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From: Newberry on 3 Jul 2010 00:37
On Jul 2, 6:06 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote:
> Newberry <newberr...(a)gmail.com> writes:
> > On Jun 27, 8:32 pm, Rupert <rupertmccal...(a)yahoo.com> wrote:
> >> In a set theory which only permits computable reals, it cannot exist,
> >> because the real which encodes the function is not computable.
>
> > Why does the function have to be encoded?
>
> It seems to me that your real question is something like this:
>
> Is there a (sensible?) set theory in which the "natural" set R of reals
> is such that:
>
> (1) Every real number is computable.
>
> (2) Not every function N -> R is computable.
>
> In such a theory, R would presumably be countable (internally as well as
> externally), so what happens with Cantor's theorem?
>
> My own guess is that there is no reasonable set theory in which R
> satisfies the above, because R would not satisfy the important
> properties of real numbers (closure under Cauchy sequences, for
> instance).

I am not sure this follows.

But let me make one observation about Cauchy sequence: it is a
decidedly potential infinity concept, rather than an actual infinity
concept. We then define the real number as this sequence unlike in
Cantor/Zermelo where the real number actually already has an infinite
number of digits, and the sequence converges to it. In our case the
sequence does converge but not to anything. It itself is the real
number.

>  If R in our theory isn't closed under limits of Cauchy
> sequences, then it isn't really much like the reals.
>
> --
> "Not all features that are found on the Security tab are designed to
> help make your documents and files more secure." --Microsoft on Office
> security features (after it was pointed out by a third party that a
> certain password setting is easily bypassed.)

From: Alan Smaill on 5 Jul 2010 05:28
"Jesse F. Hughes" <jesse(a)phiwumbda.org> writes:

> Alan Smaill <smaill(a)SPAMinf.ed.ac.uk> writes:
>
>> "Jesse F. Hughes" <jesse(a)phiwumbda.org> writes:
>>
>>> Newberry <newberryxy(a)gmail.com> writes:
>>>
>>>> On Jun 27, 8:32�pm, Rupert <rupertmccal...(a)yahoo.com> wrote:
>>>>> In a set theory which only permits computable reals, it cannot exist,
>>>>> because the real which encodes the function is not computable.
>>>>
>>>> Why does the function have to be encoded?
>>>
>>> It seems to me that your real question is something like this:
>>>
>>> Is there a (sensible?) set theory in which the "natural" set R of reals
>>> is such that:
>>>
>>> (1) Every real number is computable.
>>>
>>> (2) Not every function N -> R is computable.
>>>
>>> In such a theory, R would presumably be countable (internally as well as
>>> externally), so what happens with Cantor's theorem?
>>
>> It says that that any surjective function N -> R is not computable.
>>
>>> My own guess is that there is no reasonable set theory in which R
>>> satisfies the above, because R would not satisfy the important
>>> properties of real numbers (closure under Cauchy sequences, for
>>> instance). If R in our theory isn't closed under limits of Cauchy
>>> sequences, then it isn't really much like the reals.
>>
>> What if R is closed under something like effective Cauchy sequences
>> (where the sequence as a function N -> R is required to be computable);
>> wouldn't that be pretty much like the reals?
>
> It would be pretty similar, but if there are non-computable functions
> lying about, then it would surely lack certain features of the Reals.

It's not going to be isomprphic to the reals, of course.
But if you're focussing on computable reals, it makes sense
to think of other aspects of analysis in computable terms also,
whence effective Cauchy sequences.

You don't get intermediate value theorem for computable analysis,
except when the function is monotonic. The eneterprise still feels
like real analysis, IMHO.

--
Alan Smaill
From: Peter Webb on 5 Jul 2010 10:24
>> It would be pretty similar, but if there are non-computable functions
>> lying about, then it would surely lack certain features of the Reals.
>
> It's not going to be isomprphic to the reals, of course.
> But if you're focussing on computable reals, it makes sense
> to think of other aspects of analysis in computable terms also,
> whence effective Cauchy sequences.
>
> You don't get intermediate value theorem for computable analysis,
> except when the function is monotonic.

Why is that? What difference does it make?


> The eneterprise still feels
> like real analysis, IMHO.
>

Calculus doesn't care about non-computable Reals; it can take them or leave
them.


> --
> Alan Smaill

From: Newberry on 5 Jul 2010 22:19
On Jul 5, 2:28 am, Alan Smaill <sma...(a)SPAMinf.ed.ac.uk> wrote:
> "Jesse F. Hughes" <je...(a)phiwumbda.org> writes:
>
>
>
>
>
> > Alan Smaill <sma...(a)SPAMinf.ed.ac.uk> writes:
>
> >> "Jesse F. Hughes" <je...(a)phiwumbda.org> writes:
>
> >>> Newberry <newberr...(a)gmail.com> writes:
>
> >>>> On Jun 27, 8:32 pm, Rupert <rupertmccal...(a)yahoo.com> wrote:
> >>>>> In a set theory which only permits computable reals, it cannot exist,
> >>>>> because the real which encodes the function is not computable.
>
> >>>> Why does the function have to be encoded?
>
> >>> It seems to me that your real question is something like this:
>
> >>> Is there a (sensible?) set theory in which the "natural" set R of reals
> >>> is such that:
>
> >>> (1) Every real number is computable.
>
> >>> (2) Not every function N -> R is computable.
>
> >>> In such a theory, R would presumably be countable (internally as well as
> >>> externally), so what happens with Cantor's theorem?
>
> >> It says that that any surjective function N -> R is not computable.
>
> >>> My own guess is that there is no reasonable set theory in which R
> >>> satisfies the above, because R would not satisfy the important
> >>> properties of real numbers (closure under Cauchy sequences, for
> >>> instance).  If R in our theory isn't closed under limits of Cauchy
> >>> sequences, then it isn't really much like the reals.
>
> >> What if R is closed under something like effective Cauchy sequences
> >> (where the sequence as a function N -> R is required to be computable);
> >> wouldn't that be pretty much like the reals?
>
> > It would be pretty similar, but if there are non-computable functions
> > lying about, then it would surely lack certain features of the Reals.
>
> It's not going to be isomprphic to the reals, of course.
> But if you're focussing on computable reals, it makes sense
> to think of other aspects of analysis in computable terms also,
> whence effective Cauchy sequences.
>
> You don't get intermediate value theorem for computable analysis,
> except when the function is monotonic. The eneterprise still feels
> like real analysis, IMHO.

How does it affect measure theory? Can you do Lebesgue's integral with
computable reals?

>
> --
> Alan Smaill- Hide quoted text -
>
> - Show quoted text -

From: Jesse F. Hughes on 6 Jul 2010 08:36
Alan Smaill <smaill(a)SPAMinf.ed.ac.uk> writes:

> "Jesse F. Hughes" <jesse(a)phiwumbda.org> writes:
>
>> Alan Smaill <smaill(a)SPAMinf.ed.ac.uk> writes:
>>
>>> "Jesse F. Hughes" <jesse(a)phiwumbda.org> writes:
>>>
>>>> Newberry <newberryxy(a)gmail.com> writes:
>>>>
>>>>> On Jun 27, 8:32�pm, Rupert <rupertmccal...(a)yahoo.com> wrote:
>>>>>> In a set theory which only permits computable reals, it cannot exist,
>>>>>> because the real which encodes the function is not computable.
>>>>>
>>>>> Why does the function have to be encoded?
>>>>
>>>> It seems to me that your real question is something like this:
>>>>
>>>> Is there a (sensible?) set theory in which the "natural" set R of reals
>>>> is such that:
>>>>
>>>> (1) Every real number is computable.
>>>>
>>>> (2) Not every function N -> R is computable.
>>>>
>>>> In such a theory, R would presumably be countable (internally as well as
>>>> externally), so what happens with Cantor's theorem?
>>>
>>> It says that that any surjective function N -> R is not computable.
>>>
>>>> My own guess is that there is no reasonable set theory in which R
>>>> satisfies the above, because R would not satisfy the important
>>>> properties of real numbers (closure under Cauchy sequences, for
>>>> instance). If R in our theory isn't closed under limits of Cauchy
>>>> sequences, then it isn't really much like the reals.
>>>
>>> What if R is closed under something like effective Cauchy sequences
>>> (where the sequence as a function N -> R is required to be computable);
>>> wouldn't that be pretty much like the reals?
>>
>> It would be pretty similar, but if there are non-computable functions
>> lying about, then it would surely lack certain features of the Reals.
>
> It's not going to be isomprphic to the reals, of course.
> But if you're focussing on computable reals, it makes sense
> to think of other aspects of analysis in computable terms also,
> whence effective Cauchy sequences.
>
> You don't get intermediate value theorem for computable analysis,
> except when the function is monotonic. The eneterprise still feels
> like real analysis, IMHO.

I'll defer to your opinion, I suppose, but it seems to me that many
properties of R depends on quantifying over functions. If we have only
computable reals, but both computable and non-computable functions, then
it seems to me that things won't fit right. The nice properties of R
become messier.

For instance, in plain ZFC, we can certainly talk about the set of
computable reals and its closure properties. As you say, it's not
closed under Cauchy sequences, but only under effective Cauchy
sequences. Thus, it's similar to, but not the same as, R.

Maybe it's my category theory background (in the dark, misty past), but
it seems to me that a category of sets in which there is no set R, but
only a set of computable reals, while both computable and non-computable
functions exist, is just a weird setting. Things don't fit right.
Everything is needlessly complicated and many simple properties of R
are more complicated, due to the presence of these non-computable
functions that just shouldn't be here.

--
She move me when she get drunk, then she say I'm not nowhere
She call me a dumbbell, she tells me I'm nothing but a square
She moves me man, honey and I don't see how its done.
-- "She Moves Me", McKinley Morganfield (Muddy Waters)
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