proof that every filter is contained in an ultrafilter
Let be the set of all non-empty subsets of which are not contained in . By Zermelo’s well-orderding theorem, there exists a relation ‘’ which well-orders . Define and extend the relation ‘’ to by decreeing that for all .
We shall construct a family of filters indexed by using transfinite induction. First, set . Next, suppose that, for some , has already been defined when . Consider the set ; if and are elements of this set, there must exist an such that and ; hence, cannot be empty. If, for some there exists an element such that is empty, let be the filter generated by the filter subbasis . Otherwise is a filter subbasis; let be the filter it generates.
Note that, by this definition, whenever , it follows that ; in particular, for all we have . Let . It is clear that and that .
It is easy to see that is a filter. Suppose that . Then there must exist an such that . Since is a filter, and , hence and . Conversely, if and , then there exists an such that and . Since is a filter, , hence . By the alternative characterization of a filter, is a filter.
Moreover, is an ultrafilter. Suppose that and are disjoint and . If either or , then either or because . If and , then because . If and , there must exist such that is empty. Because is the complement of , this means that and, hence .
|Title||proof that every filter is contained in an ultrafilter|
|Date of creation||2013-03-22 14:41:44|
|Last modified on||2013-03-22 14:41:44|
|Last modified by||rspuzio (6075)|