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In article <18235534.1161365641076.JavaMail.jakarta@xxxxxxxxxxxxxxxxxxxxxx>,
Sasha P <abp83@xxxxxxxxxx> wrote:
>Thank you very much! But why ideals corresponding nonprincipal
>ultrafilters is not maximal?
Every maximal ideal in C[0,1] is m_x = {f: f(x) = 0} for some
x in [0,1].
Proof: Suppose J is a maximal ideal not of that form. For every x
in [0,1] there is some f_x in J such that f_x(x) <> 0. By compactness,
[0,1] is covered by finitely many of the open sets U_x = {t: f_x(t) <> 0},
i.e. there are f_{x_1},...,f_{x_n} in J such that for each x in [0,1],
some f_{x_j}(x) <> 0. Then sum_j f_{x_j}^2 is in J and is invertible
in C[0,1], impossible for a proper ideal.
Your next question is: why is the ideal corresponding to an ultrafilter
prime?
This is a consequence of the fact that if F is an ultrafilter and
A union B is in F, then A is in F or B is in F. Thus
if f g is in the ideal J corresponding to F, i.e.
Z(f g) = Z(f) union Z(g) is in F, then Z(f) or Z(g) is in F, and
f or g is in J.
Robert Israel israel@xxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
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