An uncountable countable set
in [Math]
|
Prev: integral problem
Next: Prime numbers
From: Virgil on 6 Aug 2006 15:27 In article <1154884004.979803.296090(a)i3g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > 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. > > Oracular statement. *All indexes are elements*. Index positions are 1's > of the number 0.111... Indices in a list are members of an inductive set, i.e., a set S containing {} and for every member, x, of S, also containing its "successor", (x union {x}). Every "1" in "0.111..." can be indexed by any inductive set. And by the "Hilbert Hotel" method, one can also index the string itself.
From: Virgil on 6 Aug 2006 15:36 In article <1154884131.211951.124080(a)75g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > > > > There is no limit. If the number of rows is infinite, the number of x's > > > in a row is also infinite. > > > > If the number of rows is unlimited then the number of x's in a row is > > equally unlimited but that does not necessitate the number of x's in any > > single row being unlimited. > > A very careful statement. A very true statement. > The number of rows is not only "unlimited" > but there are actually infinitely many rows. Which only means 'more than any finite natural number'. More precisely, given any natural number, there are more than that many rows, and there is a row with more than that number of "x"s in it. > The same holds for the x > in the rows. No row contains infinitely many of them? Where then should > infintely many x exist? A brilliant idea It goes back at least to von Nuemann, if not Peano. > > Just the right argument to obtain rows broad enough while the numbers > of x's being restricted to finite values. It conforms exactly to the Axiom of Infinity: http://en.wikipedia.org/wiki/Axiom_of_Infinity "There is a set N, such that the empty set is in N and such that whenever x is a member of N, the set formed by taking the union of x with its singleton {x} is also a member of N. Such a set is sometimes called an inductive set." > > > But, by definition, the number of x's in a > > > row is always finite (because every natural number is finite). > > > > But there is no (finite) upper bound on the number of x's. > > The upper bound is "less than infinite". The number of rows *is* omega. > The number of x's is *less than* omega. > > Regards, WM
From: Virgil on 6 Aug 2006 15:49 In article <1154884576.336501.257660(a)p79g2000cwp.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > > 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. > > It says that the complete set *does* exist with all of its elements. What the axiom of infinity says is, according to http://en.wikipedia.org/wiki/Axiom_of_Infinity "There is a set N, such that the empty set is in N and such that whenever x is a member of N, the set formed by taking the union of x with its singleton {x} is also a member of N. Such a set is sometimes called an inductive set." > The number of elements cannot be finite, but the number of elements > does exist. The axiom itself does not mention numbers at all. > > > > 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? > > By the definition of a straight line. I find no "straight lines" in a list of strings. > > > > > > > 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. > > But the number of edges does not terminate. > > > > > > > > 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. > > This is not an arument. I am claiming that these theorems lead to > contradictions. My statements may be theorems of ZF or NBG because they are capable of proof within ZF or NBG but "mueckenh"'s are not as they are not capable of proof within ZF or NBG. > Cantor's theorem may prove Card N < Card R. The binary > tree shows the opposite result. Not to anyone who knows proofs. > > > > 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. > > Set theory cranks are like other cranks: No argument or evidence can > ever be sufficient to make a crank abandon his belief. Speaking for yourself? I am convinced of statements within ZF or NBG by logically sound proofs based only the axioms and definitions of ZF or NBG. I have not seen any, at least by "mueckenh". When "mueckenh" imposes conditions not derivable from the axioms and standard definitions of ZF or NBG, I do not accept that they must hold in ZF or NBG.
From: Dik T. Winter on 6 Aug 2006 20:57 In article <1154882695.990488.7420(a)h48g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > This is illogic. What you are indexing is the list of digits of K. > > That is your definition, but it is wrong (= impossible to satisfy) You state so. See below on how to satisfy it. > > You are claiming that K is in the list of indices. That makes no sense > > at all. > > I am claiming that every index is in the list of indexes. This list is > infinite. There are infinitely many indexes. And, yes, 0.111... is not > among them and therefore cannot be indexed. No proof, yet. K = 0.111... can also be written as "for every n in N, K[n] = 1". So every digit of K has been indexed by elements of N. But K is not an element of N. Your last statement: "And, yes, 0.111... is not among them and therefore cannot be indexed" is wrong. Unless you are meaning something different with "indexed" again. So pray give *definitions* of the words you are using, because they appear to be quite extraordinary. > > > In order to save set theory with its "infinite set of finite numbers" > > > you are forced to assert that the list of all natural numbers does not > > > contain the list of all natural numbers. > > > > Yes, but that is obvious. As the list of all natural numbers is not a > > natural number, it is not in the list of all natural numbers. > > My list of all natural numbers is not a set in the pathological sense > of set theory. > My list contains all single natural numbers, and it contains all pairs > of natural numbers (though a pair is not a number) - not in one row but > in pairs of rows. And my list contains the list of all natural numbers > - not in one row but in all rows. Ah, again some confusion about terminology. I thought contain would mean "be an element of", but you meant something else, you meant: "is an element of or is a subset of". But I have not yet seen that your list contains K. And indeed, with this terminology in place, the list of all natural numbers contains the list of all natural numbers (because it is a subset). What is the problem with that? -- 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 6 Aug 2006 21:25
Are you picking up on old articles? You should be aware that I already did respond to articles that you posted in response to articles similar to this one, which is over a week old. Are you going to compare responsens? In article <1154883048.729461.122990(a)p79g2000cwp.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > In article <1153948199.656603.23130(a)p79g2000cwp.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > Dik T. Winter schrieb: > > > > In article <1153829288.635491.96140(a)m73g2000cwd.googlegroups.com> muecken= > > > h(a)rz.fh-augsburg.de writes: > > > > > Dik T. Winter schrieb: > > > > ... > > > > > I need not count to w. If a column contains one 1 then the *+sum is 1. > > > > > This is a definition like that of Cantor's list. > > > > > > > > Depends. I have no idea (and it is very dissimilar), whatever way I > > > > look at it, I find: > > > > (*+){k = 1 .. n} Ak = An > > > > and I can not substitute oo for n (so as to include all Ak), because there > > > > is no Aoo. > > > > > > Therefore you cannot need and cannot use it. It is simply not > > > available. > > > > Yes, so it is not defined for the *+ sum of all An... > > 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? > > > > > (every digit of n can be indexed) ==> (n is in the list) > > > > > > > > Where is the proof? > > > > > > It is even an equivalence by *definition*. The list contains all > > > natural numbers. All indexes *are* natural numbers and vice versa. > > > > I do not see the equivalence. The equivalence is: > > (every digit of n can be indexed) ==> (every index is in the list) > > I still see no proof that n is in the list, because I do not see a > > proof that n is an index. > > 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? > > 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. > 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. > > 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. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |