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> In article
> <18235534.1161365641076.JavaMail.jakarta@xxxxxxxxxxxxx
> forum.org>,
> Sasha P <abp83@xxxxxxxxxx> wrote:
>
> Quote the message you are replying to. Otherwise,
> there is no context
> and I don't know what you are talking about without
> doing a bunch of
> work.
>
> In Mathforum, use the "Quote Original" button.
>
>
> >Thank you very much! But why ideals corresponding
> nonprincipal
> >ultrafilters is not maximal?
>
> You are asking: is the ideal corresponding to a
> nonprincipal
> ultrafilter (perhaps with the added condition that
> the filter contains
> some closed subset of [0,1] other than the entire
> interval) prime? Is
> it maximal?
>
> First: suppose F is a principal ultrafilter on [0,1]
> (of any kind). Is
> V(F), the set of all functions in C[0,1] which vanish
> on some element
> of F, a prime ideal?
>
> The key is the following property of ultrafilters:
>
> THEOREM. If F is an ultrafilter on the set X, then
> for every subset Y
> of X, either Y is in F, or X-Y is in F.
>
> (Take an ultrafilter; if Y/\A is empty for some A in
> F, then A is
> contained in X-Y, and therefore X-Y is in F. If Y/\A
> is nonempty for
> all A in F, then you can consider the set of all
> elements of F and all
> intersections of these elements of F, and make a
> filter out of it;
> since F is an ultrafilter, this new filter will have
> to be equal to F,
> and since this new filter contains Y, that means that
> Y is in F. So,
> either X-Y is in F, or else Y is in F).
>
> (In fact, some sources ->define<- an ultrafilter as a
> filter that
> satisfies this condition, instead of as a filter
> which is maximal)
>
>
> Now, let F be an ultrafilter on [0,1], and suppose
> that f*g lies in
> V(F). That means that Z(f*g) is an element of F. Note
> that
> Z(f*g)=Z(f)\/Z(g).
>
> Now, either Z(f) lies in F (in which case f lies in
> Z(F) and we are
> done), or else [0,1]\Z(f) lies in F. In this latter
> case, the
> intersection of Z(f*g) and [0,1]\Z(f) lies in F; this
> intersection is
> Z(g)\Z(f). So Z(g)\Z(f) lies in F, hence Z(g) lies in
> F (as it
> contains an element of F), hence g is in V(F) (since
> its zero set is
> an element of F). This proves that if f*g is in V(F),
> then f is
> in V(F) or g is in V(F), and therefore V(F) is a
> prime ideal whenever
> F is an ultrafilter.
>
> Is it maximal? Depends. We know that if the
> ultrafilter is "all
> functions that vanish on {x}" (i.e., a principal
> ultrafilter), then
> V(F) is maximal. So suppose F is not principal; is
> V(F) maximal then?
>
> I believe the argument is as follows, if I did not
> screw up somewhere:
>
> Given any ideal I of C[0,1], we can consider
>
> Z(I) = {x in [0,1] : f(x) = 0 for all x in I}.
>
> If Z(I) is empty, then I=C[0,1]: for, given any x in
> [0,1],
> there exists f_x in I such that f_x(x)<>0. Find a nbd
> of U_x of x
> such that f(x) is not zero on U_x. Then {U_x : x in
> X} form an open
> cover for [0,1], which is compact. Thus there exist
> x_1,...,x_n such
> that U_{x_1},...,U_{x_n} cover [0,1]. Letting f_i =
> f_{x_i}, we have
> that f_1^2 + ... + f_n^2 is a function in I which
> does not vanish on
> [0,1], hence is a unit of C[0,1], and therefore I =
> C[0,1].
>
> So if I is a PROPER ideal of C[0,1], then Z(I) is
> nonempty.
>
> Now, if F is a nonprincipal ultrafilter, then V(F) is
> not equal to
> C[0,1]; if it contains a unit, then we would have the
> empty set in F
> (the zero set of a unit is empty), which is not the
> case. Thus,
> Z(V(F)) is nonempty. Let x be in Z(V(F)). Then every
> function in V(F)
> vanishes on {x}, so V(F) is contained in {f in C[0,1]
> : f(x) = 0}. But
> it is PROPERLY contained in V(F): for there certainly
> is a function g
> in C[0,1] which vanished only at {x} and is nonzero
> elsewhere. This g
> is in {f in C[0,1]: f(x)=0} but is not in V(F) (for
> Z(g) = {x} cannot
> be an element of F, because F is nonprincipal). So
> V(F) is properly
> contained in the maximal ideal {f in C[0,1]: f(x)=0},
> hence V(F) is a
> nonmaximal prime ideal of C[0,1].
>
> --
> ======================================================
> ================
> "It's not denial. I'm just very selective about
> what I accept as reality."
> --- Calvin ("Calvin and Hobbes" by Bill
> Bill Watterson)
> ======================================================
> ================
>
> Arturo Magidin
> magidin-at-member-ams-org
>
I'm very thankful to you. But, I have one more question:
Consider ideal
I={f \in C[0,1] : lim_{x \to 0} f(x)/x^n=0, n=0,1,2,…}
Is I prime ideal?
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