Re: Prime ideals in the ring of continuous functions

In article <4539D9C4.8070907@xxxxxx>,
Jannick Asmus <jannick.news@xxxxxx> wrote:
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.

You'll notice I talked about "ultrafilters on P([0,1])". Lucky for me,
[0,1] is infinite! (-:

But, yes, you are right; in general, the axiom of choice guarantees
the existence of nonprincipal ultrafilters on X, if and only if X is
infinite.


--
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"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
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Arturo Magidin
magidin-at-member-ams-org

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