Cantor Confusion
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From: Tonico on 29 Sep 2006 04:13 Poker Joker wrote: > "Arturo Magidin" <magidin(a)math.berkeley.edu> wrote in message > news:efhkpt$3136$1(a)agate.berkeley.edu... > > > If one proves that for EVERY function f:N->R, there exists x in R (which > > depends on f) such that x is not in f(N), then this proves that EVERY > > list of real numbers is incomplete. This proves that NO list can be > > complete, and that's what you want the proof for in the first place. > > Why the switch to functional notation? I think you are trying to make it > more complicated than it is. > > The OP already pointed out the problem with your argument. The point > is that you ->CAN'T<- prove that for ->EVERY<- function unless you > assume that ->NONE<- of them have R as their image. And since you > must assume that ->SOME MIGHT<- have R as their image, there > exists no such real number because the definition of that real is > self-referential in that case. ***************************************************************** This is huge nonsense. The functional notation is rather handy when trying to talk mathematics: mimicry and poetry not necessarily help here. The argument seems extremely simple to me...so simple that some seem to feel it MUST be wrong, for some reason (perhaps the reason is that if something ain't complex then it can't be maths...): take ANY list of real numbers (or take ANY injective mapping f:N --> R). Then it can be shown with the genial and simple diagonal argument that Cantor came up, that some real is missing there (that there's some x in R s.t. x doesn't belong to f(N)). Tell it this way, tell it that way: all is correct mathematicalwise (?!) unless proved mathematicalwise (?!) that it isn't....someone wanna buy a bridge AND the river under it? Tonio
From: Randy Poe on 29 Sep 2006 06:46 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? 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? - Randy
From: Dik T. Winter on 29 Sep 2006 07:56 In article <virgil-FDA0EC.15584628092006(a)comcast.dca.giganews.com> Virgil <virgil(a)comcast.net> writes: > In article <j%WSg.22037$SY6.17302(a)reader1.news.jippii.net>, > Aatu Koskensilta <aatu.koskensilta(a)xortec.fi> wrote: .... > > > That is a proof by contradiction, which many constructionists object to. > > > > No, they don't, presuming you mean constructivists. A direct > > constructivistic proof of ~A is a proof of contradiction from A. What > > one can't do constructively is to prove A by proving that ~A leads to a > > contradiction. > > In my understanding, a "proof by contradiction" is essentially a proof > of A by proving ~A false, which requires accepting the law of the > excluded middle. This version they indeed do object to. But a "proof by contradiction" can also be a proof of ~A by proving A false, which they do not object to. In constructivist sense, when you proof that ~A leads to a contradiction, you have proven ~~A, but not necessarily A. -- 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 29 Sep 2006 08:06 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. > > 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? > 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? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Arturo Magidin on 29 Sep 2006 08:25
In article <N_YSg.1208$3E2.403(a)tornado.rdc-kc.rr.com>, Poker Joker <Poker(a)wi.rr.com> 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. Because a direct proof is simpler than a proof by contradiction. > ANY list opens up more possiblities than >a single list. Any list does not require you to assume that there is a "single list" which some some particular property. -- ====================================================================== "It's not denial. I'm just very selective about what I accept as reality." --- Calvin ("Calvin and Hobbes" by Bill Watterson) ====================================================================== Arturo Magidin magidin-at-member-ams-org |