An uncountable countable set
in [Math]
|
Prev: integral problem
Next: Prime numbers
From: mueckenh on 13 Sep 2006 04:06 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. Correct would be to say: A{p = digit position which can be indexed} E{q = list item} But those positions are already in the list (by the axiom of infinity). > > > > > > (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). > 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). > > 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). > > > > "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. > and so there is a distinction between the > > 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. > > > > > 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. Regards, WM
From: mueckenh on 13 Sep 2006 04:09 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. > > > > 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. > > > > 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? > > > 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. Regards, WM
From: stephen on 13 Sep 2006 10:23 Dik T. Winter <Dik.Winter(a)cwi.nl> wrote: > In article <45070bcb(a)news2.lightlink.com> Tony Orlow <tony(a)lightlink.com> writes: > ... > > What about an unboundedly large but finite number? > Let me ask you to, before I can answer such a question. What is your > definition of "number"? (I asked the same from Wolfgang Mueckenheim, > but his answer was not satisfactory, also not to himself, I think, > because he never answered to questions about it.) > Once you have answered that question we can probe further into the > definition of "unboundedly large but finite number". > As far as I understand, for any kind of numbers, they are fixed, so > by definition not unbounded. The idea of numbers being something in flux seems somewhat common among the folks who rail against set theory. In standard thinking, a number, of any sort, is fixed and unchanging. The "opposition" on the other hand think of numbers as processes that can "approach" other numbers. The classic example is .99999..., which apparently is forever approaching, but never reaching, 1. Stephen
From: mueckenh on 13 Sep 2006 10:50 Dik T. Winter schrieb: > > Why don't you accept the axiom of > > infinity? There is an infinite set of natural numbers. All they are in > > the list (in unary representatition). By the axiom this is accomlished. > > And there is nothing else. > > Indeed, and 0.111... is *not* a natural number, so why should it be in > the list? Why do you think 0.111... is a natural number? There is a set which contains { }, and if it contains A then it contains A U {A}. This gives an inductive set which we can interpreted as the set of all natural numbers or the set of all indexes. > > > 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 by > > the axiom of infinity. > > It is not in the list because it is not a natural number. All indexes are in the list. All numbers which can be indexed are in the list. > > Because it is not in the list which, by definition, contains all > > numbers which can be indexed because there are all numbers which can > > index. > > This is nonsense. By the definition of your list the list contains all > finite natural numbers. That has nothing to do with indexing. Index = natural number. There is no infinite index, because indexing is identifying. > > If not all digits can be indexed, there should be a digit that can not be > indexed. But *by the definition* I gave above, all digits can be indexed. > Why is that wrong? Because all digits which can be indexed are in my infinite list. Could all digits of 0.111... be indexed then all its digits were represented in the list. But every sequence of 1's all digits of which are represented in the list, is completely in the list. > > > Indexing and covering "by all list numbers" is equivalent. Both > > is true (can be done) or both is false (cannot be done). > > You again do not clearly state what you mean. But the statement as it > stands is *false*. If a sequence can be completely indexed using any list number required for that purpose, then this sequence can be covered by a list number, and vice versa, if a sequence can be covered by a list number, then and only then can it be indexed completely by list numbers. > > "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). What can serve to distinguish > > two numbers in unary representation is a 1 at a digit position. > > You are not talking nonsense. No? > As 0.111... is different from al sequences > 0.111...1 because it is infinite in length, there is for each 0.111...1 > an digit position where 0.111... is different from that specific 0.111...1. 0.111... is different from all unary numbers which can be indexed (because they are in the list by the axiom of infinity). Hence there must be a digit which cannot be indexed. "Not to terminate" is not a property which can serve to distinguish a number from others. > > > > And if you think that that definition is wrong, please *prove* that. > > > > It has been proven by showing that 0.111... is not in the list. > > > Proof that it should be in the list. But remember: 0.111... is *not* > a natural number. That means: It is not a number which can be completely indexed. Therefore it cannot be in the list. (It is not a number at all.) Regards, WM
From: mueckenh on 13 Sep 2006 10:53
Dik T. Winter schrieb: > In article <45070bcb(a)news2.lightlink.com> Tony Orlow <tony(a)lightlink.com> writes: > ... > > What about an unboundedly large but finite number? > > Let me ask you to, before I can answer such a question. What is your > definition of "number"? (I asked the same from Wolfgang Mueckenheim, > but his answer was not satisfactory, also not to himself, I think, > because he never answered to questions about it.) Didn't you read my paper on the physical constraints of numbers? Here I give briefly my definition which is not obligatory but useful to avoid the weaknesses of set theory. It is derived from the first principle of the notion "number", i.e., a means to count. In German this becomes even more evident by the wording "Zahl" and "zaehlen". A number is an entity which: 1) either is realized by a fundamental set like the following fundamental set of 3: {III, Dik, {a,b,c}, {father, mother, child}, {sun, moon, earth}, ....} 2) or is completely determined by a series of digits. Each of these two requirements is sufficient to guarantee that the numbers satisfy the trichotomy. That is the most important property of a number. In this sense irrational numbers, complex numbers, hyperreals, transfinite numbers, vectors, etc. are not numbers but ideas which consist of numbers, like 2 + 3i, or not, like pi. > > Once you have answered that question we can probe further into the > definition of "unboundedly large but finite number". > > As far as I understand, for any kind of numbers, they are fixed, so > by definition not unbounded. O course. But the sequence or the set or the variable may be unbounded though always finite (because it consists of or takes on finite entities - in contrast to 0.111...). Regards, WM |