An uncountable countable set
in [Math]
|
Prev: integral problem
Next: Prime numbers
From: Lester Zick on 6 Oct 2006 19:20 On Fri, 06 Oct 2006 12:48:17 -0600, Virgil <virgil(a)comcast.net> wrote: >In article <ac6c7$45260f70$82a1e228$27946(a)news2.tudelft.nl>, > Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> wrote: > > >> Sure. Simply "define" something that's undefined. And create the self >> fulfilling prophecy that suits you best. > >Beats hell out of defining things that are already defined. > >Every definition worth having defines something that would be undefined >without that definition. Huh? Technically incorrect since everything defined without a worthwhile definition could certainly be worthlessly defined. Of course then the issue becomes a standard for the "worth" of definitions. True and false come to mind except in the case of intuitive definitions in modern math in which case you come to mind. ~v~~
From: Lester Zick on 6 Oct 2006 19:22 On Fri, 06 Oct 2006 12:52:43 -0600, Virgil <virgil(a)comcast.net> wrote: >In article <1160122708.108138.77770(a)e3g2000cwe.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > >> I am sure, the results "all balls in B" and "all balls not in B" are >> not to be interpreted as an actual contradiction of set theory. It is >> just counter intuitive. >> >> Regards, WM > >It seems to be intuitive in "Mueckenh" 's world, but it is not only not >intuitive, it is not true in any set theory of my acquaintance. So it's counter intuitive. What's the problem? Surely it isn't the first counter intuitive suggestion you've ever run across. ~v~~
From: Dik T. Winter on 6 Oct 2006 20:16 In article <1160127267.123550.306050(a)b28g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: .... > > No reason at all, but when you invent a new notation, it would be better > > if you did define the notation before use. > > Ok, but as we have agreement now, we can return to he main question: > Why do you think that 0.111... with the index sequences 1,2,3,... or > k+1,k+2,k+3 or -k, -k+1, -k+2, ... represents exactly *one* number > only, as you asserted? Why do you still maintain that I think it represents a number? How many times do I need to state that, without proper definition, it only is a sequence of symbols that I on occasion call a "number". Because I have not yet seen a definition of "number", and you have stated that you are not able to give one... But whatever. As a sequence of symbols, igoring the "0.", it is in bijection with N. It also is in bijection with {k+1,k+2,...} for every k. As {k+1,k+2,...} is in bijection with N. -- 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 Oct 2006 20:10 In article <1160126832.709755.186110(a)h48g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > In article <1160079778.871756.325200(a)m73g2000cwd.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: .... > > > > > The axiom of infinity does only state n+1 exists if n is given. > > > > > It is realized by he numbers of my list. > > > > > > > > That is *not* what the axiom of infinity states. The axiom states > > > > that there exists a set that contains all the successors of its > > > > elements. > > > > > > Oh I must have read always wrong texts. I never came across the word > > > "all" in connection with this axiom. > > > > What texts *have* you read where your version is stated? > > I read many texts. (Don't intermingle the quantity "all" with the > quanifyer "forall".) Well, in another posting you did show that you do read texts, and in it you were apparently aware that the formulation you gave above is a theorem that follows from the axiom of infinity. That does *not* make the two statements equivalent. A (nearly) equivalent statement would be (I dispense of the Von Neumann notation of naturals, and just just numbers, the results are equivalent): There is a set N such that 0 in N and (forall x in N, x+1 in N) (If you wish, you may replace 0 by 1, as 0 would lead to Bourbakianism, which you detest. The results remain the same.) You question whether "all x in N" does exist, apparently. Based on what? > > > > Yes, because you still do not understand that there is no "specific > > > > position" such that "every position" is the same as "upto that > > > > specific position". > > > > > > In my list there are all specific positions (if all natural numbers do > > > exist). If the required number is not in the list then it is nowhere. > > > > Again, you imply that "every position" is the same as "upto some specific > > position". There is no such "specific position". > > I imply that "every position" is equivalent with "up to evey position" > in a linear set. No, that is *not* the implication. If the number is in the list, "every position" is equivalent to "upto every position" is equivalent to "upto some specific position" (namely the last position occupied by a digit). If the "number" is not in the list, "every position" is equivalent "upto every position" is *not* equivalent to "upto some specific position". (Where "every position" uses indexing, and "upto every position" and "upto some specific position" use covering.) As I have said time and time again. > I imply that "every specific position" is equivalent with "up to evey > specific position" in a linear set. Yes, and that is false and not provable. > > > It is crazy to hear that covering up to every number is possible but > > > covering every number shall be impossible. > > > > That is nothing more than opinion and taste. But that is not a disproof of > > the proof given. > > Of course. There cannot be a disproof, because a disproof would not be > a disproof. You have no idea how to prove things. Stating "it is crazy" is not a standard form of proof. I will re-iterate. Let L be your list (tacitly assumed to be unary representations of natural numbers), define a sequence of symbols such that it starts with 0., and for each n in N the n-th symbol is 1. I state that: (1) That sequence can be indexed, because the n-th symbol is indexed by the n-th item in your list. (2) For each n that sequence can be covered by the n-th item in your list upto the n-th symbol. (3) There is no n in your list that covers the whole of that sequence. Because if there were one such n, the (n+1)-st symbol is not covered. You claim (at various times) that either the definition is false, or that (1) is false. Well, clearly (1) is not false because of the definition. So the definition must be false. But definitions are never false, they define things. There may be objects that satisfy the definition or there may be no such objects. So let's assume that you assert that there is no object that satisfies the definition. But that is contradictiory to the axiom of infinity, because using that axiom we can prove that that sequence does exist. > > What does it matter? It was shown that the set of reals has larger > > cardinality than the set of naturals. An easy consequence of that is > > the theorem stated in the first sentence: there are sets with larger > > cardinality than the reals. Sorry, I should have said naturals here (because that is what the first sentence of the paragraph states). The second paper shows that there are sets with cardinality larger than the real, in a way that easily expands to arbitrary cardinals. > > Of course. And all these sets imaginable at that time were subsets of > reals. I think you underestimate Cantor. > > > > And you assert that it is not likely that in the current > > > > paper the theorem for which a simpler proof is given is the theorem > > > > "that there are sets that have larger cardinality than the natural > > > > numbers", but something else, unstated in the current paper? > > > > > > Of course there are such sets, but these sets are numbers, namely the > > > real numbers, the irrational numbers and the transcendental numbers. > > > > So there are sets. The second paper is not concerned about what the sets > > contain, it just proves that there are (arbitrary) sets with larger > > cardinality than the naturals. And shows that with countably infinite > > sequences of two symbols. > > > > Why do you think the annotations state how the proof can be modified to > > a proof about the reals? > > Because in 1933 it had become clear that the proof without modification > was incorrect for the reals. Why do you think Cantor used 'm' and 'w', rather than the more obvious '0' and '1'? Was that because he was proving something about the reals? But I stay with my assertion. The "that theorem" can only point to the theorem stated just in the sentence before "there are sets with larger cardinality than the naturals". -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: David Marcus on 7 Oct 2006 15:54
Virgil wrote: > In article <1160124783.798885.20040(a)h48g2000cwc.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > MoeBlee schrieb: > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > Dik T. Winter schrieb: > > > > > > > > > > The axiom of infinity does only state n+1 exists if n is given. It is > > > > > > realized by he numbers of my list. > > > > > > > > > > That is *not* what the axiom of infinity states. The axiom states that > > > > > there exists a set that contains all the successors of its elements. > > > > > > > > Oh I must have read always wrong texts. I never came across the word > > > > "all" in connection with this axiom. > > > > > > What text tells you that part of the axiom of infinity is that n+1 > > > exists if n is given? > > > > If x belongs to the set then {x} belongs to it. From thia n --> n+1 > > can be proved. > > > > > > And the quantifier 'for all' is part of the axiom of infinity: > > > > > > There exists an x such that 0 is a member of x and, for all y, if y is > > > a member of x then yu{y} is a member of x. > > > > > > In symbols: > > > > > > Ex(0ex & Ay(yex -> yu{y]ex)) > > > > I know these symbols. A means "for all" y which are there, it does not > > state that "all" y are there. > > > > Regards, WM > > How does "for all y which there are" differ from "for all y there are"? He probably means the distinction between "for all y in x" versus "for all t in N". The set x that the axiom says exists need not be N. -- David Marcus |