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From: |-|ercules on 20 Jun 2010 21:28
<porky_pig_jr(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
From: porky_pig_jr on 20 Jun 2010 21:35
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.
From: |-|ercules on 20 Jun 2010 21:39
<porky_pig_jr(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

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

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