Cantor Confusion
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From: MoeBlee on 29 Sep 2006 23:26 Tony Orlow wrote: > This list must be in representation using 2 or more symbols, yes? You > cannot perform this proof in unary without the obvious result. Who, other than you, even proposes a unary representation of real numbers? Perhaps I'm wrong, but I get the impression that you think the diagonal argument is somehow hampered in its conclusiveness by the fact that some base greater than or equal to 2 is used. The conclusiveness of the argument is not at all hampered by that. We prove, using just first order logic and certain axioms, that there is no function from the set of natural numbers onto the set of real numbers (of course, as we have defined 'the set of natural numbers' and 'the set of real numbers', especially as our definitions provide that any other Peano system is isomorphic with the our defined set of natural numbers and any complete ordered field is isomorphic with our defined set of real numbers). The only options to disputing the proof are: 1. Reject first order logic. 2. Reject the axioms used in the proof (and, if I'm not mistaken, we don't need the axiom of infinity, as our proof would still go through as a statement of the impossibility of a bijection, whether or not the infinite sets in question even exist). 3. Show a step in the proof not authorized by first order logic applied to said axioms. Item 3, though, is not an option, since there are no such unauthorized steps. And, by the way, taking exception to whatever definitions are used is irrelevent since we can dispense with the definitions and use only the primitives (upon cost of convenience only). And reservations about what base is used are pointless, since we are only required to proof the impossibility of a bijection, which we have done. MoeBlee
From: William Hughes on 29 Sep 2006 23:30 Tony Orlow wrote: > William Hughes wrote: > > Poker Joker wrote: > >> "Dik T. Winter" <Dik.Winter(a)cwi.nl> wrote in message > >> news:J6CsBJ.Jys(a)cwi.nl... > >>> In article <070Tg.14143$8_5.3402(a)tornado.rdc-kc.rr.com> "Poker Joker" > >>> <Poker(a)wi.rr.com> writes: > >>>> "Randy Poe" <poespam-trap(a)yahoo.com> wrote in message > >>>> news:1159494111.724651.95600(a)i3g2000cwc.googlegroups.com... > >>>> > >>>>> That's incorrect. You don't have to assume none map onto R in order to > >>>>> prove none map onto R. > >>>>> > >>>>> The direct argument starts this way: Let f be any such function, from > >>>>> naturals to reals. > >>>> Certainly we should assume that f *MIGHT* have R as its image, right? > >>> You may assume that, but that assumption is not needed. > >> Certainly not for ostriches. > >> > >>>>> Now, are you saying that somehow that misses some possible functions > >>>>> from naturals to reals? How so? > >>>> No, but we haven't proven that the image of f can't be R in step #1, > >>>> right? > >>>> So step #2 isn't valid, right? > >>> Remember: > >>>> 1. Assume there is a list containing all the reals. > >>>> 2. Show that a real can be defined/constructed from that list. > >>>> 3. Show why the real from step 2 is not on the list. > >>>> 4. Conclude that the premise is wrong because of the contradiction. > >>> Why is step 2 invalid? > >> Do you always accept steps that have questionable validity? > >> > >>>> Under the most general assumption, we can't count out that > >>>> R is f's image, so defining a real in terms of the image of > >>>> f *MIGHT* be self-referential, and it certainly is if the image > >>>> of f is R. > >>> What is the problem here? > >> I assume you accept this proof that there are no complete lists > >> of reals: > >> > >> Let r be a real number between 0 and 1. Let r_n denote the nth digit > >> in r's decimal expansion. Let r_n = 5 if r_n = 4, otherwise let r_n = 4. > >> r isn't on any list of reals. Therefore there isn't a complete list of > >> reals. > > > > Hardly, but this "proof" does not reflect what is being done. > > > > You start out with a set of real numbers A, with a certain > > property, there is a surjective function from the natural > > numbers to A. In other words A is a list of real numbers. > > > > You define a process D(A) which gives you a real number. > > D depends only on the fact that a surjective function > > from the natural numbers exists, it does not depend on > > any other property of A. Thus D can be applied to any > > list . In particular, if we assume there is a list > > containing all the real numbers, D can be applied to this > > list. That this application will lead to a contradiction > > does not change the fact that D can be applied to the > > list. So step 2 is valid. > > > > Of course, we need not make this assumption. In this case > > the proof goes > > > > 1 Let A be a list > > 2 Use D to contruct a real number r > > 3 Show that r in not an element of A > > 4 Conclude that A does not contain all the > > real numbers > > > > Again, since D can be applied to any list > > step 2 is valid. > > > > - William Hughes > > > > This list must be in representation using 2 or more symbols, yes? You > cannot perform this proof in unary without the obvious result. Twaddle. This list is not in any representation. A list is a set of real numbers and a surjective function f from the naturals Not real numbers in unary notation, not real numbers in hexadecimal notation, not real numbers using a Cauchy sequence notation. Real numbers. The real numbers are independent of any notation. True one obvious way of defining D makes use of a notation (only some notations are useful here) but since the real numbers are independent of notation, we can choose any notation we find convenient. Given any real number we can express this number using binary notation, hexadecimal notation, etc. etc. The real number does not change and we can choose any notation (or more than one notation) we see fit. - William Hughes P.S. How would you go about describing the real numbers with a unary notation. If we only have one symbol we can only describe a countable number of distinct things. But no countable set of reals can contain all the reals.
From: Poker Joker on 30 Sep 2006 00:28 "MoeBlee" <jazzmobe(a)hotmail.com> wrote in message news:1159581066.478532.212600(a)e3g2000cwe.googlegroups.com... > Poker Joker wrote: > >> No its not. Its like if you give me a proof that zero isn't a real >> number: >> >> For ANY x, there is procedure to construct a y, such that x = 1/y. >> When x = 0, there is no y. >> Therefore x is not a real number. >> >> If everybody neglects the fact that the construction isn't valid >> for x=0, then the proof is flawless. > > The point is that we are NOT be able to prove that for any x there is y > such that x=1/y. > > Whatever we argue about x, if there is special case blocking the > generalization from x, then then our argument will NOT go through for > an arbitrary real number x. > > If we prove that for any x there is a y such that x=1/y, then would > have had to use the assumption that x not= 0. So that would NOT be an > argument regarding an arbitrary real number x. > > Look, here's what we have: > > Mr. Mathematician says: Let p be an arbitrary prime number. [Insert > proof here that p has the property that there is a prime number greater > than p.] Therefore, for ALL prime numbers p, we have that p has the > property that there is a prime number greater than p. > > Goober Boober says: But what if p is even? What if p is greater than > 100? What if p is a successor of a prime? Indeed, what if p is such > that there is NOT a prime greater than p? Hey, Mr. Mathematician, you > didn't take into consideration all these special cases, did you? No, > you didn't. So your argument is invalid. > > MoeBlee MoeBlee always uses the Goober Boober proof in all his work.
From: Poker Joker on 30 Sep 2006 00:40 "Alan Morgan" <amorgan(a)xenon.Stanford.EDU> wrote in message news:efklq5$t5q$1(a)xenon.Stanford.EDU... > But if the construction isn't valid in that special case then it > isn't valid in the general case either. Thus I *can't* have proven > it in the general case. That being the case, there should be a flaw > in the construction in general. Which is.... ? Forget about the OP and all other posts and consider your question. You think that if there is a problem with a special case, that alone isn't enough for there to be a problem in general. You want more proof that there is a problem in the general case. One specific problem isn't good enough for you to realize that there's a problem? How many problems does there need to be before there is a problem in general?
From: Poker Joker on 30 Sep 2006 00:45
"Virgil" <virgil(a)comcast.net> wrote in message news:virgil-9C1609.21071129092006(a)comcast.dca.giganews.com... > It is in mathematics. Once a proof for any list is established, it > covers every list. This list doesn't contain 4: 1 2 3 Proof: The 1st number isn't 4. The 2nd number isn't 4. The 3rd number isn't 4. That list does't contain 4 Therefore, Virgil believes that in mathematics, no list contains 4. |