number aleph-0 and ordinal number w. A proof please. > > > > > > I have already shown how the set of bit positions in the binary > > > naturals has no cardinal or ordinal that can be assigned to it. > > > > Yes, but your showing was not a proof. > > Yes, it is. A finite number of positions is too few, so it must be > infinite, but the smallest infinite is too large. Indeed. The set of all natural numbers is just sufficient. > So, you have a > contradiction when trying to assign a cardinality to this sequence of > bits. There is no cardinality for K which does not produce a contradiction. But there is. It is the cardinality of the naturals, and that is something that *can* be assigned. > > Sorry, that is not current understanding, that is using the definitions > > and terminology. Understanding may change, but within the definitions > > and the terminology any unbounded set will remain infinite. > > FIne, the current "system" then. Right. You might come up with a different system where that is not the case. > > > Using that system, the phrase is senseless, but that system is not the > > > real universe. It's a concoction. > > > > Well, that is just a statement that (in my opinion) is senseless. > > Only in the context of the Dedekind definition of infinity, which one is > not obligated to consume wholesale. Until one can prove that transfinite > set theory is "correct", no one is obligated to accept the theory at all. No one is obligated to accept the theory at all. Whether it is proven to be "correct" or not, as I have no idea what "correct" in this context means. Is Euclidean geometry "correct"? Is hyperbolic geometry "correct"? Is elliptic geometry "correct"? But what is the case is that if you accept the axioms, you also have to accept what follows from the axioms. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |