Cantor Confusion
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From: Arturo Magidin on 29 Sep 2006 08:28 In article <8i_Sg.25507$QT.3131(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: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? To make clear and explicit was has up to now been implicit: that a list is a function from the natural numbers to the reals. > I think you are trying to make it >more complicated than it is. You are free to think what you want. Reality notwithstanding. >The OP already pointed out the problem with your argument. No, the original poster pointed out what he thinks is a problem with the proof. > The point >is that you ->CAN'T<- prove that for ->EVERY<- function unless you >assume that ->NONE<- of them have R as their image. Using capital letters does not make the objection true. Your inability to prove it does not affect my ability to prove that for every function function f:N->R, assuming only that it is a function from N to R, and no other properties, there exists x such that x is not in f(N). Since the proof assumes nothing at all about f other than the fact that it is a function, the proof shows that I can prove it for every function, and that I do not assume anything like what you assert is being assumed. >And since you >must assume that ->SOME MIGHT<- have R as their image, I assume nothing about the range of f, other than that it is contained in R. Whether or not it has R as the image is irrelevant. -- ====================================================================== "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
From: georgie on 29 Sep 2006 08:35 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.
From: georgie on 29 Sep 2006 08:38 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. Sorry, I mean x = sqrt(x).
From: Jesse F. Hughes on 29 Sep 2006 09:03 "Dave L. Renfro" <renfr1dl(a)cmich.edu> writes: > Each sequence of real numbers omits at least one real > number. (Sequence being a 1-1 and onto function from the positive > integers to the real numbers.) Er, are you sure about that definition of sequence? Seems to me that a sequence is any function from the positive integers to the reals. With your definition, the conclusion would be: every onto function N->R is not onto. This is true, of course, but is more likely to confuse the "Anti-Cantorians" rather than enlighten them. -- Jesse F. Hughes "I guess it's a passable day to die." -- Lt. Dwarf, /Star Wreck:In the Pirkinning"
From: Jesse F. Hughes on 29 Sep 2006 09:06
"david petry" <david_lawrence_petry(a)yahoo.com> writes: > The argument of the "anti-Cantorians" is that "real" mathematics is > computationally testable. Bullshit. That's *your* argument. Perhaps some other so-called "anti-Cantorians" have agreed with you, but that is not the motivation of every anti-Cantorian. -- "So, at this time, I'd like to assure you that I am not interested in making sure mathematicians worldwide get fired."--JSH Apr 28, 2003 "I'll have prosecutors knocking on your doors. I have no problem with any number of mathematicians spending time in jail."--JSH Jun 10, 2003 |