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On 20.10.2006 17:32, Arturo Magidin wrote:
> In article <18749928.1161357471779.JavaMail.jakarta@xxxxxxxxxxxxxxxxxxxxxx>,
> Sasha P <abp83@xxxxxxxxxx> wrote:
>>> No, it clearly is maximal. Take an ultrafilter F on
>>> P([0,1]). Then the
>>> set of functions f such that the zero set of f is in
>>> F is a prime ideal.
>> Why the set of such functions is ideal?
>
>
> An ultrafilter F on P([0,1]) is a collection of subset of [0,1] such
> that:
>
> (i) If A is in F, and B is a subset of [0,1] that contains A, then B
> is in F.
>
> (ii) If A and B are in F, then A/\B is in F.
>
> (iii) F does not contain the empty set.
>
> (iv) F is maximal among collection of subsets of [0,1] satisfying
> (i)-(iii).
>
> (A collection that satisfies (i)-(iii) is called a filter).
>
> Examples of ultrafilters are "all subsets that contain {x}", where x
> is an element of [0,1]. These are called "principal
> ultrafilters". However, if we assume the Axiom of Choice, then there
> are nonprincipal ultrafilters.
... on non-finite sets only, since any ultrafilter containing a finite
set is principal.
> Now: let F be any ultrafilter on [0,1], and let S be the collection
> of all continuous functions f on [0,1] such that the zero set of f is
> an element of F.
>
> If f and g are two continuous functions on [0,1], and Z(f),Z(g) are
> both in F, then Z(f)/\Z(g) is also in F. It is also clear that
> Z(f-g) contains Z(f)/\Z(g), hence Z(f-g) is also in F, since it is a
> subset of [0,1] that contains a set in F. Thus, if f and g are in S,
> then f-g is in S.
>
> Now let f be a continuous function on [0,1] such that Z(f) lies in F,
> and let g be any continuous function on [0,1]. Then Z(g*f) contains
> Z(f), hence lies in F. Thus, g*f is also in S.
>
> So the set S = {f in C[0,1] | Z(f) is in F}
>
> is an ideal of C[0,1].
Bit nitpicking: S is non-void, since [0,1] is in F implying the the
zero-function is in S.
> Notice that we never used the fact that F is an ->ultra<-filter, just
> the fact that it is a ->filter<-. So, if F is any filter on [0,1],
> then Z(F) = {f in C[0,1] : Z(f) is in F} is an ideal of C[0,1].
>
> If F is a principal ultrafilter, then we get the usual maximal ideals
> of all continuous functions that vanish on a point.
>
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