An uncountable countable set
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From: Virgil on 4 Aug 2006 17:43 In article <1154721998.226924.42100(a)i3g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > > Let's state that more proper. The index of every digit that can be > > indexed is in the true list (of natural numbers). But later you use > > the word "index" to mean something completely different. You are > > stating there that if every digit of a number can be indexed the number > > *itself* is in the list. But that does not follow. > > 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. 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. 1/9 only needs one 1. It is its decimal representation that needs more that any finite natural but not more that the set of finite naturals provides. > > "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. Question: Which digit(s) of the decimal expansion of 1/9 cannot be indexed by some finite natural? Answer: there is no digit in that expansion that cannot be indexed by a finite natural. To maintain the existence of such a digit requires proof that it exists, and "mueckenh" has no such proof because no such digit exists. > > > 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. What holds for every member of a list need not be true of the list itself. Every digit in the binary expansion of a proper rational fraction is a member of {0,1}, but no proper rational fraction is a member of {0,1}.
From: Virgil on 4 Aug 2006 17:46 In article <1154722206.554768.237310(a)p79g2000cwp.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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. 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. Or you > must claim that infinity is not larger than finity. What you claim does not follow. What is true for each element of a set need not be true of the set itself. They symmetry that "mueckenh" claims exists only in his head, and not in mathematics itself.
From: Virgil on 4 Aug 2006 17:56 In article <1154722450.815196.312700(a)75g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > > > > > > > The proof is given by the list: > > > > > > > > > > > > > > 1 0.1 > > > > > > > 2 0.11 > > > > > > > 3 0.111 > > > > > > > ... ... > > > > > > > w 0.111... > > > > > > > > > > > > That is neither a proof, nor a list. > > > > > > > > > > The list is given in the upper part. Call the whole an extended > > > > > list. > > > > > It shows that w is not the number of the naturals, because w > > > > > appears in > > > > > the sequence of ordinal numbers only once. > > > > > > > > That is only a statement, *not* a proof. > > > > > > Why is the ordinality of the left column of the extended list another > > > than the ordinality of the right column? > > > > I do not state so. The ordinality of both "lists" is w+1. Why do you > > think > > they are different? > > > > In general, the ordinal number of any initial segment of ordinal numbers > > is the smallest ordinal number that is larger than all ordinal numbers in > > that segment. (This can be done because the ordinal numbers form a > > well-ordered set.) And note that 0 is also an ordinal number (of the empty > > set). So the ordinal number of {} is 0, of {0, 1, 2} is 3, etc. > > Finally, the ordinal number of {0, 1, 2, 3, 4, ..., w} is w+1. And so > > is the ordinal number of {1, 2, 3, 4, 5, ..., w}, because there is an > > order preserving bijection between the two. > > > > You apparently did not appreciate that 0 is an ordinal number. (And that > > is independent of Bourbaki.) > > I find that in the *true list* every number n gives the cardinality of > the sequence of *positive* ordinal numbers from 1 to n inclusive and > the cardinality of the 1's of the corresponding unary representation. > > 0 = 0.0 > 1 = 0.1 > 2 = 0.11 > 3 = 0.111 > ... > n = 0.111...1 (with n 1's) > > I find that this is no longer true for positive ordinal w of the > extended list > > 0 = 0.0 > 1 = 0.1 > 2 = 0.11 > 3 = 0.111 > ... > w = 0.111... > > This shows that w is not the cardinal number of the natural numbers > alone. If by w, "mueckenh" means the ordinal we call omega, it is not a cardinal at all, though it has one, namely aleph_0, which IS the cardinal of the set of all finite naturals, with or without 0.
From: Virgil on 4 Aug 2006 18:16 In article <1154722892.397252.6590(a)b28g2000cwb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > > > (every digit of n can be indexed) ==> (n is in the list) > > > > I ask for a proof, not for a restatement. Pray prove the above using > > common > > mathematical and logic methods. It is false. > > 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. If 1/9 is not in the list, this symmetry is > broken, because: What means "not in the list"? I means that 1/9 has > more 1's than every list number. Which 1's in the decimal expansion of 1/9 do NOT have a finite natural as index, "mueckenh"? Unless you can find one and identify it unambiguously, you are wrong to posit its existence. > > 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. As only the character "x" is involved, no natural numbers are needed. For every row, the number of preceding rows and the number of preceding x's before the last one are finite and equal. > But according to the axiom of infinity only (1) is true and (2) is > false. Of course both statements are equivalent That claim about what the AoI says is itself false. The AoI says that there is at least one set which contians {} as a member and which for every member x also contains (x union {x}) as a member. And that is all it says. > That means: an actually infinitely long triangle is an actually > infinitely broad triangle. is the limiting case a triangle at all? By what definition of triangle? > Therefore there are either natural numbers > with infinite magnitude or there are not infinitely many natural > numbers. Or "mueckenh" is out of his mind! > > But in this case neither the paths, nor the edges are terminated. So while > > in all cases the number of edges is less Actually in any tree all edges terminate in nodes. in an infinite tree (with no terminal or leaf nodes) maximal paths will not terminate, but every edge still terminates. > > That is correct. But in no case the number of paths can be larger than > that of edges (which is countable). That is provably false and has been proved false. The set of edges of an infinite binary tree has been shown to biject with the infinite set of finite natural numbers. The set of paths of an infinite binary tree has been shown to biject with the power set of the infinite set of finite natural numbers. So that "mueckenh" is claiming a bijection between a set and its power set. which is a no-no. > > > > Yes, did I contradict that? My statement was: there are infinitely many of > > finite numbers. You state that is impossible. Pray show some *proof* of > > your statements. > > Infinitely many naturals have infinitely many differences of 1 which > accumulate to an infinite number. That doesn't prove anything. At least not in ZF or NBG.
From: Dik T. Winter on 4 Aug 2006 18:11
In article <1154721998.226924.42100(a)i3g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > Let's state that more proper. The index of every digit that can be > > indexed is in the true list (of natural numbers). But later you use > > the word "index" to mean something completely different. You are > > stating there that if every digit of a number can be indexed the number > > *itself* is in the list. But that does not follow. > > 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...) > 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. > > 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 (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. > 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. > Indexing up to every position means covering > up to every position. And what is wrong with that? > > > (every digit of n can be indexed) ==> (n is in the list) > > > > Where is the *proof* of this? > > > > The left hand states: > > for all n in N, K[p] = 1 > > your conclusion reads: > > there is an n such that An = K > > As I said before, this is true *only* if N is finite. > > "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. > > 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). I see not yet a contradiction. Now if we take the columnwise or of digits along the list, we come at 0.111..., but that number is not in the list (it is not a natural number, but something different). Now if we add that thing to the list, we no longer have a list of natural numbers. If we add it in front we still have a list, but not of natural numbers, and so I have no idea what indexing with it would mean. If we add it at the end, we no longer have a list. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |