An uncountable countable set
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From: Virgil on 14 Aug 2006 14:10 In article <1154722450.815196.312700(a)75g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > This shows that w is not the cardinal number of the natural numbers This shows that "Mueckenh" has no idea what he is talking about.
From: Virgil on 14 Aug 2006 14:17 In article <1154722206.554768.237310(a)p79g2000cwp.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > > > > The last conclusion requires proof (and it is false). What is the > > > > case > > > > is that for every Ai the indexing of digits will terminate at some > > > > point, > > > > and the indexing of digits of 0.111... will never terminate when we > > > > start > > > > at the first digit and go on. Because there will always be a further > > > > digit. (As there will always be a further An.) > > > > > > But each one will be finite. > > > > You mean the index of the digit and the An. Yes. > > All you agree with for finite numbers applies to all index numbers, > because each one is finite. > > > > > (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. Wrong! The set of all naturals can index the set of all even naturals and vice- versa. The critical issue for the indexing of any countable set is whether the set of indices has a unique first index and for each index has a unique successor index distinct from those before it. Absolutely nothing else is required. 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. It is not necessarily "symmetric", at least in "Mueckenh"'s sense.
From: Virgil on 14 Aug 2006 14:27 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: Then if each digit can be indexed by a unique natural number, "Mueckenh"requires that each natural number be indexed by a unique decimal digit. Also note that the set of even naturals can be indexed by the set of all naturals so it would follow that the set of all naturals can be indexed by the even naturals leaving the odd naturals to index the set of all naturals a second time without breaking into a sweat. > "Every digit of n can be indexed" And can be 'indexed" by using only those digits whose position number is a prime, leaving most of them unused as indices but still indexing all of them. > This is a *symmetric* relation. Not at all. when one turns it about, only those digits in prime positions are indexed leaving infinitely many un-indexed. "Mueckenh"'s notions about infinity are centuries out of date.
From: Virgil on 14 Aug 2006 22:28 In article <1155067333.259873.193700(a)p79g2000cwp.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > Again, you are arguing with a particular model in mind. But when you > > *do* argue in a particular model, pray do it correctly. Especially your > > sentence: "Each union of finite segments is a natural number" is false. > > 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. Nor quite. Every FINITE union of finite segments is finite, but infinite unions need not be. In fact the union of all naturals in NBG equals the set of all naturals, and by the axiom of infinity does not have any largest member. > You don't like set theoretic models if they are uncomfortable for your > current arguing? Every model of NBG contradicts "Mueckenh"'s arguing, so he must be infinitely uncomfortable with it.. > > Again, a model, with an unfounded statement. I was talking about sets, > > not about specific models of natural numbers. > > 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. "Mueckenh" repeats what is false in NBG (the von Neuman-Bernays-Goedel model). The union of all von Neumann naturals is not finite. > You don't like set theoretic models if they are uncomfortable for your > current arguing? "Mueckenh" must hate NBG then since his arguing is wrong in every model of nNBG.
From: Dik T. Winter on 14 Aug 2006 22:57
In article <1155485742.529974.141050(a)i3g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > Further: Indexing the digit number n is equivalent to covering the > > > string up to digit number n. A finite string can never cover an > > > infinite string. > > > > But *that* is irrelevant. > > Wrong. Why is it wrong> > > Consider K = 0.111... . For each n we can > > index digit number n of K and so cover K up to digit number n. > > Of course you can index each n. But your "each n" stems from the true > list. And we know that 0.111... is not in the true list, because it is > distinguished from any element of the true list. Pray read what I state. Again, what you state is irrelevant. > Therefore your assertion is correct but void of power to prove that any > digit of 0.111... can be indexed. Eh? This tells me exactly nothing. My assertion "each digit of K can be indexed, but K can not be covered" is void of power to prove that any (why any here?) digit of 0.111... can be indexed? If each digit of K can be indexed, any digit of K can be indexed. > > > Don't forget: The infinite set of natural numbers contans only natural > > > numbers, i.e., finite strings of 1's in unary representation. > > > > Yes, but there is no bound on the index positions, and hence the number > > of index positions is infinite. > > The number of digit positions is irrelevant. The digit positions are > all finite - in the true list. All numbers with merely finite digit > positions are in the true list. 0.111... is not there. Yes, I never said otherwise. Why do you always come back with this statement that I do not contradict? > > > Why can't you learn that the > > > asserted infinity of the number of numbers is completely irrelevant. > > > > That is just opinion. > > That is not opinon. In oder to index or to cover, only the digit > postions of the numbers are relevant. It is completely irrelevant how > many othe numbers are there. Because we always consider only one > special number. Again, makes no sense to me. *If* there are infinitely many natural numbers, there are also infinitely many digit positions. To cover a number by a natural it is required that the number to be covered has only finitely many digit positions. So a number with infinitely many digit positions can not be covered by a natural number. > > > What counts (in the true sense of the word) is how many digits a number > > > has. And every natural number has only finitely many digits, regardless > > > of how many numbers there are. > > > > I never contradicted that. Why are you always saying that I claim > > something which I do not claim? > > You claimed just above that this fact is not relevant. Yes, I stated that it was irrelevant, not that it was wrong. K does *not* have finitely many digits (and is not a natural number), so your statement is irrelevant, but not wrong. > > > Please stop your current intermingling of infinity in sizes of numbers > > > and number of numbers. > > > There is no ifinite size. from hat I conclude that there is no infinite > > > set. > > > > Please stop claiming you found an inconsistency with the axiom of infinity > > when your basic assumption already is in contradiction with it. > > My basic assumption is the existence of this axiom. I *derive* a > contradiction. You do not. In all your derivations somewhere you assume that that axiom is false. > > > > But if you claim that the set if finite numbers is finite, there is > > > > a largest number. > > > > > > There is no largest number. The set of of natural numbers is infinite, > > > > Eh? Above you wrote that there is *not* an infinite set. Contradicting > > yourself? > > Important: There is no largest number. The axiom says only that the set > is infinite. I said, there is no largest number. The axiom says the set does exist. In and of itself the axiom does *not* state that the set is infinite. It is a theorem that the set is not finite, and so must be infinite. > > > There is no infinite natural number. Because there is no infinite > > > finite number - and every natural number is finite. Even the axiom of > > > infinity does not state the contrary. (But if the axiom of infinity had > > > to be satisfied, then a finite infinite number would be required. > > > Compare the staircase.) > > > > Again an assertion, without proof. > > The staircase is a proof. No. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |