An uncountable countable set
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From: Dik T. Winter on 7 Aug 2006 22:01 In article <1154967573.942270.13030(a)p79g2000cwp.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > 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. It is a strange formulation to state that the third 1 can index the third 1 of something else, because it is *not* the 1 that indexes, it is its position number. I would rather formulate it as: the index number of the third 1 can index the third 1 of something else. Your formulation only confuses matters. > General: Instead of third, 3, and 5 you can choose n-th, n, and m > n. > This shows the symmetry of the relation "indexing". But there is no symmetry as you appear to assume. > > > 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? How do you conclude that from what I write? > Compare Cantor: Trotz > wesentlicher Verschiedenheit der Begriffe des potentialen und aktualen > Unend=AClichen, indem ersteres eine ver=E4nderliche endliche, ber alle What is the "not sign" doing between "Unend" and "lichen"? > Grenzen hinaus wach=ACsende Gre, letztere ein in sich festes, And here again one? > konstantes, jedoch jenseits aller endlichen Gren liegendes Quantum > bedeutet, tritt doch leider nur zu oft der Fall ein, da 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. Why is that forbidden? The number of natural numbers is also aleph-0, so it appears to me there is a perfect match possible. nevertheless, you state it is impossible. By the axiom of infinity, the set of naturals does exist, and by Cantor, the cardinality is aleph-0. Also, that set does not have a last element. > > 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). Yes, you keep stating that, but the axiom of infinity asserts that there is. But if you claim that the set if finite numbers is finite, there is a largest number. What is the successor of that largest number? I would state that successor is also finite. A contradiction, so the set of finite numbers is not finite. > >(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. Yes, and so what? There are nevertheless infinitely many of them. > > > > > 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. Not so. For each n a finite segment of 0.111... is covered, but there is no n that covers 0.111... completely. So we can cover each finite segment of 0.111..., and we can index each finite position of 0.111... . But there is *no* n that indexes the last position of 0.111... (because there is none) and so there is no n that covers all of 0.111... . Could we use 0.111... to index a digit of itself (assuming it is some representation of w)? The answer is *no*, 0.111... has no w-th digit. > > > > > > "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. I would like that you prove your assertion first. But let me have a go at my proof. Assume [p] (with p in N) to mean the p'th digit in the unary representation of a number, and assume K to be the (non-natural) number with a 1 in each position (i.e., for each p in N, K[p] = 1) and no other 1's. What do I have to prove? Clearly, all digits of K can be indexed, by the definition of K. What remains is, I think, to pro
From: Dik T. Winter on 7 Aug 2006 22:11 In article <1154967767.420316.33010(a)p79g2000cwp.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > 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. Yes, where I did state otherwise? > > > > > (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. And you answer only one of my two questions. > > 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. And two more questions you do not answer. > > 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. Using terminology I asked you to explain, but you refrain to do so. > > > 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... . Again a claim without proof. Why not? Can you state a position in 0.111... that can *not* be indexed by a natural number? > > 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. Eh? What natural number infinity are you talking about? -- 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 7 Aug 2006 22:57 In article <1154968120.846531.209650(a)75g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > In article <1154722892.397252.6590(a)b28g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > Dik T. Winter schrieb: > > ... > > > > > So the triangle gets infinitely long but not infinitely broad. This > > > > > means the Diagonal must bend. > > > > > > > > Eh? Where do you conclude *that* from? The triangle gets infinitely long > > > > and infinitely broad. Your conclusion is based on something unstated. > > > > But this is, again, a distraction. > > > > > > 1) Infinitely long means that there are infinitely many natural > > > numbers. > > > 2) Infinitely broad means that there are infinite natural numbers. > > > > Eh? Why? Can you provide a proof? > > You agreed above that the triangle gets infinitely long and infinitely > broad. Yes, I ask for a proof that infinitely broad means infinite natural numbers. I may note that in *both* cases, infinite has been used losely, meaning that there is no bound. I have already stated that elsewhere. I think you are confusing actual infinity (the infinite list exist) with potential infinity (no member of the list is infinite). At each finite stage, the length and width of the triangle are equal. And the width of the triangle is the width of the last element. If you complete the triangle (conforming to the axiom of infinity) the length of the list is infinite, as is the width, but there is *no* element of the list that has infinite width, because there is no last element. (If there were a last element, indeed, that element would have infinite width, but as there is none...) > > > Of course both statements are equivalent. > > > > There is no "of course" here. Such a statement needs proof. > > You want a proof that 3 = 3 or 4 = 4 or oo = oo? The last one would be interesting. What does it mean? But I want to see a proof of "infinitely many natural numbers is equivalent to infinite natural numbers". That is what you asserted. > You see here the natural numbers in unary representation: > 0.1 > 0.11 > 0.111 > ... > > Can't you see that width and length are the same? Yes, see above. Still your assertion that "infinitely many natural numbers is equivalent to infinite natural numbers". Where is the *proof*? Note that the axiom of infinity asserts otherwise. And still assuming that you want to prove that that axiom is consistent, you should prove that assuming that axiom. > > > That means: an actually infinitely long triangle is an actually > > > infinitely broad triangle. Therefore there are either natural numbers > > > with infinite magnitude or there are not infinitely many natural > > > numbers. > > > > Why? > > Length = number of numbers > Width = size of numbers > You agreed above that the triangle gets infinitely long and infinitely > broad. See above. > > But that is not mathematical thinking. This is the thinking that leads > > users of (non-symbolic) calculators to complain that when they > > calculate sin(pi) they do not get 0. Mathematics is not concerned with > > finite space, time or whatever. But, again, this is a philosophical > > standpoint, *not* a mathematical one. By the same reasoning sqrt(2) > > does not exist. > > Of course it does not. You cannot distinguish sqrt(2) and the same > number with the digit of position number 10^100^100 replaced by 5. So we are factoring numbers using things that do not exist. Now what? You are still focusing on decimal representations of numbers. But they are not all and everything. sqrt(2) is the unique positive number that, when it is squared, gives 2. You may state that it does not exist (what does it mean that a number exists or not?), but it comes in pretty useful. > > > That is correct. But in no case the number of paths can be larger than > > > that of edges (which is countable). > > > > No, the number of edges will also be uncountable. > > The set of edges is easily enumerated. > o > /1 \2 > o o > /3\4 /5 \6 > o o o o > /7... Yes, as long as you remain finite. So the edges leading to final nodes are countable. What is the natural number that can be mapped to the final edge in 0.111... ? > > What is the argument here? If you state that the number of paths is > > countable you need to provide a method that gives the index number when > > a path is given. > > No! I only need a method to count a set which contains provably more > elements. This set is the set of all edges. there is no doubt that it > is countable. What is the number of the final edge in 0.111... ? Actually a trick question, because there is no final edge. > > > > Where do I state such, or how can you conclude that? > > > > > > Because all naturals have to be finite. But infinitely many naturals > > > have infinitely many differences of 1 which accumulate to an infinite > > > number. > > > > By what reasoning do you conclude that that "infinite number" is a natural > > number? > > I conclude it from your saying that there are infinitely many naturals. > In fact this is impossible, because the sum of infinitely many 1's is > not a natural number. Yes. Your conclusion is only valid if there is a last natural, and that we can complete the adding of infinitely many 1's. Neither is true. > > Nope, the plain statement above "that is impossible" is a contradiction > > of the axiom of infinity. So as such there is no basis for that statement. > > So your statement in itself is already clearly in contradiction with the > > axiom as infinity and so can not show that the axiom of infinity is in > > contradiction with itself. > > It is not in contradiction with itself but in contradiction with > others, for instance in contradiction with the fact that every natural > number is finite. Again, you state that. Without proof. You are arguing using the argument that a set of finite numbers can not be infinite. And not once, but each and every time. Pray give a proof of that fact, using the axiom of infinity. If you succeed, you have shown that the axiom of infinity is inconsistent. And again, I state "proof". Not an example. > > That is just a repeat of your
From: mueckenh on 8 Aug 2006 16:07 Dik T. Winter schrieb: > > But the *+ sum is defined for all digit positions which can be indexed > > by a natural number. > > You still fail to see. There is a definition for (*+){k = 1 .. n} Ak, > I still do not see a definition for (*+){k = 1 .. oo} Ak. Can you, > please, once give a proper definition? There is no natural number oo. Therefore {k = 1 .. oo} is nonsense. There is, as you say, a definition for (*+){k = 1 .. n} Ak. More is not possible. > > n is a natural number. An index is a natural number. "Index" and > > "natural number" are synonymous. > > You are assuming that only digits of natural numbers can be indexed. > But 0.111... is not a natural number. And you are stating (in fact): > (every digit of 0.111... can be indexed) ==> (0.111... is in the list). > so, why does this follow? Because: index = natural number = "number that can be indexed" > > > > Yes. This does *not* proof that the *+ sum of them all (once defined) > > > is in the list, because there is no proof that it is an index. Rather, > > > there is an easy proof that it is *not* an index. > > > > Then it cannot be indexed. Then it contains digit positions which are > > not indexed by natural numbers. > > You state so, and remain stating so, without any proof. It is a definition. > > > Indexing is a symmetric operation: > > A digit position n can be indexed by a natural number 0.111...1 (with n > > digits) if and only if the natural number (in unary representation) > > can be indexed by the digit position of said number. > > Assuming again n natural I think, and unary representation? But this > tells *nothing* about the indexing of digits of non-natural numbers. It is impossible, because index number = number which can be indexed. > > > > Except in his later times, as you state so eloquently just above. Or > > > do you *not* see a contradiction between these two statements? > > > > Between which two statements should I see a contradiction? > > To quote you: > > > > Their cardinality in newer set theory (and already by the late Cantor) > > > > is expressed by the least ordinal. The cardinality of sets w , 2w, w^2, > > > > ... is expressed by aleph_0 or by omega. > And: > > > > On the contrary! Cantor distinguished very clearly in all his written > > > > papers. > > > > Fact is: > > 1) Cantor distinguished neatly between omega and aleph_0 in all his > > written work. > > Except in his later days (see the first quote and below). > > > 2) Modern set theorists (and the late Cantor) use omega to denote the > > order type of |N as well as the cardinality of countable sets. > > So the late Cantor did *not* clearly distinguish cardinal and ordinal > numbers. Or he recognized that the first ordinal of a number class may well serve as the cardinal of this class. But this topic is not very interesting to me because neither of these numbers are numbers. Regards, WM
From: Virgil on 8 Aug 2006 16:11
In article <1155067333.259873.193700(a)p79g2000cwp.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > The v. Neumann model of natural numbers is just the model of segments. > n = {0, 1, 2, ..., n-1} > Every natural number is finite. > Hence every union of finite segments is a finite segment. Not necessarily. You assume that every union of finite segments is a finite union, which begs the question. > You don't like set theoretic models if they are uncomfortable for your > current arguing? When "mueckenh" assumes his conclusion, we do not like his arguments. |