Cantor Confusion
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From: MoeBlee on 29 Sep 2006 20:21 Poker Joker wrote: > "Arturo Magidin" <magidin(a)math.berkeley.edu> wrote in message > news:efhkpt$3136$1(a)agate.berkeley.edu... > > > If one proves that for EVERY function f:N->R, there exists x in R (which > > depends on f) such that x is not in f(N), then this proves that EVERY > > list of real numbers is incomplete. This proves that NO list can be > > complete, and that's what you want the proof for in the first place. > > Why the switch to functional notation? I think you are trying to make it > more complicated than it is. No, he's not making it more complicated than it is. The functional notation emphasizes that is FUNCTIONS we're talking about. A list is an enumeration is a certain kind of function. > > you ->CAN'T<- prove that for ->EVERY<- function unless you > assume that ->NONE<- of them have R as their image. Wrong. We consider an arbitrary function from N into R and show that N is not onto R. We do not assume that no function from N has R as the range; rather we PROVE that no function from N has R as the range. > And since you > must assume that ->SOME MIGHT<- have R as their image, there > exists no such real number because the definition of that real is > self-referential in that case. Whatever you mean by "self-referential", there is no step in the argument that uses anything other than first order logic applied to the axioms of set theory. And we don't need to assume anything as to what "might" be the case. We simply show that for an arbitrary function f whose domain is N, there is some real number not in the range of f. MoeBlee MoeBlee
From: MoeBlee on 29 Sep 2006 20:24 Poker Joker wrote: > Whether the proof is by-contradiction or not is immaterial. Either > way the real number from step #2 is defined in terms of itself under > the assumption that the list *MIGHT* contain all the reals. No, it is not. Either you are completely confused about a simple mathematical argument, or you're just belching smoke that you know to be smoke. MoeBlee
From: Poker Joker on 29 Sep 2006 20:28 "Poker Joker" <Poker(a)wi.rr.com> wrote in message news:WHiTg.25583$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.
From: Poker Joker on 29 Sep 2006 20:30 "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.
From: MoeBlee on 29 Sep 2006 20:31
Poker Joker wrote: > 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, we can't say that the definition is not > self-referential. Therefore it is meaningless. It's an arbitrary enumeration. We assume nothing about it other than that it is an enumeration. We don't assume that it is an enumeration of all real numbers and we don't assume that it is not an enumeration of all real numbers. On that basis, we prove that there there is a real number that is not in the range of the enumeration. There is nothing used except first order logic applied to the axioms of set theory. To reject the argument you have three choices: 1. Reject first order logic. 2. Reject at least one of the axioms of set theory used for the argument. 3. Show a step in the argument that uses anything not authorized by first order logic applied to the axioms of set theory.. And item 3 is not on the table, since there is no step in the argument that is not authorized by first order logic applied to the axioms of set theory. MoeBlee |