An uncountable countable set
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From: Dik T. Winter on 13 Sep 2006 20:51 In article <1158133913.582392.219700(a)e3g2000cwe.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > In article <1157743126.475610.189460(a)d34g2000cwd.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > Dik T. Winter schrieb: > > > > > > > A law is derived from the natural properties of arithmetics. > > > > > > > > > > > > Oh. What law derived from the natural properties of > > > > > > arithmetics is he talking about when he gets the > > > > > > completely determined set of all integral finite numbers? > > > > > > > > > > That is but his conviction. > > > > > > > > I ask you what law, but you are not willing to answer? Again, > > > > what law is he using (note that in English law generally refers > > > > to Theorem). > > > > > > Cantor held the opinion that there are some natural "laws" or "rules" , > > > not theorems, but Grundwahrheiten, so say truths, which are valid for > > > the natural numbers and which cannot be changed without leading to > > > rubbish. > > > > Yes, in English such things are called axioms. > > An axiom can be chosen arbitrarily, either the axiom or its negation. > Cantor's truths cannot be chosen but exist prior to any mathematics. Still you have not answered my question: "what law is he using"? On the other hand, you should familiarise yourself with the meaning of the English word "axiom". One of the meanings from Merriam-Webster: an established rule or principle or a self-evident truth Note the latter part, which (I think) entirely conforms with Cantor's "Grundsatz". > > Although I disagree that > > changing them leads to rubbish. > > because you have not yet understood what Cantor's truths are. Educate me. What are they? Pray provide sources. But your rubbish is not my rubbish... > > Changing them leads to at most a > > different kind of mathematics. > > Therefore, the truths are *not* axioms. Changing these truths leads to > rubbish. Yes. Just like the change of the parallel axiom lead to rubbish. Is that what you are intending to state? > > In current set theory that is an axiom, and so I think my > > translation with the word "axiom" is quite correct. > > Then you should try to improve your thinking. You wanted to translate > Cantor and you have completely neglected his words and his view. A > worse translation is impossible. Well. "law" is much worse. And that is the translation *you* provided, because in mathematics that means a theorem (unless you go in something like statistics, and even there it has a specific meaning). You might have stated "self evident truth", but that is covered by the word "axiom". -- 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 13 Sep 2006 21:23 In article <1158134793.927816.37420(a)d34g2000cwd.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > You are running in circles. You have to show that *using my definition* > > it should be in the list. > > > > > > If it does not hold you should be able to give an index position > > > > such that it is false. > > > > > > It is not in the list, although we cannot give an index position which > > > is responsible for that fact. All possible indexes are in the list. > > > > Yes, it is not in the list. So what is the problem? All possible indexes > > are in the list, so: > > (1) A{p = digit position} E{q = list item} {such that q indexes p} > > which you deny. > > That is not an argument. *Why* is it not an argument? You state: "all possible indexes are in the list". What is *wrong* about the argument? You always state "it is wrong", but never state what part of the argument is wrong. > Correct would be to say: > A{p = digit position which can be indexed} E{q = list item} In what way is that different when all possible indexes are in the list? And in what way does that not cover K = 0.111... defined as: whenever p in N, the p-th digit of K is 1, there are no other digits. Clearly all digit positions can be indexed. > > > > > > (2) A{p = digit position} E{q = list item} {such that q > > > > > > covers p} > > > > > > > > > > If your definition could be satisfied, the construction of the > > > > > list would imply this, yes. What we can safely say, however, > > > > > is only: > > > > > (2') A{p = digit position of list item} E{q = list item} > > > > > {such that q covers p} > > > > > > > > The same here. But apparently you think my definition of 0.111... > > > > can not be satisfied. Why not? > > > > > > Because it is not in the list which, by definition, contains all > > > numbers which can be indexed and which can index. > > > > That is *not* the definition of your list. Your list contains *by > > definition* all finite sequences of 1. > > That is all natural numbers (in unary representation). Yes. So there is nothing in the definition that it contains all numbers that can be indexed and which it can index. > > > There is nothing in that definition that it > > contains all numbers that can be indexed and that can index. > > That is the same as all natural numbers (here in unary representation). No. There are "numbers" (exactly one) that can be indexed by that list but that is not in that list. > > > Indexing and covering "by all list numbers" is equivalent. > > > Both is true or both is false. > > > > Your statement is false. > > If you deny that indexing and 'covering "by all list numbers" ' are > equivalent, then give me a natural number of the list (no other > naturals do exist) which can be indexed by all list numbers but which > cannot be covered by all list numbers (or the reverse). But there is a single "number" that can be indexed by your list but not covered by your list. 0.111... is one (as I defined it above). But it is *not* a natural number. You are switching continuously between number and natural number to confuse the issue. > > > "Not to terminate" is not a property which can serve to distinguish a > > > number from others, because we can never observe the end (because it > > > does not exist) neither the non-end (because it is nothing but a > > > negation of an unobservable property). > > > > Again. We are arguing within the realm of the axiom of infinity. You > > are ignoring that axiom with that statement. With that axiom the set > > of natural numbers does exist, and so your list (which is also > > non-terminating) does exist, > > and no number does exist which has more digits than the list numbers > have. There is a "number" that has more digits than all list numbers. This is similar to the statement that there is a set that contains more naturals as each finite set of naturals. But you always want to make 0.111... a number. It is not (in principle), we can talk about it as being a number, but as such it is only a sequence of symbols. I restore context: > > > > Consider them as decimal fractions. The list consists > > > > of the numbers (1-10^(-n))/9, 0.111... is 1/9. > > > > > > There is no n in the list (and in N) which yields 1/9. Therefore 1/9 = > > > 0.111... cannot completely be indexed. > > > In that case there is an n in N such that the n-th digit of 1/9 does not > > exist. Or else, there is an n in N such that the n-th digit of 1/9 does > > exist, but that n does not exist. What do you mean? > > The n-th digit does exist. Only digits which cannot be indexed by > natural numbers do not exist. What digit of 0.111... = 1/9 can not be indexed? > > > > > because 0.111... does not have any other component. > > > > > > > > It does not have any other component, but also it does not terminate, > > > > in contrast to all numbers on the list. > > > > > > This cannot occur other than by more digits 1 than available by list > > > numbers. > > > > Why? Your list is non-terminating, so it can index non-terminating > > numbers. On the other hand, each list element is terminating. > > Correct. The list is non-terminating, so it can index all existing > numbers. All are in the list by the axiom of infinity. But none is > non-terminating. So a non-terminating number cannot be indexed. Again a non-answer. Can your non-terminating list index non-terminating numbers? And it the answer is no, why? To state that it is not in the list is not an answer because you can not prove that all numbers that can be indexed are in the list. That is just circular reasoning. Do you agree that the non-terminating list can index non-terminating numbers? If the answer is no, why? -- 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 13 Sep 2006 21:46 In article <1158134972.282052.7360(a)b28g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > In article <1157836470.071804.254270(a)i3g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > Dik T. Winter schrieb: > > > > In article <virgil-411702.16371406092006(a)comcast.dca.giganews.com> Virgil <virgil(a)comcast.net> writes: > > ... > > > A set of n elements is *a number*. It need no necessarily be a latin or > > > arabic symbol. > > > > Bizarre. What number is the set of all natural numbers? > > That does not exist. At least it is not described by n. III is a > representation of 3. Now you shift back to representation, pray remain consistent. > > > > I count with my fingers. Does that count? > > > > > > The positions of your fingers are numbers. And you know what to do in > > > order to get from 32 to 33. > > > > I know what is the successor of the finger positions for 32. The position > > the 32 is all fingers (including the thumb) of the right hand up and two > > fingers raised on the left hand. So I just raise the ring-finger of the > > left hand to get the next number. No addition involved at all. > > And in both cases, the set contains either seven or eight raised fingers. > > If you know the positions by heart, then you need no addition actually. > You had already used it or the person who devised that technique had > used it. But earlier or later your knowing by heart will end and you > will have to count +1. Nope. I will only to need to know successors. Anyhow, you read much more in the successor of the Peano axioms than is present. The successor is defined without even any knowledge of addition at all. So succ(George V) = Edward VIII, succ(Edward VIII) = George VI and succ(George VI) = Elizabeth II within the set of British kings and queens. I do not think what way of addition you would propose for that. Of course, this successor function does not satisfy all of Peano's axioms, but I hope you get the idea. > > > > And I can go to 35 with my > > > > fingers, and I use my feet (not toes) to extend to 143. I never > > > > really managed to do it base 2, but 4 and 128 are interesting > > > > numbers in that case. > > > > > > Every position of your fingers and feet is an expression of a number. > > > > Perhaps. > > You cannot say yes? Indeed, no. I cannot say yes. > > > Counting without numbers would be nonsense. A boy like Virgil may count > > > 1, 2, 3, 3, 3, ... > > > > a, b, c, d, e, f, ... > > The old Greek and othe cultures have just used their letters as numbers > too. Yes, every culture had their representation of numbers. Sometimes base 10, sometimes other bases. Base 20 is quite common in Europe. Mixed bases are also used. Sometimes they were used to count from 1. Sometimes from 0. I may note that the decimal system you appear to like most is, in Europe, mostly due to Simon Stevin, and at that time it included 0 as a digit. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Virgil on 13 Sep 2006 22:00 In article <45085cbb(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > > Well, isn't this "IFR" thing basically Tony's semicomprehension of > > limiting density, which applies to some infinite subsets of the > > integers (and to some other sets, but not all). AFAICS, Tony simply > > ignores everything known about limiting densities, to assume that his > > version gives a "size" of - presumably - any set. > > > > Brian Chandler > > http://imaginatorium.org > > > > IFR is a generalization of limiting density. Except that it doesn't work .
From: Virgil on 13 Sep 2006 22:02
In article <45085d46(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Mike Kelly wrote: > >>> > >> If the string is unbounded but finite, then 2) follows. > > > > What's a finite but unbounded infinite string? > > > > One with all finite bit positions but no greatest. That contradicts all standard definitions of finite versus infinite. So absent a complete system u=in which such contradictions are allowed, it is not allowed. And TO has no such system. |