Cantor Confusion
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From: Alan Morgan on 29 Sep 2006 22:46 In article <vOjTg.25590$QT.1504(a)tornado.rdc-kc.rr.com>, Poker Joker <Poker(a)wi.rr.com> wrote: > >"Alan Morgan" <amorgan(a)xenon.Stanford.EDU> wrote in message >news:efkegr$6d9$1(a)xenon.Stanford.EDU... > >>>if its true for ANY list, then it must be >>>true for a specific list. So if considering a single specific list >>>shows a flaw, then looking at ANY (ALL of them) list doesn't >>>help. >> >> But if it's true for ANY list then it must be true for a specific >> list. So if considering a single specific list shows a flaw then >> perhaps that list doesn't really exist. > >That's true, but that's not the entire story. > >Suppose I claim that I have a list that contains all the reals. >You claim you can take that list and construct a real not >on the list. You procede to show the construction. I would >claim that your construction is flawed because it is >self-referential, which it must be if I truly gave you a list of >all the reals. So in that *SPECIAL CASE*, unlike the >general case, your construction isn't valid. 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.... ? >> It's as if, after hearing the proof that sqrt(2) is irrational, >> you reply "But what if I give you a and b, both integers, such >> that a/b = sqrt(2)? Your argument is demolished!". Yes, if >> you provide what the proof shows can't exist then the proof is >> wrong. But you can't just assume that that thing exists. > >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. This analogy is flawed (although I think it is a subtle point). It's not that the construction is invalid for x=0, it's invalid in general. The construction of y is done by taking 1/x, but the properties of the real numbers don't permit dividing 1 by an arbitrary real number (and you don't have to provide a case where it doesn't work. You can just say "Please show me where it says that 1/x is always a real number"). The properties of lists, however, do permit identifiying particular entries in the list. If an argument is flawed for a specific case then it *must* be flawed in the general case as well. Must be. It should be possible to point out the flaw in the general case without reference to any specific cases. If the specific case is valid then there must be something wrong with the general case. But if the general case is valid then there must be something wrong with the specific case. Alan -- Defendit numerus
From: Virgil on 29 Sep 2006 22:51 In article <WHiTg.25583$QT.7978(a)tornado.rdc-kc.rr.com>, "Poker Joker" <Poker(a)wi.rr.com> 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. > > 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? Except that mathematicians can't prove "any" when there are counterexamples. And mathematicians can prove "any" in the list of reals case. > > Or are you just saying that you need not look at special > cases when we don't want to? Or is it that if a special > case is overlooked enough, then it no longer counts? I am saying that when one can prove something for "any" case then special cases are irrelevant.
From: Virgil on 29 Sep 2006 23:00 In article <eTiTg.25584$QT.11590(a)tornado.rdc-kc.rr.com>, "Poker Joker" <Poker(a)wi.rr.com> wrote: > "Poker Joker" <Poker(a)wi.rr.com> wrote in message > news:WHiTg.25583$QT.7978(a)tornado.rdc-kc.rr.com... > > > > By analogy, what you're saying is: > > > > For ANY x > > there is a procedure to find a y such that x/y = 1. > > > > Because we are using the verbage "ANY", we don't > > have to worry about special cases like when x = 0. > > That's how mathematicians work? > > > > Or are you just saying that you need not look at special > > cases when we don't want to? Or is it that if a special > > case is overlooked enough, then it no longer counts? > > Better yet: > > For ANY real number x there is a procedure to find a real number y, such > that x/y = 1. > y isn't a real number when x = 0. > Therefore, 0 isn't a real number. When Joker can prove "For ANY real number x there is a procedure to find a real number y, such that x/y=1." only then need we consider any special cases. When one proves something for "any" list of real numbers, one has proved it for "every" list of real numbers, and there is no longer any point in arguing special cases.
From: Virgil on 29 Sep 2006 23:01 In article <YUiTg.25585$QT.8397(a)tornado.rdc-kc.rr.com>, "Poker Joker" <Poker(a)wi.rr.com> 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. Or he is too polite to say what PJ's arguments are really worth.
From: Tony Orlow on 29 Sep 2006 23:00
Randy Poe 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? > > 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? Duh. Sounds like PJ's constructing an anti-diagonal. > >> 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 > It bears much resemblance to Cantor's second regarding uncountability of...a set. The original proof was regarding a complete language using at least two symbols, m and w, no? That was later conflated to a proof about the reals. Numbers are representations of quantity. It's related to powerset, via N=S^L. Tony |