CANTOR DISPROOF <<<<<<<<<<<<<<<<
in [Logic]
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From: porky_pig_jr on 20 Jun 2010 21:51 On Jun 20, 9:39 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: > <porky_pig...(a)my-deja.com> wrote ... > > > > > On Jun 20, 9:28 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: > >> <porky_pig...(a)my-deja.com> wrote > > >> > On Jun 20, 9:12 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: > >> >> <porky_pig...(a)my-deja.com> wrote ... > > >> >> > On Jun 20, 7:15 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: > >> >> >> <porky_pig...(a)my-deja.com> wrote > > >> >> >> > On Jun 20, 7:07 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: > > >> >> >> >> I disproved Turing, Halt, Godel and Cantor > > >> >> >> > Don't Halt here. > > >> >> >> You on the other hand cannot answer how wide this set is! > > >> >> >> 3 > >> >> >> 31 > >> >> >> 314 > >> >> >> ... > > >> >> >> You um and arr and cannot parse words like CONTAIN, BELOW, BIG, SET > >> >> >> and accuse me of shifting language to make nonsense claims. > > >> >> >> How wide is the set porky, George is really stumped on this one! > > >> >> >> Herc > > >> >> > This is the first time I see someone's using the adjective "wide" with > >> >> > the set. I assume you're asking about the cardinality of a set. Or may > >> >> > be you're asking what happens with the number of digits in each > >> >> > element of the set, as we generate the successive approximations of > >> >> > the pi? > > >> >> > Well, pi is computable real. We're all agree on that. Generating > >> >> > successive approximations of pi is a counting process, so the number > >> >> > of such approximations is countably infinite. Each entry of such list > >> >> > contains some finite number of digits, we can call it a finite prefix > >> >> > of a pi. > > >> >> > OK, I guess I see your point. By width you mean "the largest such > >> >> > prefix", right? And you want to imply that it is also countably > >> >> > infinite. So the "width" of this set is countably infinite. Well, > >> >> > there's a subtlety here. On one hand, certainly the width of this list > >> >> > (using your terminology) is not bounded by any natural n, right? We > >> >> > can always make one more iteration and create a prefix with the width n > >> >> > +1. So, since the width of this list is not bounded by any natural n, > >> >> > it must be infinite. Does it follow then that the list actually > >> >> > *contains* one entry which is the infinite string (and, of course, > >> >> > that infinite string *is* pi)? No, it does not follow at all, and this > >> >> > is where I believe you get off the track. > > >> >> > I don't think I can help you with that. But now I'm thinking of the > >> >> > question someone asked me once. Suppose we prove something by > >> >> > induction. That is, we prove something for any natural n. Suppose we > >> >> > have some sequence with some limiting behavior. So we can think of > >> >> > some limit point. Does proof of induction include that limit point? > >> >> > The answer is categorical "No, it does not!" You can certainly think > >> >> > of a limit as a point, but it's a special point, never to be accessed > >> >> > in a finite number of steps. This is what Cantor (your hero, as I > >> >> > know), called "transfinite ordinals", the first such limit point is > >> >> > normally designated as little omega. Each predecessor of it is a > >> >> > finite number, yet it has no immediate predecessor and can't be > >> >> > accessed in a finite number of steps. > > >> >> > Well, think of that little omega as a horizon. You sort of know it's > >> >> > there but you can never reach it in finite number of steps. One may > >> >> > say that that little omega-entry in our list is indeed our goal, our > >> >> > dream, the infinite string representing Pi. But it's not reachable. We > >> >> > can never generate such a string in a finite number of steps. In this > >> >> > respect, even if "the width of this set is infinite", the list does > >> >> > not contain that infinite string. Exactly for the same reason the list > >> >> > (1/2, 2/3, 3/4, 4/5, ...) will never contain 1. 1 is what happens at > >> >> > the limit, but we never reach that limit in a finite number of steps. > >> >> > We can stack that 1 on a top of the list, it will correspond to little- > >> >> > omega entry. But it's not reachable by our algorithm we use to > >> >> > generate the list (1/2, 2/3, 3/4 ...). So the infinite string > >> >> > representing pi won't be on the list (3, 31, 314, ...) for exactly the > >> >> > same reason. Because the limit is an imaginary never-reachable-in- > >> >> > finite-number-of-steps point. The fact that the width of such list is > >> >> > not bounded by any n, and hence infinite does not imply that the > >> >> > infinite string representing pi *is* on that list. Just we can stack 1 > >> >> > on a top of a sequence (1/2, 2/3, 3/4 ...), we can stack your infinite > >> >> > string on a top of (3, 31, 314 ...), but that's not a part of that > >> >> > sequence, it does not belong to it. > > >> >> > That's as much as I can say. A while ago I wasn't clear on that, but > >> >> > after taking a few courses on real analysis, doing lots of exercises > >> >> > involving limits, and then a little bit involving transfinite > >> >> > induction vs the regular induction, I finally started getting the hold > >> >> > of that. Your mistake is thinking of a limiting behavior as an actual > >> >> > point, or actual location in that list. It isn't there. As I've said, > >> >> > I'm thinking of such entry as little omega-entry. You can mentally > >> >> > stack it on a top of your list, but it's *not* part of the list. It's > >> >> > never reachable. It's not there. The fact that the width of the list > >> >> > is not bounded by any real n *still* does not mean that it's there. > >> >> > Gee whiz, my answer is a bit long. And, as I said, at some point I > >> >> > wasn't that clear on that whole thing. The clarity came when I started > >> >> > reading about transfinite induction. > > >> >> > PPJ. > > >> >> so it's not as wide as 3.14.. ? > > >> >> Herc > > >> > Using your terminology, the "width" of the entries in a list (a number > >> > of digits in each entry) is not bounded by any natural number, hence > >> > it's (countably) infinite. > > >> > Now, the decimal representation of pi, generated by some algorithm is > >> > a counting process. You can think of number pi as some limiting value > >> > of that process. Hence it's also countably infinite. > > >> > So, answering your question, and using your terminology, it *is* as > >> > wide as 3.14 ... . > > >> > And if your next question is gonna be "then how come that list does > >> > not contain 3.14 ...", please, don't bother, for it means that you are > >> > still not getting it. > > >> > Regards, > > >> > PPJ. > > >> So how many digits of PI in order are in > > >> 3 > >> 31 > >> 314 > >> ... > > >> ? > > >> Herc > > > In the list you have provided, you can request any number n of digits > > (any finite prefix) of Pi, and that will be nth entry in a list. Since > > the list is not bounded by any natural n. So I guess the answer is: As > > many as you wish. Say, we are running the algorithm to generate the > > successive approximations of Pi. We can run as many iterations as we > > want to. > > How many digits of PI in order are in 3.14.. ? > > Herc Uhm, I'm sorry, don't you just keep repeating the same questions? Well, OK, I am the game. We say that Pi is computable real, hence can be thought of as a limit of successive approximations given some algorithm. Each such approximation contains a finite number of digit. The limit, therefore, contains countably infinitely many digits. We run algorithm n times and then take n to infinity.
From: |-|ercules on 20 Jun 2010 21:59 <porky_pig_jr(a)my-deja.com> wrote > On Jun 20, 9:39 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: >> <porky_pig...(a)my-deja.com> wrote ... >> >> >> >> > On Jun 20, 9:28 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: >> >> <porky_pig...(a)my-deja.com> wrote >> >> >> > On Jun 20, 9:12 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: >> >> >> <porky_pig...(a)my-deja.com> wrote ... >> >> >> >> > On Jun 20, 7:15 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: >> >> >> >> <porky_pig...(a)my-deja.com> wrote >> >> >> >> >> > On Jun 20, 7:07 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: >> >> >> >> >> >> I disproved Turing, Halt, Godel and Cantor >> >> >> >> >> > Don't Halt here. >> >> >> >> >> You on the other hand cannot answer how wide this set is! >> >> >> >> >> 3 >> >> >> >> 31 >> >> >> >> 314 >> >> >> >> ... >> >> >> >> >> You um and arr and cannot parse words like CONTAIN, BELOW, BIG, SET >> >> >> >> and accuse me of shifting language to make nonsense claims. >> >> >> >> >> How wide is the set porky, George is really stumped on this one! >> >> >> >> >> Herc >> >> >> >> > This is the first time I see someone's using the adjective "wide" with >> >> >> > the set. I assume you're asking about the cardinality of a set. Or may >> >> >> > be you're asking what happens with the number of digits in each >> >> >> > element of the set, as we generate the successive approximations of >> >> >> > the pi? >> >> >> >> > Well, pi is computable real. We're all agree on that. Generating >> >> >> > successive approximations of pi is a counting process, so the number >> >> >> > of such approximations is countably infinite. Each entry of such list >> >> >> > contains some finite number of digits, we can call it a finite prefix >> >> >> > of a pi. >> >> >> >> > OK, I guess I see your point. By width you mean "the largest such >> >> >> > prefix", right? And you want to imply that it is also countably >> >> >> > infinite. So the "width" of this set is countably infinite. Well, >> >> >> > there's a subtlety here. On one hand, certainly the width of this list >> >> >> > (using your terminology) is not bounded by any natural n, right? We >> >> >> > can always make one more iteration and create a prefix with the width n >> >> >> > +1. So, since the width of this list is not bounded by any natural n, >> >> >> > it must be infinite. Does it follow then that the list actually >> >> >> > *contains* one entry which is the infinite string (and, of course, >> >> >> > that infinite string *is* pi)? No, it does not follow at all, and this >> >> >> > is where I believe you get off the track. >> >> >> >> > I don't think I can help you with that. But now I'm thinking of the >> >> >> > question someone asked me once. Suppose we prove something by >> >> >> > induction. That is, we prove something for any natural n. Suppose we >> >> >> > have some sequence with some limiting behavior. So we can think of >> >> >> > some limit point. Does proof of induction include that limit point? >> >> >> > The answer is categorical "No, it does not!" You can certainly think >> >> >> > of a limit as a point, but it's a special point, never to be accessed >> >> >> > in a finite number of steps. This is what Cantor (your hero, as I >> >> >> > know), called "transfinite ordinals", the first such limit point is >> >> >> > normally designated as little omega. Each predecessor of it is a >> >> >> > finite number, yet it has no immediate predecessor and can't be >> >> >> > accessed in a finite number of steps. >> >> >> >> > Well, think of that little omega as a horizon. You sort of know it's >> >> >> > there but you can never reach it in finite number of steps. One may >> >> >> > say that that little omega-entry in our list is indeed our goal, our >> >> >> > dream, the infinite string representing Pi. But it's not reachable. We >> >> >> > can never generate such a string in a finite number of steps. In this >> >> >> > respect, even if "the width of this set is infinite", the list does >> >> >> > not contain that infinite string. Exactly for the same reason the list >> >> >> > (1/2, 2/3, 3/4, 4/5, ...) will never contain 1. 1 is what happens at >> >> >> > the limit, but we never reach that limit in a finite number of steps. >> >> >> > We can stack that 1 on a top of the list, it will correspond to little- >> >> >> > omega entry. But it's not reachable by our algorithm we use to >> >> >> > generate the list (1/2, 2/3, 3/4 ...). So the infinite string >> >> >> > representing pi won't be on the list (3, 31, 314, ...) for exactly the >> >> >> > same reason. Because the limit is an imaginary never-reachable-in- >> >> >> > finite-number-of-steps point. The fact that the width of such list is >> >> >> > not bounded by any n, and hence infinite does not imply that the >> >> >> > infinite string representing pi *is* on that list. Just we can stack 1 >> >> >> > on a top of a sequence (1/2, 2/3, 3/4 ...), we can stack your infinite >> >> >> > string on a top of (3, 31, 314 ...), but that's not a part of that >> >> >> > sequence, it does not belong to it. >> >> >> >> > That's as much as I can say. A while ago I wasn't clear on that, but >> >> >> > after taking a few courses on real analysis, doing lots of exercises >> >> >> > involving limits, and then a little bit involving transfinite >> >> >> > induction vs the regular induction, I finally started getting the hold >> >> >> > of that. Your mistake is thinking of a limiting behavior as an actual >> >> >> > point, or actual location in that list. It isn't there. As I've said, >> >> >> > I'm thinking of such entry as little omega-entry. You can mentally >> >> >> > stack it on a top of your list, but it's *not* part of the list. It's >> >> >> > never reachable. It's not there. The fact that the width of the list >> >> >> > is not bounded by any real n *still* does not mean that it's there. >> >> >> > Gee whiz, my answer is a bit long. And, as I said, at some point I >> >> >> > wasn't that clear on that whole thing. The clarity came when I started >> >> >> > reading about transfinite induction. >> >> >> >> > PPJ. >> >> >> >> so it's not as wide as 3.14.. ? >> >> >> >> Herc >> >> >> > Using your terminology, the "width" of the entries in a list (a number >> >> > of digits in each entry) is not bounded by any natural number, hence >> >> > it's (countably) infinite. >> >> >> > Now, the decimal representation of pi, generated by some algorithm is >> >> > a counting process. You can think of number pi as some limiting value >> >> > of that process. Hence it's also countably infinite. >> >> >> > So, answering your question, and using your terminology, it *is* as >> >> > wide as 3.14 ... . >> >> >> > And if your next question is gonna be "then how come that list does >> >> > not contain 3.14 ...", please, don't bother, for it means that you are >> >> > still not getting it. >> >> >> > Regards, >> >> >> > PPJ. >> >> >> So how many digits of PI in order are in >> >> >> 3 >> >> 31 >> >> 314 >> >> ... >> >> >> ? >> >> >> Herc >> >> > In the list you have provided, you can request any number n of digits >> > (any finite prefix) of Pi, and that will be nth entry in a list. Since >> > the list is not bounded by any natural n. So I guess the answer is: As >> > many as you wish. Say, we are running the algorithm to generate the >> > successive approximations of Pi. We can run as many iterations as we >> > want to. >> >> How many digits of PI in order are in 3.14.. ? >> >> Herc > > Uhm, I'm sorry, don't you just keep repeating the same questions? Only until I get an answer. > Well, OK, I am the game. We say that Pi is computable real, hence can > be thought of as a limit of successive approximations given some > algorithm. Each such approximation contains a finite number of digit. > The limit, therefore, contains countably infinitely many digits. We > run algorithm n times and then take n to infinity. So 3.14.. is countable infinity wide, and has infinitely many digits of PI in order. 3 31 314 ... is countable infinity wide and has "as many as we wish" digits of PI in order. BOY who said Porky Pig isn't still worth a laugh! Herc
From: porky_pig_jr on 20 Jun 2010 22:09 On Jun 20, 9:59 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: > <porky_pig...(a)my-deja.com> wrote > > > > > On Jun 20, 9:39 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: > >> <porky_pig...(a)my-deja.com> wrote ... > > >> > On Jun 20, 9:28 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: > >> >> <porky_pig...(a)my-deja.com> wrote > > >> >> > On Jun 20, 9:12 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: > >> >> >> <porky_pig...(a)my-deja.com> wrote ... > > >> >> >> > On Jun 20, 7:15 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: > >> >> >> >> <porky_pig...(a)my-deja.com> wrote > > >> >> >> >> > On Jun 20, 7:07 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: > > >> >> >> >> >> I disproved Turing, Halt, Godel and Cantor > > >> >> >> >> > Don't Halt here. > > >> >> >> >> You on the other hand cannot answer how wide this set is! > > >> >> >> >> 3 > >> >> >> >> 31 > >> >> >> >> 314 > >> >> >> >> ... > > >> >> >> >> You um and arr and cannot parse words like CONTAIN, BELOW, BIG, SET > >> >> >> >> and accuse me of shifting language to make nonsense claims. > > >> >> >> >> How wide is the set porky, George is really stumped on this one! > > >> >> >> >> Herc > > >> >> >> > This is the first time I see someone's using the adjective "wide" with > >> >> >> > the set. I assume you're asking about the cardinality of a set.. Or may > >> >> >> > be you're asking what happens with the number of digits in each > >> >> >> > element of the set, as we generate the successive approximations of > >> >> >> > the pi? > > >> >> >> > Well, pi is computable real. We're all agree on that. Generating > >> >> >> > successive approximations of pi is a counting process, so the number > >> >> >> > of such approximations is countably infinite. Each entry of such list > >> >> >> > contains some finite number of digits, we can call it a finite prefix > >> >> >> > of a pi. > > >> >> >> > OK, I guess I see your point. By width you mean "the largest such > >> >> >> > prefix", right? And you want to imply that it is also countably > >> >> >> > infinite. So the "width" of this set is countably infinite. Well, > >> >> >> > there's a subtlety here. On one hand, certainly the width of this list > >> >> >> > (using your terminology) is not bounded by any natural n, right? We > >> >> >> > can always make one more iteration and create a prefix with the width n > >> >> >> > +1. So, since the width of this list is not bounded by any natural n, > >> >> >> > it must be infinite. Does it follow then that the list actually > >> >> >> > *contains* one entry which is the infinite string (and, of course, > >> >> >> > that infinite string *is* pi)? No, it does not follow at all, and this > >> >> >> > is where I believe you get off the track. > > >> >> >> > I don't think I can help you with that. But now I'm thinking of the > >> >> >> > question someone asked me once. Suppose we prove something by > >> >> >> > induction. That is, we prove something for any natural n. Suppose we > >> >> >> > have some sequence with some limiting behavior. So we can think of > >> >> >> > some limit point. Does proof of induction include that limit point? > >> >> >> > The answer is categorical "No, it does not!" You can certainly think > >> >> >> > of a limit as a point, but it's a special point, never to be accessed > >> >> >> > in a finite number of steps. This is what Cantor (your hero, as I > >> >> >> > know), called "transfinite ordinals", the first such limit point is > >> >> >> > normally designated as little omega. Each predecessor of it is a > >> >> >> > finite number, yet it has no immediate predecessor and can't be > >> >> >> > accessed in a finite number of steps. > > >> >> >> > Well, think of that little omega as a horizon. You sort of know it's > >> >> >> > there but you can never reach it in finite number of steps. One may > >> >> >> > say that that little omega-entry in our list is indeed our goal, our > >> >> >> > dream, the infinite string representing Pi. But it's not reachable. We > >> >> >> > can never generate such a string in a finite number of steps. In this > >> >> >> > respect, even if "the width of this set is infinite", the list does > >> >> >> > not contain that infinite string. Exactly for the same reason the list > >> >> >> > (1/2, 2/3, 3/4, 4/5, ...) will never contain 1. 1 is what happens at > >> >> >> > the limit, but we never reach that limit in a finite number of steps. > >> >> >> > We can stack that 1 on a top of the list, it will correspond to little- > >> >> >> > omega entry. But it's not reachable by our algorithm we use to > >> >> >> > generate the list (1/2, 2/3, 3/4 ...). So the infinite string > >> >> >> > representing pi won't be on the list (3, 31, 314, ...) for exactly the > >> >> >> > same reason. Because the limit is an imaginary never-reachable-in- > >> >> >> > finite-number-of-steps point. The fact that the width of such list is > >> >> >> > not bounded by any n, and hence infinite does not imply that the > >> >> >> > infinite string representing pi *is* on that list. Just we can stack 1 > >> >> >> > on a top of a sequence (1/2, 2/3, 3/4 ...), we can stack your infinite > >> >> >> > string on a top of (3, 31, 314 ...), but that's not a part of that > >> >> >> > sequence, it does not belong to it. > > >> >> >> > That's as much as I can say. A while ago I wasn't clear on that, but > >> >> >> > after taking a few courses on real analysis, doing lots of exercises > >> >> >> > involving limits, and then a little bit involving transfinite > >> >> >> > induction vs the regular induction, I finally started getting the hold > >> >> >> > of that. Your mistake is thinking of a limiting behavior as an actual > >> >> >> > point, or actual location in that list. It isn't there. As I've said, > >> >> >> > I'm thinking of such entry as little omega-entry. You can mentally > >> >> >> > stack it on a top of your list, but it's *not* part of the list. It's > >> >> >> > never reachable. It's not there. The fact that the width of the list > >> >> >> > is not bounded by any real n *still* does not mean that it's there. > >> >> >> > Gee whiz, my answer is a bit long. And, as I said, at some point I > >> >> >> > wasn't that clear on that whole thing. The clarity came when I started > >> >> >> > reading about transfinite induction. > > >> >> >> > PPJ. > > >> >> >> so it's not as wide as 3.14.. ? > > >> >> >> Herc > > >> >> > Using your terminology, the "width" of the entries in a list (a number > >> >> > of digits in each entry) is not bounded by any natural number, hence > >> >> > it's (countably) infinite. > > >> >> > Now, the decimal representation of pi, generated by some algorithm is > >> >> > a counting process. You can think of number pi as some limiting value > >> >> > of that process. Hence it's also countably infinite. > > >> >> > So, answering your question, and using your terminology, it *is* as > >> >> > wide as 3.14 ... . > > >> >> > And if your next question is gonna be "then how come that list does > >> >> > not contain 3.14 ...", please, don't bother, for it means that you are > >> >> > still not getting it. > > >> >> > Regards, > > >> >> > PPJ. > > >> >> So how many digits of PI in order are in > > >> >> 3 > >> >> 31 > >> >> 314 > >> >> ... > > >> >> ? > > >> >> Herc > > >> > In the list you have provided, you can request any number n of digits > >> > (any finite prefix) of Pi, and that will be nth entry in a list. Since > >> > the list is not bounded by any natural n. So I guess the answer is: As > >> > many as you wish. Say, we are running the algorithm to generate the > >> > successive approximations of Pi. We can run as many iterations as we > >> > want to. > > >> How many digits of PI in order are in 3.14.. ? > > >> Herc > > > Uhm, I'm sorry, don't you just keep repeating the same questions? > > Only until I get an answer. > > > Well, OK, I am the game. We say that Pi is computable real, hence can > > be thought of as a limit of successive approximations given some > > algorithm. Each such approximation contains a finite number of digit. > > The limit, therefore, contains countably infinitely many digits. We > > run algorithm n times and then take n to infinity. > > So 3.14.. is countable infinity wide, and has infinitely many digits of PI in order. > > 3 > 31 > 314 > .. > > is countable infinity wide and has "as many as we wish" digits of PI in order. > > BOY who said Porky Pig isn't still worth a laugh! > > Herc I guess this is you everyone at sci.math laughs at (with exception of WM, of course). And you keep wondering why no one understands your point. In fact this is you who don't understand anything what people are trying to tell you. Apparently you have no math background, you don't understand the concepts of limits, infinity, have no clue what Cantor's proof is all about. You are functionally illiterate. Yet you believe that you are qualified to claim that you disproved Cantor, etc etc. But in fact you're not even qualified to be in this group at all. PPJ.
From: |-|ercules on 20 Jun 2010 22:15 <porky_pig_jr(a)my-deja.com> wrote ... > On Jun 20, 9:59 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: >> <porky_pig...(a)my-deja.com> wrote >> >> >> >> > On Jun 20, 9:39 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: >> >> <porky_pig...(a)my-deja.com> wrote ... >> >> >> > On Jun 20, 9:28 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: >> >> >> <porky_pig...(a)my-deja.com> wrote >> >> >> >> > On Jun 20, 9:12 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: >> >> >> >> <porky_pig...(a)my-deja.com> wrote ... >> >> >> >> >> > On Jun 20, 7:15 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: >> >> >> >> >> <porky_pig...(a)my-deja.com> wrote >> >> >> >> >> >> > On Jun 20, 7:07 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: >> >> >> >> >> >> >> I disproved Turing, Halt, Godel and Cantor >> >> >> >> >> >> > Don't Halt here. >> >> >> >> >> >> You on the other hand cannot answer how wide this set is! >> >> >> >> >> >> 3 >> >> >> >> >> 31 >> >> >> >> >> 314 >> >> >> >> >> ... >> >> >> >> >> >> You um and arr and cannot parse words like CONTAIN, BELOW, BIG, SET >> >> >> >> >> and accuse me of shifting language to make nonsense claims. >> >> >> >> >> >> How wide is the set porky, George is really stumped on this one! >> >> >> >> >> >> Herc >> >> >> >> >> > This is the first time I see someone's using the adjective "wide" with >> >> >> >> > the set. I assume you're asking about the cardinality of a set. Or may >> >> >> >> > be you're asking what happens with the number of digits in each >> >> >> >> > element of the set, as we generate the successive approximations of >> >> >> >> > the pi? >> >> >> >> >> > Well, pi is computable real. We're all agree on that. Generating >> >> >> >> > successive approximations of pi is a counting process, so the number >> >> >> >> > of such approximations is countably infinite. Each entry of such list >> >> >> >> > contains some finite number of digits, we can call it a finite prefix >> >> >> >> > of a pi. >> >> >> >> >> > OK, I guess I see your point. By width you mean "the largest such >> >> >> >> > prefix", right? And you want to imply that it is also countably >> >> >> >> > infinite. So the "width" of this set is countably infinite. Well, >> >> >> >> > there's a subtlety here. On one hand, certainly the width of this list >> >> >> >> > (using your terminology) is not bounded by any natural n, right? We >> >> >> >> > can always make one more iteration and create a prefix with the width n >> >> >> >> > +1. So, since the width of this list is not bounded by any natural n, >> >> >> >> > it must be infinite. Does it follow then that the list actually >> >> >> >> > *contains* one entry which is the infinite string (and, of course, >> >> >> >> > that infinite string *is* pi)? No, it does not follow at all, and this >> >> >> >> > is where I believe you get off the track. >> >> >> >> >> > I don't think I can help you with that. But now I'm thinking of the >> >> >> >> > question someone asked me once. Suppose we prove something by >> >> >> >> > induction. That is, we prove something for any natural n. Suppose we >> >> >> >> > have some sequence with some limiting behavior. So we can think of >> >> >> >> > some limit point. Does proof of induction include that limit point? >> >> >> >> > The answer is categorical "No, it does not!" You can certainly think >> >> >> >> > of a limit as a point, but it's a special point, never to be accessed >> >> >> >> > in a finite number of steps. This is what Cantor (your hero, as I >> >> >> >> > know), called "transfinite ordinals", the first such limit point is >> >> >> >> > normally designated as little omega. Each predecessor of it is a >> >> >> >> > finite number, yet it has no immediate predecessor and can't be >> >> >> >> > accessed in a finite number of steps. >> >> >> >> >> > Well, think of that little omega as a horizon. You sort of know it's >> >> >> >> > there but you can never reach it in finite number of steps. One may >> >> >> >> > say that that little omega-entry in our list is indeed our goal, our >> >> >> >> > dream, the infinite string representing Pi. But it's not reachable. We >> >> >> >> > can never generate such a string in a finite number of steps. In this >> >> >> >> > respect, even if "the width of this set is infinite", the list does >> >> >> >> > not contain that infinite string. Exactly for the same reason the list >> >> >> >> > (1/2, 2/3, 3/4, 4/5, ...) will never contain 1. 1 is what happens at >> >> >> >> > the limit, but we never reach that limit in a finite number of steps. >> >> >> >> > We can stack that 1 on a top of the list, it will correspond to little- >> >> >> >> > omega entry. But it's not reachable by our algorithm we use to >> >> >> >> > generate the list (1/2, 2/3, 3/4 ...). So the infinite string >> >> >> >> > representing pi won't be on the list (3, 31, 314, ...) for exactly the >> >> >> >> > same reason. Because the limit is an imaginary never-reachable-in- >> >> >> >> > finite-number-of-steps point. The fact that the width of such list is >> >> >> >> > not bounded by any n, and hence infinite does not imply that the >> >> >> >> > infinite string representing pi *is* on that list. Just we can stack 1 >> >> >> >> > on a top of a sequence (1/2, 2/3, 3/4 ...), we can stack your infinite >> >> >> >> > string on a top of (3, 31, 314 ...), but that's not a part of that >> >> >> >> > sequence, it does not belong to it. >> >> >> >> >> > That's as much as I can say. A while ago I wasn't clear on that, but >> >> >> >> > after taking a few courses on real analysis, doing lots of exercises >> >> >> >> > involving limits, and then a little bit involving transfinite >> >> >> >> > induction vs the regular induction, I finally started getting the hold >> >> >> >> > of that. Your mistake is thinking of a limiting behavior as an actual >> >> >> >> > point, or actual location in that list. It isn't there. As I've said, >> >> >> >> > I'm thinking of such entry as little omega-entry. You can mentally >> >> >> >> > stack it on a top of your list, but it's *not* part of the list. It's >> >> >> >> > never reachable. It's not there. The fact that the width of the list >> >> >> >> > is not bounded by any real n *still* does not mean that it's there. >> >> >> >> > Gee whiz, my answer is a bit long. And, as I said, at some point I >> >> >> >> > wasn't that clear on that whole thing. The clarity came when I started >> >> >> >> > reading about transfinite induction. >> >> >> >> >> > PPJ. >> >> >> >> >> so it's not as wide as 3.14.. ? >> >> >> >> >> Herc >> >> >> >> > Using your terminology, the "width" of the entries in a list (a number >> >> >> > of digits in each entry) is not bounded by any natural number, hence >> >> >> > it's (countably) infinite. >> >> >> >> > Now, the decimal representation of pi, generated by some algorithm is >> >> >> > a counting process. You can think of number pi as some limiting value >> >> >> > of that process. Hence it's also countably infinite. >> >> >> >> > So, answering your question, and using your terminology, it *is* as >> >> >> > wide as 3.14 ... . >> >> >> >> > And if your next question is gonna be "then how come that list does >> >> >> > not contain 3.14 ...", please, don't bother, for it means that you are >> >> >> > still not getting it. >> >> >> >> > Regards, >> >> >> >> > PPJ. >> >> >> >> So how many digits of PI in order are in >> >> >> >> 3 >> >> >> 31 >> >> >> 314 >> >> >> ... >> >> >> >> ? >> >> >> >> Herc >> >> >> > In the list you have provided, you can request any number n of digits >> >> > (any finite prefix) of Pi, and that will be nth entry in a list. Since >> >> > the list is not bounded by any natural n. So I guess the answer is: As >> >> > many as you wish. Say, we are running the algorithm to generate the >> >> > successive approximations of Pi. We can run as many iterations as we >> >> > want to. >> >> >> How many digits of PI in order are in 3.14.. ? >> >> >> Herc >> >> > Uhm, I'm sorry, don't you just keep repeating the same questions? >> >> Only until I get an answer. >> >> > Well, OK, I am the game. We say that Pi is computable real, hence can >> > be thought of as a limit of successive approximations given some >> > algorithm. Each such approximation contains a finite number of digit. >> > The limit, therefore, contains countably infinitely many digits. We >> > run algorithm n times and then take n to infinity. >> >> So 3.14.. is countable infinity wide, and has infinitely many digits of PI in order. >> >> 3 >> 31 >> 314 >> .. >> >> is countable infinity wide and has "as many as we wish" digits of PI in order. >> >> BOY who said Porky Pig isn't still worth a laugh! >> >> Herc > > I guess this is you everyone at sci.math laughs at (with exception of > WM, of course). And you keep wondering why no one understands your > point. In fact this is you who don't understand anything what people > are trying to tell you. > > Apparently you have no math background, you don't understand the > concepts of limits, infinity, have no clue what Cantor's proof is all > about. You are functionally illiterate. Yet you believe that you are > qualified to claim that you disproved Cantor, etc etc. But in fact > you're not even qualified to be in this group at all. > > PPJ. I studied logic at Uni, and I disproved Turing, Halt, Godel and Cantor. Let's examine your answers again, are you SERIOUS? >> So 3.14.. is countable infinity wide, and has infinitely many digits of PI in order. >> >> 3 >> 31 >> 314 >> .. >> >> is countable infinity wide and has "as many as we wish" digits of PI in order. Uhuh uhuh uhuh That's all folks! Herc
From: Sylvia Else on 20 Jun 2010 22:56
On 21/06/2010 2:45 AM, |-|ercules wrote: > "Math-a-nator" <MorePornLips(a)example.com> wrote... >>> Hypothesis: Ah have nothin to day and ah am sayin it. >> >> > > > > > it's worth thinking what "anti-diagonals" entail. > > You're not just constructing 0.444454445544444445444.. a 4 for every non > 4 digit and a 5 for a 4. > > You're constructing ALL 9 OTHER DIGITS to the diagonal digits. > > And it's not just the diagonal, it's the diagonal of ALL PERMUTATIONS OF > THE LIST. Why? Suppose you have your list, and you label each line. 1 0.xxxxxx 2 0.yyyyy 3 0.zzzzzz 4 0.aaaaa 5 0.bbbbb Now choose your anti-diagonal. For this purpose, X is anti-x, and so on. So the anti-diagonal is 0.XYZAB. X != x in line labelled 1, Y != y in line labelled 2, and so on. Clearly, it's not in the list. Now permute your list. Note that the lines retain their labels. 4 0.aaaaa 2 0.yyyyy 1 0.xxxxxx 3 0.zzzzzz 5 0.bbbbb It is still true that X != x in line labelled 1, Y != y in line labelled 2, and so on. Clearly, the 0.XYZAB is still not in the list. We can also immediately see that 0.AYXZB is not in the list either. So now we have two numbers that are not in the list. Sylvia. |