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Unformatted text preview: additive subgroup of R and that c R G, prove that there exists an additive
subgroup S of R such that G S and c R S and such that every additive subgroup of R that
includes S must also contain the number c.
We look at the family of all additive subgroups of R that fail to contain the number c and give
it the partial order . Since the union of all the groups in any chain in must belong to , each
chain in has an upper bound in .
c. Given that S is an additive subgroup of R with the properties just described and that y R S then it
possible to find a member x of S and an integer n such that ny x c.
The set ny x n Z and x S is clearly a group that properly includes S and therefore
c ny x n Z and x S 3. A nonempty family of nonempty subsets of a set S is said to be a filter in S if the intersection of any two
members of belongs to and any subset of S that includes a member of must belong to . A filter that is
not properly included in any other filter is called an ultrafilter.
a. Prove that every filter in a set S is included in an ultrafilter.
If we give the family of all filters in S the partial order then it is clear that the union of any
chain of filters is a filter and so every chain has an upper bound.
b. Given that is a family of subsets of a set S and that the intersection of any finite number of members of
is nonempty, prove that is included in an ultrafilter in S.
We define to be the family of all those subsets of S that include the intersection of finitely
many members of . Since
and is a filter, the desired result follow from part a.
c. Suppose that is a filter in a set S and that for every subset E of S, either E
or S E
. Prove
that must be an ultrafilter in S.
Whenever a subset E of S fails to belong to we know that E is disjoint from the member S E
of and that, consequently, E can’t belong to any filter that includes .
d. Prove that if is an ultrafilter in a set S and E S, then either E
or S E
.
Suppose that is a filter in a set S and that E S and that neither of the sets E and S E belong
to . Then, since neither of the sets E and S E can include a member of , neither of these
sets can be disjoint from a member of and it is clear that the family of all those subsets of S
that include an intersection E F with F
must be a filter that properly includes . Therefore
if is an ultrafilter, one of the sets E and S E must belong to .
e. Prove that there is an ultrafilter in Z such that for every positive integer n we have 70 Z m
Prove that if for each E m n . Z we define
1 if E E 0 if E 1, E 0 whenever E is a finite set, and
E1 Þ E2 E1 E2
whenever the sets E 1 and E 2 are disjoint from each other.
The existence of an ultrafilter that contains every set of the form m Z m n with n a
positive integer follows from part b.
The desired properties of the function follow at once because, if E 1 and E 2 are subsets of Z
and E 1 E 2 then the condition E 1 Þ E 2
will hold if and only if exactly one of the sets E 1...
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 Fall '08
 STAFF
 Math, Calculus

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