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From: porky_pig_jr on 20 Jun 2010 19:11
On Jun 20, 7:07 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote:

>
> I disproved Turing, Halt, Godel and Cantor

Don't Halt here.

From: |-|ercules on 20 Jun 2010 19:15
<porky_pig_jr(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
From: porky_pig_jr on 20 Jun 2010 20:05
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.
From: |-|ercules on 20 Jun 2010 21:12
<porky_pig_jr(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
From: porky_pig_jr on 20 Jun 2010 21:23
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.
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