Cantor Confusion
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From: Randy Poe on 28 Sep 2006 21:41 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? To make the statement more compact. > I think you are trying to make it > more complicated than it is. It was one sentence. How complicated is that? > 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. 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. Now, are you saying that somehow that misses some possible functions from naturals to reals? How so? 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. What function from naturals to reals is missed by that argument? - Randy
From: cbrown on 28 Sep 2006 22:01 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. 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. Simpler is the following: 1) Let n be any even number. 2) n is then of the form 2*n. 3) It follows that n cannot be odd. 4) Therefore, there is no such thing as an oddven number. In the case of the original argument, substitute "is a list of real numbers" for "even", "is a complete mapping" for "odd", and "is a complete list of the reals" for "oddven". Cheers - Chas
From: Poker Joker on 28 Sep 2006 22:39 "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.
From: Poker Joker on 28 Sep 2006 22:51 <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, we can't say that the definition is not self-referential. Therefore it is meaningless.
From: William Hughes on 28 Sep 2006 23:07
Poker Joker 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, we can't say that the definition is not > self-referential. Therefore it is meaningless. Nope. The steps are as follows. Consider a list of real numbers L. (At this point we do not know if L contains all real numbers or not). In your terminology, L might contain all real numbers. 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 Apply the procedure D to the list L to get a real number r=D(L). Note that D(L) exists because we can apply D to any list. The fact that L *might* contain all the real numbers is not a problem because D was defined in such a way that it can be applied to any list. We now have a list L and a real number r. We argue that by the way r was constructed r cannot be an element of L. At this point we conclude that L does not contain all the real numbers. There is nothing self referential here. - William Hughes |