Cantor Confusion
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From: Virgil on 2 Oct 2006 23:16 In article <j8iUg.28330$8_5.24730(a)tornado.rdc-kc.rr.com>, "Poker Joker" <Poker(a)wi.rr.com> wrote: > "Virgil" <virgil(a)comcast.net> wrote in message > news:virgil-B233B9.12324230092006(a)comcast.dca.giganews.com... > > In article <93uTg.1860$3E2.1673(a)tornado.rdc-kc.rr.com>, > > "Poker Joker" <Poker(a)wi.rr.com> wrote: > > > >> "Virgil" <virgil(a)comcast.net> wrote in message > >> news:virgil-4F0272.01064530092006(a)comcast.dca.giganews.com... > >> > >> > Is that is the best PJ can do? Personal attacks are the last refuge of > >> > the incompetent. > >> > >> That's been your ONLY attack. > > > > I have, to the satisfaction of anyone competent in mathematics, repeated > > other's proofs that, despite PJ's objections, no list of reals can > > contain all reals. > > The NG is hardly "anyone competent in mathematics" that's why > Virgil is so well accepted. In addition to the NG, there are all those textbooks and reference works that agree that the Cantor "diagonal" proof is valid. Which is where I found those proofs that P.J. does not accept.
From: Arturo Magidin on 3 Oct 2006 08:43 In article <l1iUg.28279$8_5.20941(a)tornado.rdc-kc.rr.com>, Poker Joker <Poker(a)wi.rr.com> wrote: [...] >I see no proof here. The proof was not here. The proof was presented earlier. But, since you can't be bothered: (i) Let f:N->R be any list of real numbers. (ii) For each n in n, let d_n be the n-th decimal digit of f(n). (iii) Define e:N->{5,6} as follows: if d_n = 5, then e(n)=6. if d_n is not 5, then e(n)=5. (iv) Let r = sum_{n=1}^{infty} e(n)/10^n. (v) r is a real number, being defined via a Caucy sequence. (vi) Claim: r is not in the image f(N). Proof of claim: For all n in N, the n-th decimal digit of f(n) is different from e(n). Therefore, |f(n)-r|>=1/10^n >0, so f(n) is different from r. As this holds for all n in N, r is not in f(N) = {f(n) : n in N}. (vii) As the only property of f that was used is the fact that it is a function from N to R, the argument holds for all functions from N to R. In particular, for every list f:N->R of real numbers, there exists (at least one) number r, which may depend on f, such that r is not in the range of f. Therefore, no list can contain all real numbers. > What am I to make of your statement that you did? You are to make whatever you want. >You are not sincere. And you aren't honest. >> No doubt, your understanding of that is about as accurate as your >> understanding of Cantor's proof or of mathematical arguments in >> general. I wouldn't try to borrow against its value. > >I understand it much better than you think. I'm sure you think you do. Or at least, I'm sure you want to pretend you think you do. -- ====================================================================== "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 3 Oct 2006 08:54 Arturo Magidin wrote: > In article <l1iUg.28279$8_5.20941(a)tornado.rdc-kc.rr.com>, > Poker Joker <Poker(a)wi.rr.com> wrote: > > [...] > > > >I see no proof here. > > The proof was not here. The proof was presented earlier. But, since > you can't be bothered: > > (i) Let f:N->R be any list of real numbers. > (ii) For each n in n, let d_n be the n-th decimal digit of f(n). > (iii) Define e:N->{5,6} as follows: if d_n = 5, then e(n)=6. > if d_n is not 5, then e(n)=5. > (iv) Let r = sum_{n=1}^{infty} e(n)/10^n. > (v) r is a real number, being defined via a Caucy sequence. > (vi) Claim: r is not in the image f(N). > Proof of claim: For all n in N, the n-th decimal digit > of f(n) is different from e(n). Therefore, > |f(n)-r|>=1/10^n >0, so f(n) is different from r. > As this holds for all n in N, r is not in > f(N) = {f(n) : n in N}. > > (vii) As the only property of f that was used is the fact that it is a > function from N to R, the argument holds for all functions from > N to R. In particular, for every list f:N->R of real numbers, > there exists (at least one) number r, which may depend on f, > such that r is not in the range of f. Therefore, no list can > contain all real numbers. Let's see, the first case I can think of is a special case. Your proof doesn't work for an empty list. I'm sure you are now going to make modifications or excuses, but you obviously aren't paying attention to special cases. I think the jury votes that you ignore special cases. DONE.
From: Randy Poe on 3 Oct 2006 09:20 georgie wrote: > Arturo Magidin wrote: > > In article <l1iUg.28279$8_5.20941(a)tornado.rdc-kc.rr.com>, > > Poker Joker <Poker(a)wi.rr.com> wrote: > > > > [...] > > > > > > >I see no proof here. > > > > The proof was not here. The proof was presented earlier. But, since > > you can't be bothered: > > > > (i) Let f:N->R be any list of real numbers. > > (ii) For each n in n, let d_n be the n-th decimal digit of f(n). > > (iii) Define e:N->{5,6} as follows: if d_n = 5, then e(n)=6. > > if d_n is not 5, then e(n)=5. > > (iv) Let r = sum_{n=1}^{infty} e(n)/10^n. > > (v) r is a real number, being defined via a Caucy sequence. > > (vi) Claim: r is not in the image f(N). > > Proof of claim: For all n in N, the n-th decimal digit > > of f(n) is different from e(n). Therefore, > > |f(n)-r|>=1/10^n >0, so f(n) is different from r. > > As this holds for all n in N, r is not in > > f(N) = {f(n) : n in N}. > > > > (vii) As the only property of f that was used is the fact that it is a > > function from N to R, the argument holds for all functions from > > N to R. In particular, for every list f:N->R of real numbers, > > there exists (at least one) number r, which may depend on f, > > such that r is not in the range of f. Therefore, no list can > > contain all real numbers. > > Let's see, the first case I can think of is a special case. > Your proof doesn't work for an empty list. That doesn't meet the qualifications of a function f:N->R. How are you defining this "empty list" in terms of f:N->R? What is f(1)? What is f(1000)? - Randy
From: Arturo Magidin on 3 Oct 2006 10:03
In article <1159880065.516975.222930(a)m7g2000cwm.googlegroups.com>, georgie <geo_cant(a)yahoo.com> wrote: > >Arturo Magidin wrote: >> In article <l1iUg.28279$8_5.20941(a)tornado.rdc-kc.rr.com>, >> Poker Joker <Poker(a)wi.rr.com> wrote: >> >> [...] >> >> >> >I see no proof here. >> >> The proof was not here. The proof was presented earlier. But, since >> you can't be bothered: >> >> (i) Let f:N->R be any list of real numbers. >> (ii) For each n in n, let d_n be the n-th decimal digit of f(n). >> (iii) Define e:N->{5,6} as follows: if d_n = 5, then e(n)=6. >> if d_n is not 5, then e(n)=5. >> (iv) Let r = sum_{n=1}^{infty} e(n)/10^n. >> (v) r is a real number, being defined via a Caucy sequence. >> (vi) Claim: r is not in the image f(N). >> Proof of claim: For all n in N, the n-th decimal digit >> of f(n) is different from e(n). Therefore, >> |f(n)-r|>=1/10^n >0, so f(n) is different from r. >> As this holds for all n in N, r is not in >> f(N) = {f(n) : n in N}. >> >> (vii) As the only property of f that was used is the fact that it is a >> function from N to R, the argument holds for all functions from >> N to R. In particular, for every list f:N->R of real numbers, >> there exists (at least one) number r, which may depend on f, >> such that r is not in the range of f. Therefore, no list can >> contain all real numbers. > >Let's see, the first case I can think of is a special case. >Your proof doesn't work for an empty list. Notice where I wrote "Let f:N->R be any list of real numbers"? Do you know what the notation "f:N->R" means? It means "a function, whose name is f, whose domain is N, and whose range is contained in R." Since list, then, means in this case "a function from N to R", there is no such thing as "an empty list". A function from N to R is defined on each and every natural number, because it is a function with domain N. > I'm sure you >are now going to make modifications or excuses, No. I'm just going to point out you were unable to understand even the first line of the proof. > but you >obviously aren't paying attention to special cases. Or you aren't paying attention to definitions. -- ====================================================================== "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 |