Robert Shaw wrote:
>
>> But this is the problem that we are trying to discuss. Or,
>> there are two problems:
I'm glad you didn't correct me and mention that there
is an uncountable number of such problems! :-)
>> which numbers can be constructed
>> in an exact way, and which numbers are the accumulation
>> points of the set of numbers that can be constructed in an
>> exact way. The first set is enumerable, maybe finite. The
>> second set might be anything.
>
> We know the second set is uncountable, and includes the
> rationals.
>
> Their are five sets involved, so for clarity they need names.
> We've got the set of calculator functions Ca={cos,sin,tan,exp etc)
>
Ok, but maybe we might extend this to the "binary" functions.
> Second, the set of all finite sequences of calculator functions, S.
> S is countable, with elements s of the form (fn,....f3,f2,f1) with
> each fm being an element of Ca.
> We have a mapping G from S to K=R U{_error_} given
> by G(s) =fn(...f3(f2(f1(0)..)
> Define fn(_error_)=_error_ for all fn in Ca
>
> G(S) is the set of all finitely constructable number starting
> with 0.
>
Maybe it might be better to "extend" this definition
to something like Constr(Ca, K0), the set of all
finitely constructable numbers starting with an element
of K0 and using functions from the set of functions Ca.
> For arbitary set Ca G(S) is not automatically countable. If
> c={1/x,x^2}then G(S)={0,_error_}
>
> For this C we know Q is a subset of G(S), as you proved,
^
Ca, isn't it?
> so G(S) is countably infinite and dense.
> We don't know what G(S) is, though we think the
> algebraic numbers A are not a subset of G(S)
>
> Next we've got the set Si of all infinite sequences of functions
> in Ca. By Cantor's diagonal argument we know Si is
> an uncountable set if at least one pair of functions in Ca does
> not commute.
>
Hmmmm... Commuting is irrelevant, since we are just taking
the functions, and not operating them - because it's
meaningless to write an expression like x = f0 f1 f2 ... for
an infinite sequence of functions.
> We also have the set G(Si), which includes all the accumulation
> points of G(s). For sequences si in Si such that G(si) isn't
> in R (e.g 1/x,1/x,1/x ....) we can reasonably define G(si) as
> being _error_ For arbitary sets of functions G(Si) needn't
> be uncountable, and can even be finite nor does G(S) have to be
> a subset of G(Si).
>
Ok - because there might not be any "loop" (f1 f2 ... fn)(x) = x
> We don't know much about G(S) or G(Si) for the particular set
> Ca of calculator functions, and less for arbitary sets of functions.
> Even for Ca we don't know if G(Si) includes Q.
>
Uh? Wasn't this proven before?
> The obvious way to prove G(Si) uncountable is to look at
> how many elements of Si map to each element of G(Si).
> If only a countable number do so then we know G(Si)
> can't be countable else Si would be a countable union
> of countable sets, which is not possible.
>
Ok.
> Because a subset of Ca maps the positive reals to themeselves
> by considering this subset, which produces a subset of G(Si)
> which doesn't include _error_ , we can place a lower bound on
> the cardinality of G(Si) without having to consider _error_.
> If the restricted subset can be proved uncountable then G(Si)
> is also uncountable.
>
Ok.
>
> Is anyone but Alberto reading this?
>
:-))))) I think they aren't bored enough to read this
Alberto Monteiro
