An uncountable countable set
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From: mueckenh on 6 Jul 2006 07:04 Virgil schrieb: > > All sequences, even all single transpositions which I apply can be > > enumerated by natural numbers, just like the lines of Cantor's list. > > OK? And as long as I can enumerate, the number reached is finite. OK? > > But the numbers which are reachable include values larger than any given > natural. How should that become possible??? All transpositions can be enumerated by *finite* natural numbers. (Perhaps larger than all you can write down, but that is not he question.) > To assume a limit on what is achievable is counterfactual. To assume that the realm of finite numbers is left and that somewhere "infinity" is entered is counterfactual. > > > Hence I do not need more than a finite set of transpositions, though > > its cardinality cannot be given. > > You, in fact, need more than any finite number can supply. I may need more than any fixed number of transpositions. But the notion of countability includes the meaning that all elements can be enumerated by natural numbers. > > > > This case is in general denoted by > > "countably infinite" - but the number oo does *not* appear. Please > > consider these facts before you demand something to be proved for "oo". > > WE do not require that it be proved for "oo", which is not a natural, > but that it be proved for "for all n in N". So we agree on this point. > > You method produces exceptions to "for all n in N", by implying that > there is some remote member of N beyond which it is not necessary to > proceed in order to establish "for all n in N". Completely wrong. All I say is that every natural number is finite. Regards, WM
From: Dik T. Winter on 6 Jul 2006 07:51 In article <1152182749.926627.310460(a)a14g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > All sequences, even all single transpositions which I apply can be > > > enumerated by natural numbers, just like the lines of Cantor's list. > > > > Yup, so there are infinitely many. > > Are yo sure? How could an enumeration by natural numbers supply > infinity? But the axioms. > > I do not know what you intend to say here. But you can enumerate, yes, > > and when you stop you have done a finite number of transpositions you > > have done. > > And when I do not stop I nevertheless have done a finite numer of > transpositions anyhow. At any time you have done a finite number of transpositions. Yes. But you never will have done them all. > > Yes. When you continue, you do not stop, so you do not reach a > > number, OK? > > I can never reach the number oo or omega or aleph_0 if you mean that. Any number. > > Moreover, this way you will only do the transpositions in > > the first sequence of transpositions, so you will never even start with > > the second sequence of transpositions. > > If I do not get finished with this first infinity, then also Cantor > does get get finished with his list. That is the consequnce I wnated to > point out. This is nonsense. A list is given, not created by Cantor. > > > Hence I do not need more than a finite set of transpositions, though > > > its cardinality cannot be given. > > > > Nope. > > There are only finite natural numbers. Indeed each natural number is finite. Nevertheless, there are infinitely many natural numbers. If you disagree, you disagree with one of the axioms, and you should start your own set theory and mathematics without that axiom or with the negation of that axiom. > > In addition to cardinality we have to do here with ordinality. The > > sequence of transpositions you give has order type w * w. > > No. There is a first element and there remains a first element. You stated in one of your articles: for n = 1 to oo: (q_2n-1, q_2n) --> etc. for n = 1 to oo: (q+2n, q_2n+1) --> etc. what is the predecessor of (2, 3) -->...? > > That makes no sense, again. The transpositions have to be execute in the > > order given. > > Of course. But that does not exclude that these transposition can be > executed and finished (if Cantor's list can be finished). They can not be executed and finished. What you have to do is *define* the meaning of an infinite sequence of transpositions. The meaning of an (countably) infinite sequence of decimal digits *has* a standard definition. > > If you do them in any other order the result can be different. > > So that you *can* enumerate them does not mean that the result is > > independent from the order in which you perform them. > > That is true but has no relevance. It has some relevance. > > > > Apparently you do not understand what Cauchy involves. > > > > > > I understand that Cauchy requires precision on a positive epsilon, > > > arbitrarily small. > > > > In fact: given sum{i = 1, oo} d_i/10^i, where 0 <=3D d_i < 10, by Cauchy > > it follows that that denotes a real number. What the number is is > > irrelevant, but it *is* a real number. > > That is the theory of Meray and Cantor. By Cauchy it follows that it has a limit. By Meray and Cantor that limit is a real number. > > > It is completely differenet from Cantor's requirement. Cantor > > > does not define any limit process (because he considers finite natural > > > numbers for enumeration purposes only). > > > > Perhaps. Please use my definition of the diagonal number above, which > > does not definie a "limit process", but uses the definition of limit. > > Cauchy requires the epsilon. Yes, and with a sequence of decimal digits that is easy enough to satisfy. > > > > There are. The first sequence of transpositions defines the first > > > > "infinite permutation". Indeed, the transpositions do not overlap. > > > > The second sequence of transpositions defines the second "infinite > > > > permutation", but that one overlaps the first "infinite permutation". > > > > > > No problem. Everything is determined. > > > > If you change the order of those permutations the result will be different > > after a finite number of such infinite permutations... > > But why should I change the order??? Well, you claim that the transpositions alone define the result, I state that also the *order* of the transpositions is relevant. > > > > By representing a list it is assumed that all members are known. > > > > If they are not all known, it is not a list. > > > > > > And that all are enumerated by finite numbers. The same is valid for > > > the set of all positive rationals. > > > > Yes, you can enumerate them. But enumerating does *not* mean ordering. > > Enumerating by 1,2,3,... does mean well-ordering with order type > omega. If at the same time you re-order them. > > It is true that when you order them consistent with the enumeration you > > get a well-ordered set (of type w). This does *not* mean that any > > ordering gives a well-ordered set (of type w). A straight counter-example > > is the integers ordered in reverse order. That is not a well-ordered set. > > From that example we see that it is impossible to maintain the > well-order and we see why: Because the actual infinite does not exist. > Would it exist, there was always a first element, a second, an so on. But you claimed that any number of transpositions would retain well-ordering. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: mueckenh on 6 Jul 2006 08:36 Dik T. Winter schrieb: > But the order type of your set of transpositions is w * w. Sorry, I do not see that. Every sequence of transpositions leaves a first, a second, a third element. And there is no element number oo. By definiton, that is order-type omega. That is a well-ordered set with ordinal number omega. > Moreover, > different orderings of the same subset of transpositions will yield > different results. In the formula I gave for the diagonal number, > calculation of one digit does *not* depend on the calculation of > another digit. And what is the consequence of this? > > It is Cauchy's theorem that proves that the limit exists, using the > epsilon argument. But Cantor's arguing is that without epsilon. > > > > > No problem. For a given well-order to start with, every result is > > > > determined. > > > > > > When you use finitely many overlapping sequences of "infinite permutations". > > > > When I stay within the domain of finite numbers. > > You do not. Or you are re-ordering a subset of the rationals, not all > rationals. Which part do I not re-order? > > If the transpositions are well-ordered, they are indexed by finite > > numbers - up to the last one. > > Wrong. In the first place, there is not necessarily a last transposition, OK. So let's better say: The transpositions are indexed by finite numbers - up to every one. > and in the second place they are not necessarily indexed by finite numbers. > Consider the following (well-ordered) sequence of transpositions (where I > use, as you do, index numbers): > (1, 2)(3, 4)(5, 6)(7, 8)...(1, 3)(2, 4)(5, 6)(7, 8)... > what is the finite index number of the transposition (1, 3)? I don't know it because the set of transpositions has not the order-type omega. But if all the transpositions are countable (which is correct without doubt) and if countability means that every member of the set (of transpositions) can be enumerated, then (1, 3) has a finite number. But recently you raised the problem that the ordinal number of the set of all transpositions is not omega. This is in fact a problem, as your above question shows again, because we cannot count up to or even beyond the first limit-ordinal number. But I think I can remedy this disadvantage in the following way. Start with transpositions of the indices: (1, 2) (2, 3) (1, 2) This orders the set of the first three elements by size. Then order the set of the first 4 elements by size. (1, 2), (3, 4) (2, 3) (1, 2) And so on. After the set of the first n elements has been ordered by size, order the set of the first n+1 element by size. At most n-1 transpositions are required for a set of n elements. Now, the complete set of transpositions is of order type omega. And it maintains the well-order by index while it achieves well-order by size. Regards, WM
From: mueckenh on 6 Jul 2006 09:11 Virgil schrieb: > In article <1152133235.172340.249300(a)v61g2000cwv.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > > What I think is that there has to be a function exactly defined before > > > K exits and when you have defined that you already have that function, > > > so K is not in the range of the function. The P(N) is different because > > > it contains all possible K:s. > > > > Insn't that a bit too much naive intuition? > > That is hard logic. | o / \ o o / \ / \ There are more paths than edges. That is hard logic. Regards, WM
From: mueckenh on 6 Jul 2006 09:12
Virgil schrieb: > In article <1152133409.564175.47720(a)p79g2000cwp.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Virgil schrieb: > > > > > > > > > > 0.1 > > > > > > 0.11 > > > > > > 0.111 > > > > > > ... > > > > > > 0.111...1 > > > > > > ... > > > > > > > > > > > > But in this list the number 0.111... is not contained. Hence not all > > > > > > of > > > > > > its digits can be identified. > > > > > > > > > > You have just proved that your example supports Cantor's theorem by > > > > > producing a number, 0.111..., not in your own list. > > > > > > > > All digits of a number must be indexed by natural numbers. Otherwise > > > > they cannot be identified. All digits which can be identified are > > > > pesent in the list which contains all unary representations of naturlal > > > > numbers. > > Those are not natural numbers at all in the list, but decimal, or other > base, fractions. These are unary representations of natural numbers: 1 = 0.1, 2 = 0.11, 3 = 0.111, ... > It is the digit positions in those numbers that are > indexed by naturals and there is no more an end to the naturals used to > index them than there is to the digits being indexed. > > > > > > > > > That presumes that the set of naturals is finite, which in ZFC and NBG > > > is false. > > > > It presumes only that *every* natural is finite. > > It presumes a last natural to be usable as an index, which presumes that > the set of naturals is finite. No, no, no. Of course not! The only condition is and remains: Every natural is finite. Therefore 0.111... is not an unary representation of a natural number. Regards, WM |