An uncountable countable set
in [Math]
|
Prev: integral problem
Next: Prime numbers
From: mueckenh on 7 Aug 2006 12:19 Dik T. Winter schrieb: > > I does. Indexing, or let's say "indexibility", is a symmetric > > relation: If a digit position of a number like 1/9 = 0.111... can be > > indexed by a list number, then also the list number can be indexed by > > the corresponding digit position of 1/9. > > Let us state it more proper (and I will use "list" only to refer to the > list of natural numbers). > If a digit position of a number like 1/9 = 0.111... can be indexed by > a number of the list, than also the number of the list can be indexed > by that number of the list. > (I have no idea what indexing by positions means...) Example: The third 1 of 0.111... can index the third 1 of 3 = 0.111 as well as the third one of 5 = 0.11111. 3 = 0.111 can index the third 1 of 0.111... as well as the third one of 5 = 0.11111 (as well as the third 1 of itself). General: Instead of third, 3, and 5 you can choose n-th, n, and m > n. This shows the symmetry of the relation "indexing". > > If 1/9 is not in the list but > > can be indexed completely by list numbers, this symmetry is broken, > > because: What means "not in the list"? It means that 1/9 has more 1's > > than every list number. > > Yes, right, and that is the case. 1/9 has "infinitely many" digit > positions, each of them indexable by one of the "infinitely many" > natural numbers. But each and every natural number has only finitely > many digit positions. So infinitely is not more than finitely? Compare Cantor: Trotz wesentlicher Verschiedenheit der Begriffe des potentialen und aktualen Unend¬lichen, indem ersteres eine veränderliche endliche, über alle Grenzen hinaus wach¬sende Größe, letztere ein in sich festes, konstantes, jedoch jenseits aller endlichen Größen liegendes Quantum bedeutet, tritt doch leider nur zu oft der Fall ein, daß das eine mit dem andern verwechselt wird. According to that the number of 1's of 0.111... is aleph_0 which is larger than the number of 1's of any finite number. This forbids complete indexibility. > > > You need a proof > > > for that. And there is no such proof. As I stated earlier, every > > > *terminating* number of which all digits can be indexed through numbers > > > in the list is itself in the list. That is trivial. The last digit > > > index position is the number itself. But that is *not* true for a > > > non-terminating list, as it does not have a last index position. > > > > But it is true for every number which can be indexed by list numbers, > > because all of them are finite. > > Not if you need *all* list numbers to index You say "not". That "not" is understandable but nevertheless it is wrong. Even if all list numbers are needed, none of them is infinite (by definition each one is finite; that is why there is no infinite set of finite numbers). >(i.e. when you have a non- > terminating sequence of digits). Each and every index of a digit is > finite, but there are "infinitely many" of them. Absolutely uninteresting how many there are. Each one is finite. > > > And indexing up to position n means > > covering up to position n. > > I have still not seen a clear definition of "covering". But I have > some vague idea about it. If n indexes the position number n of m > n, then n covers the first n positions of m. > > > Indexing up to every position means covering > > up to every position. > > And what is wrong with that? Indexing all positions of 0.111... means covering all positions, i.e., covering the whole number 0.111... . If this were possible, then the list would contain at least one number 0.111... with at least aleph_0 positions filled with 1's. That, however, is impossible. > > > "Every digit of n can be indexed" This property is true for every > > number which is in the list, i.e. for every natural number in binary > > representation. It is true with no doubt because every list number does > > index (and cover) itself. This is a *symmetric* relation. Now you claim > > that a number with more 1's than every list number can also be indexed > > by the list numbers. This would destroy the symmetry. It is wrong. > > Prove it. In the first place, if the list contains unary representations > of natural numbers, K is *not* a natural number, and so is not in the list. > Nevertheless, all its digit positions can be indexed by natural numbers. That is squared insanity. Prove it. "Jede transfinite Zahl der zweiten Zahlenklasse z. B. hat ihrer Definition nach dieselbe Bestimmtheit, dieselbe Vollendung in sich wie jede endliche Zahl. Der Begriff omega bei¬spielsweise enthält nichts Schwankendes, nichts Unbestimmtes, nichts Veränderli¬ches, nichts Potenzielles [Cantor p. 390] .... wobei " omega größer zu nennen ist als jede der Zah¬len nü" [Cantor, p. 195] Greater than every natural number is the omega (completed, defined, unchanging) only then, if it contains a digit position which is not reached by any natural number. > > > But N is not finite > > > by the axiom of infinity. > > > > Every index number = natural number is finite. Hence what you claim for > > finite numbers of the list, holds for every number of the list. > > It is no longer clear to me what list we are talking about. If you > are talking about the list of natural numbers, that is obvious. My > claim for numbers in that list is that there are for each of them > only finitely many non-zero digit positions (and in unary, only > finitely many 1's). That is correct (for the true list without omega). As omega = 0.111... has no trailing zeros but every list number has infinitely many trailing zeros, there is at least one digit position of 0.111... which cannot be indexed by a natural number, namely one of the infinitely many trailing zeros. Regards, WM
From: mueckenh on 7 Aug 2006 12:22 Dik T. Winter schrieb: > > All you agree with for finite numbers applies to all index numbers, > > because each one is finite. > > Yes, each index number is finite, I *never* did argue otherwise. But > I do *not* agree that the set of index numbers is finite, because it > isn't. That is completely irrelevant. For indexing purposes we have only finite numbers, and, therefore, we can index only finite positions. > > > > > (every digit of n can be indexed) ==> (n is in the list) > > > > > > Where is the proof? There is an easy proof of: > > > (n is in the list) ==> (every digit of n can be indexed) > > > but not of the converse. > > > > Indexing is a symmetric relation. A position n of number N can index a > > position m of number M, if and only if n = m. > > What do you *mean* with "a position n of number N"? What is N here? Just a natural number in unary representation. > What do you *mean* with "a position ... can index"? How do you index > with positions? As far as I know you index with natural numbers in > this case. > > The only thing I see is: a natural number n indexes digit position m > of M when n = m. I see no symmetry at all. m is also a natural number. Hence, m indexes n if and only if m = n. There is a complete symmetry. > > > As 0.111... has more > > digit positions than the binary representation of any natural number, > > your claim implies that indexing is not a symmetric relation. > > Let's assume that we use indexing to mean a bijective mapping from one > set to another set. And further in this case, a bijective mapping from > the set of naturals to the other set. What is the other set? Well, > that is the set of index positions. And indeed, the set of index > positions *also* is the set of natural numbers. That 0.111... has > more index positions than each and every natural number in unary > (not binary) notation is irrelevant. No. As 0.111... has more index positions than each and every natural number in unary notation, the natural indexes are not sufficient to index 0.111... . > Just as N has more elements > than each and every indivual natural number is irrelevant. This assertion is impossible. Compare the differences of 1 between the naturals which would sum up to a natural number infinity if there were infinitely many differences possible. Regards, WM.
From: Virgil on 7 Aug 2006 16:28 In article <1154967573.942270.13030(a)p79g2000cwp.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Example: > The third 1 of 0.111... can index the third 1 of 3 = 0.111 as well as > the third one of 5 = 0.11111. Since the third 1 and the fourth 1 are identical, absent their predecessors, neither can index anything by itself. If "mueckenh" wants the string of the first three 1's as index for the third 1, that is possible. > > > > If 1/9 is not in the list but > > > can be indexed completely by list numbers Each digit in the decimal expansion can be indexed by a natural number but the 'completed' expansion can only be "indexed" by the 'completed' set of all naturals. Which is perfectly symmetric. , this symmetry is broken, > > > because: What means "not in the list"? It means that 1/9 has more 1's > > > than every list number. It does not mean that 1/9 has a decimal expansion with "more" 1's that there are naturals in the set of all naturals. > According to that the number of 1's of 0.111... is aleph 0 which is > larger than the number of 1's of any finite number. This forbids > complete indexibility. AS omega is an ordinal and any ordinal can be used as an index, "mueckenh" is wrong again. > Indexing all positions of 0.111... means covering all positions, i.e., > covering the whole number 0.111... . To say that each member of a set is "covered" does not require that the set be "covered" whatever "covered" means.
From: Virgil on 7 Aug 2006 21:58 In article <1154967767.420316.33010(a)p79g2000cwp.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > For indexing purposes we have only > finite numbers, and, therefore, we can index only finite positions. For indexing purposes we have arbitrary ordinals as potential indices, including those which prove you wrong.
From: Virgil on 7 Aug 2006 22:06
In article <1154967767.420316.33010(a)p79g2000cwp.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > No. As 0.111... has more index positions than each and every natural > number in unary notation, the natural indexes are not sufficient to > index 0.111... . > > > Just as N has more elements > > than each and every indivual natural number is irrelevant. > > This assertion is impossible. Compare the differences of 1 between the > naturals which would sum up to a natural number infinity if there were > infinitely many differences possible. That makes no more sense that to say that the sum of infinitely naturals being infinite prevents existence of infinitely many naturals. Since the sum of more than two naturals must be a natural larger than any of them, there are more than any finite number of naturals. If "mueckenh" wishes to claim that "more than any finite number" is a finite number, he should not try to do it in sci.math, as it only makes him look foolish. |