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In article <31617721.1161416029645.JavaMail.jakarta@xxxxxxxxxxxxxxxxxxxxxx>,
abp83@xxxxxxxxxx says...
>> [..]
>
>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,â?¦}
>Is I prime ideal?
No. In fact one can easily find a pair f, g such that fg=0 but neither
f nor g is in I. Just let f be a continuous function that vanishes at
zero and on [1/2n, 1/(2n-1)] but reaches a maximum of 1/2n on
(1/(2n+1), 1/2n). Then there's a sequence x_n -> 0 such that f(x_n)>x_n
so f is not in I. Then let g be a similar function that vanishes on
[1/(2n+1),1/2n].
However you can combine this idea with the ultrafilter idea to get a
prime ideal:
Let E ={ (0,1/n) : n = 1,2,..} E has the finite intersection property
so it extends to an ultrafilter F (which is necessarily nonprincipal).
Let I = {f : for every n there exists A_n in F such that |f(x)| < x^n on A_n}
I is an ideal, in fact it is a proper subideal of m_0. (And it's not the
zero ideal since f(x) = x^(1/x) is in it.) To see that I is prime,
suppose fg in I. Then exists A_n such that f(x)g(x) < x^n on A_n. Let
B_n = {x : |f(x)| < x^n}
C_n = {x : |g(x)| < x^n}
We have that A_{2n} is a subset of B_n \cup C_n, hence B_n \cup C_n is
in F, hence for each n at least one of B_n, C_n is in F. So either there
are infinitely many n such that B_n is in F, or infinitely many n such
that C_n is in F. In the former case f is in I, in the latter g is in I.
Robert Sheskey
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