Cantor Confusion
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From: Virgil on 29 Sep 2006 00:18 In article <sH%Sg.14138$8_5.3593(a)tornado.rdc-kc.rr.com>, "Poker Joker" <Poker(a)wi.rr.com> wrote: > "Virgil" <virgil(a)comcast.net> wrote in message > news:virgil-1201BC.19311828092006(a)comcast.dca.giganews.com... > > In article <C2ZSg.1209$3E2.571(a)tornado.rdc-kc.rr.com>, > > "Poker Joker" <Poker(a)wi.rr.com> wrote: > > > >> "Peter Webb" <webbfamily-diespamdie(a)optusnet.com.au> wrote in message > >> news:451b5130$0$28952$afc38c87(a)news.optusnet.com.au... > >> > >> > You produce ANY list of all Reals, I can show you a missing real. > >> > Therefore I can do it for ALL lists, and hence there is no complete > >> > list > >> > of Reals. > >> > >> If he shows you a list of all reals you can't. > > > > The point is that if there is no such list he cannot show you one, and > > the proof show that there is no such list. > > The point is that if there is such a list the proof that there isn't one > isn't valid, but even more, under the assumption that there *MAY* > be such a list, the proof is not valid because it would entail a > self-referential definition. > > 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. Maybe in Jokerland, but not in any real world. If, as has been proved, EVERY list leaves out at least one real, then NO list can possible list all of them. Your objections have no logic behind them, and in, fact, contradict the status quo.
From: Virgil on 29 Sep 2006 00:22 In article <SS%Sg.14140$8_5.8299(a)tornado.rdc-kc.rr.com>, "Poker Joker" <Poker(a)wi.rr.com> wrote: > <cbrown(a)cbrownsystems.com> wrote in message > news:1159495306.456537.79590(a)e3g2000cwe.googlegroups.com... > > > Apply your logic to "oddven". > > > > A natural number is even if it is of the form 2*n, and odd if it is not > > of this form. A number is oddven if it is both even and odd. > > > > 1) Assume that n is oddven. > > 2) If it is oddven, then it is even; so it is of the form 2*n. > > 3) But if it is oddven, then it is odd; and therefore it is not of the > > form 2*n. > > 4) Contradiction; therefore no such n exists. > > First, the OP isn't saying anything is wrong with "proofs > by contradiction". He's saying that there is a meaningless > definition like "This statement is false." > > There is no meaningless definition in your step 2. > > Defining a real number in terms of all real numbers in a > set is self referential if the set contains all the real > numbers. The same is true for lists. So under the > assumption that the list *MIGHT* contain all the real > numbers One need not make that assumption. One need only refrain from assuming it to be impossible.
From: Virgil on 29 Sep 2006 00:24 In article <070Tg.14143$8_5.3402(a)tornado.rdc-kc.rr.com>, "Poker Joker" <Poker(a)wi.rr.com> wrote: > "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? Why make any assumption at all?
From: Virgil on 29 Sep 2006 00:26 In article <ad0Tg.14144$8_5.737(a)tornado.rdc-kc.rr.com>, "Poker Joker" <Poker(a)wi.rr.com> wrote: > "William Hughes" <wpihughes(a)hotmail.com> wrote in message > news:1159499239.811924.305030(a)c28g2000cwb.googlegroups.com... > > > Define a procedure D, that takes a list M and produces > > a real number r=D(M). This procedure D will work on any > > list M > > If M contains all reals, it contains r, so: > > r = D( M : r in M) > > See the self reference? No!
From: Eric Schmidt on 29 Sep 2006 02:38
Poker Joker wrote: > "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? Why? What does it even mean to assume something might be true? I could "assume" that Goldbach's Conjecture might be true, but that's really not assuming anything substantial. >>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? It's only *VALID* if R isn't the image of f. > But without step #2 being valid under *ALL* conditions, the proof > as a whole breaks down. Nothing in step 2 (or anything else) assumes or requires that R isn't the image of f. >>Given only that assumption about f (which includes ALL POSSIBLE >>SUCH FUNCTIONS) it follows that f misses some reals. It follows >>just from assuming f maps naturals to reals. > > > 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. So, in general, the best we can say is that the > real from step #2 might be self-referential and hasn't been > shown not to be so without assuming the conclusion is true. > I don't usually accept proofs that *REQUIRE* you to assume > the conclusion is true for the proof to be valid. At best, the proof > might show the conclusion to be independent of the premises. Defining a real from a set of reals is not self-referential, even if the set contains all reals. The fact that the real may be in the set does not make the definition self-referential. For example, consider a function g whose domain is the set of all subsets of R. For S a subset of R, g(S) is the supremum of S if S is non-empty and bounded above. Otherwise, g(S) = 0. Do you assert that this definition is self-referential if S = R? If S = {x in R : x <= 0}? >>What function from naturals to reals is missed by that argument? > > > I'm not missing any, because I'm considering functions with R as > their image. It seems that you are the one not considering > those functions and their implication for step #2. You have not given any coherent explanation of why such functions are a special case. -- Eric Schmidt -- Posted via a free Usenet account from http://www.teranews.com |