An uncountable countable set
in [Math]
|
Prev: integral problem
Next: Prime numbers
From: mueckenh on 18 Aug 2006 07:30 Virgil schrieb: > In article <1155885500.337316.94880(a)b28g2000cwb.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Virgil schrieb: > > > > > In article <1155725007.690845.21360(a)i42g2000cwa.googlegroups.com>, > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > The representation of infinitely many natural numbers by > > > > the stairs requires infinitely many stairs. Infinitely many stairs > > > > require infinite height. > > > > > > Which in mathematics, a purely imagined world, is achievable, though not > > > so in any physical world. > > > > But it is disliked very much in mathematics, because the set of only > > *finite* numbers is pretended to be infinite. > > Does the set of points in a line have to "pretend" to be infinite? It has to pretend to exist (while it is only a line). > > That infinite sets are disliked by "Mueckenh" does not in anyway > indicate that they are disliked by mathematicians, nor that they cause > any problems for mathematicians that have to been adequately dealt with. > > And "mathematics" itself, being inanimate, has no opinion on the issue > at all. Fortunately for you. Otherwise it would curse many of those today's "mathematicians" and kick them out of the temple. Regards, WM
From: Franziska Neugebauer on 18 Aug 2006 08:36 mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: >> mueckenh(a)rz.fh-augsburg.de wrote: >> > Franziska Neugebauer schrieb: >> >> mueckenh(a)rz.fh-augsburg.de wrote: >> >> >> >> > An infinite sum of 1's is not infinite? >> >> >> >> n >> >> lim sum 1 = lim n =def L >> >> n -> oo i = 1 n -> oo >> >> >> >> There is no such L in N. >> > >> > Correct. >> >> The antecedent is true. >> >> > Therefore there are not infinitely many difference[s] of 1 >> > between natural numbers. >> >> Your consequent is proven false (see below). Therefore your >> implication is false, too. > > You are in error. Where precisely is the error? > You just proved it to be true. You may "just" as well prove the opposite. Please do so. > The set of natural numbers (i.e., finite numbers n, i.e., numbers > with finitely many differences of 1 between 1 and n) does not yield > infinitely many differences of 1. This is a reiteration not a proof. >> A difference of two numbers b and a is usually denoted as b - a. We >> introduce the difference operator "-" action upon ordered pairs: >> >> -(a, b) def= b - a >> >> "How many differences there are" means the cardinality of the set >> of all pairs {(a, b)}. >> >> Restricting a and b to omega and to "difference[s] of 1" one gets >> >> P def= {(a, b) | a, b e omega & -(a, b) = 1} >> = {(a, a + 1) | a e omega } >> >> Since there is a bijection between P and omega, namely >> >> B: P x omega def= {((a, a + 1), a) | a e omega}, >> >> it follows that P ~ omega, meaning P is of same cardinality as omega. >> >> Thus there are "as many difference[s] of 1 between natural numbers as >> there are natural numbers". Since the cardinality omega is infinite >> there *are* "infinitely many difference[s] of 1 between natural >> numbers". > > Correct, but your final conclusion is the typical mistake of set > theorists. Do you agree that P ~ omega? Do you agree that card (omega) /e omega? > 1) The set exist. > 2) The elements do not exist. What type of comment do you expect here? > You state it in the form: > > 1) There are infinitely many differences of 1. Do you agree that P ~ omega is the correct formalization of this statement? If not: What _is_ the correct formalization of the statement? > 2) The sum of these is not infinite. Why not? You have agreed to that this limit does not exist ("There is no L in N"). So the sum is at least _not_ _finite_ (not in omega). > You should try to comprehend what "existence" means. Do not confuse me with your therapist. F. N. -- xyz
From: Dik T. Winter on 18 Aug 2006 08:38 In article <1155885445.248401.9340(a)p79g2000cwp.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > In article <1155725133.570350.44280(a)74g2000cwt.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > mike4ty4(a)yahoo.com schrieb: > > ... > > > The set of all constructible numbers including all real numbers in > > > Cantor-lists and all of their diagonal numbers is countable. > > > Nevertheless most people assert that the construction of a diagonal > > > number would show the uncountability of this countable set of > > > constructible numbers. > > > > But can the list of constructable numbers be constructed? > > No, of course it cannot, because only finite sets can be constructed, > at least if there is not a law connecting infinitely many numbers. When talking with people, speak their language. I meant constructed in a mathematial sense. > > That they > > are countable comes from other considerations. > > It doesn't matter where that comes from. In fact all numbers which can > be constructed form a countable set. Proof? And pray use the mathematical sense of constructed. Not what you think it should mean. > (Unconstructible numbers were not at all taken into account > when Cantor's proof was published.) Yes, right. Constructable numbers come from considerations in complexity theory. -- 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 18 Aug 2006 09:02 In article <1155885821.815144.187270(a)m73g2000cwd.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: .... > > To be more correct. The first proof is not relying on the representation > > of reals, the second (diagonal) proof is not about reals at all. I think > > it was Zermelo who first modified the second proof to a proof about the > > reals. I do not know whether you follow all, but at one time you asked > > Mueckenheim whether Cantor considered dual repersentations in his diagonal > > proof. The answer was: no. To clarify, in Cantor's diagonal proof there > > are no dual representations, so there was no need for Cantor to consider > > them. It is about infinite sequences of symbols. But Mueckenheim was > > insidious there, as he answered no, while not giving the clarification. > > Cantor considered his 2nd proof as being /valid/ for real numbers. But > he did not use irrational numbers: "Aus dem in =C2=A7 2 Bewiesenen folgt > n=C3=A4mlich ohne weiteres, da=C3=9F beispielsweise die Gesamtheit aller > /reellen Zahlen/ eines beliebigen Intervalles sich nicht in der > Reihenform=EF=80=A0=EF=80=AE=EF=80=AE=EF=80=AE=EF=80=A0darstellen l=C3=A4= > =C3=9Ft. > Es l=C3=A4=C3=9Ft sich aber von /jenem Satze/ ein viel einfacherer Beweis > liefern, der unabh=C3=A4ngig von der Betrachtung der Irrationalzahlen ist." > (The italics are mine) Where can I find that quote? I have problems with reading utf-8. > > > The diagonal proof certainly shows that the set of constructible > > > numbers cannot be listed in its entirety. > > > > That is wrong (as I see it). There exists a list of constructible numbers > > (countability proves that). But that list is not constructible, so the > > diagonal number is not constructible. > > No infinite list (of independent numbers) is constructible, because no > one can construct infinitely many different real numbers. This argument > shows that the whole Cantor diagonal proof is void. > > But if there are any lists of constructible numbers, then their > diagonal numbers are constructible too. And this construction does > *not* show that there are uncountably many constructible numbers. This > argument shows again that the whole Cantor diagonal proof is void. Well, let me be clear. Constructable numbers is used in two senses in mathematics. The sense I meant is "computable numbers". So let me refrase with this word: The set of computable numbers is countable, but the diagonal is not computable. -- 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 18 Aug 2006 09:31
In article <1155899470.573313.313470(a)m73g2000cwd.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > > Now you may claim, that the mathematics of finity does not hold in > > > infinity, but that there everything is different. But that is not > > > mathematics in my opinion. May be that it is mathematics in some one > > > other's opinion. Nevertheless it is not useful to continue the > > > discussion on this topic between us. > > > > I do not only *claim* that 0.111... can be indexed by the natural numbers; > > I can *prove* it, as I have done again and again. > > You supposed that all digits of 0.111... could be indexed, and than > you proved that they can be indexed. No. I *defined* 0.111... such that all digits could be indexed. > > The only things you > > are doing is handwaving and circular reasoning. Indeed, the discussion > > is useless if you are not able to formulate definitions and proofs. > > I have formulated the definition according to which my list is > constructed. You try to convince us that > "there is no n such that 0.111... can be indexed by the numbers 1 to n. > This does *not* mean that 0.111... can not be indexed by the complete > list." > > And that is simply nonsense. All finite digit positions are contained > in the list, most several times, but every one at least one time. I say no to the first statement and yes to the second. There is no contradiction. > > > > Each finite segment can be > > > > covered, and each finite index can be covered. But the total is not > > > > finite, but contains only finite digit positions. > > The digit positions constitute the number. You cannot build a house of > bricks and claim the total house is wooden. I do not claim such a thing. > > So, in your opinion the axiom of infinity is purest nonsense. > > No. It leads to nonsense in the framework of mathematics, for instance > in contradicts the theorem that 0.111... is not a member of the > sequence of sequences of 1's. It states no such thing. It gives the theorem that 0.111... is not a member of the sequece of finite sequences of 1's. > > >You have > > a right to opiniate that. But do *not* claim that you have found an > > inconsistency with the axiom of infinity, because you have not. > > Otherwise I had to accept your handwaving argument that all list > sequences could index a sequence which does not belong to the list. Why > should I? Because there is a proof that a number K that I *define* as K[p] = 1 for all p in N, and without any other digit is not in the list (because it is not a natural number), but can be indexed (because all the index positions are natural numbers)? What is the handwaving here. > > > > > Because your assertion that K could be completely indexed and covered > > > > > is equivalent to K being in the true list. > > > > > > > > My assertion is that K can be completely indexed but not completely > > > > covered. Again you misquote my assertion. > > > > > > Indexing n implies covering all digit positions m =< n. Indexing all n > > > implies covering all m =< n. > > > > What is that 'n' in the last equation? It can not be the 'n' in 'all n' > > because that is an arbitrary 'n'. > > n e |N is a whole number. All n e |N are whole numbers. Yes, I know that. See again what you wrote: "Indexing all n implies covering all m <= n." ^ ^ the first n and the last n can not mean the same thing, so what does the last n mean. > > > Do you agree that every natural number which can index a digit position > > > is in the list? > > > > Yes. > > > > > If a number is not in the list, what does this fact say about its digit > > > positions? > > > > Nothing. If all digit positions are natural numbers, they can all be > > indexed by numbers from the list. > > That is correct. But all numbers with those digit positions are already > in the list. All finite digit positions are contained in the list, most > several times, but every one at least one time. No, there is *no* number in the list that has *all* finite digit positions occupied. That number can be indexed. The only numbers in the list are those that have finitely many finite digit positions occupied. And if *all* finite digit positions are occupied, there are infinitely many digit positions occupied. > > > > Yes. When you can add all those numbers. But the result is not > > > > a natural number. > > > > > > Exactly. Therefore there cannot exist an infinite number of natural > > > numbers. > > > > Why the 'therefore'? Can you provide a proof? > > Because the sum of infinitely many differences of 1 would make up an > infinite number. That is not a proof. Indeed, the sum of infinitely many differences of 1 make up an infinite number, but it is not a natural number. I see no contradiction. > > > > > Why should a set which is not finite be actually infinite > > > > > (i.e. have a cardinal number)? > > > > > > > > A set that is not finite is infinite, that is the definition > > > > of the word infinite. > > > > > > No. Every (potentially) infinite set is finite. > > > > That is quite non-standard terminology. In standard terminology something > > that is not finite is infinite and the reverse. However, I have no idea > > what a 'potentially infinite set' is. Can you provide a definition? > > Peano axioms, inductive set. If there is n, then there is n+1. But this > does not mean that an infinite set does actually "exist". The axiom of infinity asserts that it does exist. > > > And all extremely false too. > > > > Elementary mathematics with equivalence classes is false? What is false > > about it? > > The assumption that an infinite set had a cardinal number. The > assumption that the limit omega would exist. There is no assumption, there is an axiom that asserts it. You keep stating that the axiom of infinity is false, without ever showing that it leads to a contradiction. > > > > > You may deny this, nevertheless it is true. If there is an actual > > > > > number of steps, then the width is equal to the lenghts is aleph_0. > > > > > > > > You keep asserting that. > > > > > > If |