Cantor Confusion
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From: MoeBlee on 29 Sep 2006 20:52 Poker Joker wrote: > "MoeBlee" <jazzmobe(a)hotmail.com> wrote in message > news:1159575840.752320.304250(a)h48g2000cwc.googlegroups.com... > > Poker Joker wrote: > >> Whether the proof is by-contradiction or not is immaterial. Either > >> way the real number from step #2 is defined in terms of itself under > >> the assumption that the list *MIGHT* contain all the reals. > > > > No, it is not. Either you are completely confused about a simple > > mathematical argument, or you're just belching smoke that you know to > > be smoke. > > Or you can't come up with anything better to say. No, I refuted you directly in previous posts. The problem here is very basic: You don't understand what a mathematician means when he argues such as this: "Let x be an arbitrary even number. [Then he proves some property P about x.] So, since x is arbitrary, property P holds for all even numbers." MoeBlee
From: Poker Joker on 29 Sep 2006 20:56 "MoeBlee" <jazzmobe(a)hotmail.com> wrote in message news:1159576895.349562.212010(a)i42g2000cwa.googlegroups.com... > Poker Joker wrote: >> "MoeBlee" <jazzmobe(a)hotmail.com> wrote in message >> news:1159574564.041788.85490(a)k70g2000cwa.googlegroups.com... >> > Poker Joker wrote: >> >> "Arturo Magidin" <magidin(a)math.berkeley.edu> wrote in message >> >> news:efgfhd$261u$1(a)agate.berkeley.edu... >> >> > In article <1159410937.013643.192240(a)h48g2000cwc.googlegroups.com>, >> >> > <the_wign(a)yahoo.com> wrote: >> >> >>Cantor's proof is one of the most popular topics on this NG. It >> >> >>seems that people are confused or uncomfortable with it, so >> >> >>I've tried to summarize it to the simplest terms: >> >> >> >> >> >>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. >> >> > >> >> > This is hardly the simplest terms. Much simpler is to do >> >> > a ->direct<- >> >> > proof instead of a proof by contradiction. >> >> > >> >> > 1. Take ANY list of real numbers. >> >> > 2. Show that a real can be defined/constructed from that list. >> >> > 3. Show that the real from step 2 is not on the list. >> >> > 4. Conclude that no list can contain all reals. >> >> > >> >> >> >> How can it be simpler if the list can be ANY list instead of a >> >> particular one. ANY list opens up more possiblities than >> >> a single list. >> > >> > By 'any' we mean an arbitrary one. The way we talk about an arbitrary >> > object is to choose a variable not free in any previous line in the >> > argument nor free in the conclusion we will eventually draw and then >> > use the rule of universal generalization to draw our eventual >> > conclusion. So if we want to prove something about an arbitrary >> > enumeration of denumerable sequences of digits, we say, "Suppose f is >> > an enumeration whose range is a subset of the set of denumerable >> > sequences of digits" (and, of course, we will presume NOTHING about f >> > other than that it is an enumeration whose range is a subset of the set >> > of denumerable sequences of digits. >> > >> >> Also, if its true for ANY list, then it must be >> >> true for a specific list. >> > >> > If the property holds for any list, then, if there exists a list, then >> > the property holds of any such list that exists. >> > >> >> So if considering a single specific list >> >> shows a flaw, then looking at ANY (ALL of them) list doesn't >> >> help. >> > >> > If there is a specific list that does not have the property, then we >> > will not be able to prove that the property holds of an arbitrary list. >> > >> > But I don't know what specific list you think is "flawed". Nor do I see >> > what your point has to do with Arturo's point that we don't have to >> > adopt a reductio ad absurdum assumption, since we can just show for an >> > arbitrary list that it does not list every real number. >> > >> > MoeBlee >> >> By analogy, what you're saying is: >> >> For ANY x >> there is a procedure to find a y such that x/y = 1. > > I said no such thing by analogy or otherwise. I did. >> Because we are using the verbage "ANY", we don't >> have to worry about special cases like when x = 0. >> That's how mathematicians work? > > No, you've got it reversed. No, I don't. You neglected to point out the difference in the analogy. I assume you might not understand analogy so I provided a link to a good site for you to start learning: http://en.wikipedia.org/wiki/Analogy > If we say 'any' to mean an arbitrary one, > then we don't assume that it falls under a special case or that it > doesn't fall under a special case. If there are special cases such that > the property does not hold of those objects that are a special case, > then the universal genearlization WON'T go through. It is the VERY > NATURE of the rule of universal generalization that it won't go through > if there is even a single special case that precludes the > generalization. After you figure out the analogy, then maybe you should come back and discuss this some more. It's quite obvious you need to learn the analogy before you'll understand. > You seem completely unfamiliar with such basic principles of > mathematical reasoning. Why don't you read a book on the subject? I'm sure from a person who has no understanding of what an analogy is, my argument may be incomprehensible and therefore beyond your grasp. I'm sorry about that, but I'm sure the link I gave you will help you in the future.
From: Randy Poe on 29 Sep 2006 21:00 Poker Joker wrote: > "Randy Poe" <poespam-trap(a)yahoo.com> wrote in message > news:1159548960.704514.113100(a)m7g2000cwm.googlegroups.com... > > > > georgie wrote: > >> Randy Poe wrote: > >> > > >> > Uh, no. We just assume that f is a function that takes naturals > >> > and produces reals. > >> > > >> > For instance, let f(x) = sqrt(x). > >> > > >> > Do I have to assume that f(x) "might have R as its image" in order > >> > to talk about f(x) = sqrt(x)? Can't I just talk about f(x) = sqrt(x) > >> > and examine what properties is has rather than speculating about > >> > what it might have? > >> > >> No, but f(x) = sqrt(x) isn't true in general. We don't take it as > >> being valid for all x. > > > > What do you mean "valid"? That's my definition of f(x). > > > > For any x, what I mean by f(x) is the value of sqrt(x). > > Thats not the same as defining x in terms of the set of all reals. What in the world are you talking about? When I define f:N->R, I am defining a function that takes naturals as input and produces reals as output. How the heck does sqrt(x) fail that test? What do you mean by "defining x in terms of the set of all reals"? f maps naturals to reals. x is a natural. - Randy
From: Poker Joker on 29 Sep 2006 21:03 "MoeBlee" <jazzmobe(a)hotmail.com> wrote in message news:1159577565.810365.174360(a)m7g2000cwm.googlegroups.com... > Poker Joker wrote: >> "MoeBlee" <jazzmobe(a)hotmail.com> wrote in message >> news:1159575840.752320.304250(a)h48g2000cwc.googlegroups.com... >> > Poker Joker wrote: >> >> Whether the proof is by-contradiction or not is immaterial. Either >> >> way the real number from step #2 is defined in terms of itself under >> >> the assumption that the list *MIGHT* contain all the reals. >> > >> > No, it is not. Either you are completely confused about a simple >> > mathematical argument, or you're just belching smoke that you know to >> > be smoke. >> >> Or you can't come up with anything better to say. > > No, I refuted you directly in previous posts. > > The problem here is very basic: You don't understand what a > mathematician means when he argues such as this: "Let x be an arbitrary > even number. [Then he proves some property P about x.] So, since x is > arbitrary, property P holds for all even numbers." There also needs to be assurance that the property P actually holds for all x. Just because some people can't come to grips with the fact that there are special cases to consider, doesn't mean they aren't there. I understand FULLY your thought process. The ONLY difference between your understanding of the proof and mine is that I understand the special case and you don't.
From: Randy Poe on 29 Sep 2006 21:04
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? Why does step 2 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. That doesn't make sense. You are saying that every digit of r both is equal to 4 and is equal to 5. Consider r = 0.00000000... So you're saying the first digit of r is 4 because the first digit of r isn't 4? What the hell are you talking about? > r isn't on any list of reals. Therefore there isn't a complete list of > reals. That bears no resemblance at all to a proof. - Randy |