Cantor Confusion
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From: Herman Jurjus on 29 Sep 2006 14:53 Dik T. Winter wrote: [snip] > 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. With interest i observe that you people here on sci.math talk about constructivists in terms of 'us' and 'them'. ;-) -- Cheers, Herman Jurjus
From: A N Niel on 29 Sep 2006 16:44 In article <451d6c54$0$2031$ba620dc5(a)text.nova.planet.nl>, Herman Jurjus <h.jurjus(a)hetnet.nl> wrote: > > With interest i observe that you people here on sci.math talk about > constructivists in terms of 'us' and 'them'. > And I observe that Herman talks about "you people here" as though he were not here as well.
From: Poker Joker on 29 Sep 2006 17:58 "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.
From: Poker Joker on 29 Sep 2006 19:45 "Arturo Magidin" <magidin(a)math.berkeley.edu> wrote in message news:efj3bk$120f$1(a)agate.berkeley.edu... > 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) > ====================================================================== We all noticed you neglected this logic: 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.
From: Poker Joker on 29 Sep 2006 19:52
"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. |