Trouble understanding filter and ultrafilter

Question

A filter F on S is a collection of subsets of S in which two conditions hold:

If A and B belong to the collection F then A∩B also belongs to the collection. If A belongs to the collection F and A is a subset of B then B also belongs to the collection. (If A⊂B then B is said to be a superset of A.) For S={a, b, c} one filter is the collection : { {a}, {a, b}, {a, b, c} } Why is {a,c} not in the filter collection here, {a} is a subset of {a,c} so it must be in the example given above . Can someone make me understand ultrafilter with respect to Boolean Algebra ?

Answer 1

Indeed $\{a,c\}$ should be in this collection, in fact all subsets that contain $a$. On a finite set we only have filters that are generated by their intersection, as is easy to check.

A filter $\mathcal{F}$ on a Boolean algebra $B$ is a subset of $B$ obeying the conditions:

• $\emptyset \neq \mathcal{F} \neq B$
• $\forall b_1,b_2 \in B: b_1, b_2 \in \mathcal{F} \implies b_1 \land b_2 \in \mathcal{F}$.
• $\forall b_1, b_2 \in B: (b_1 \in \mathcal{F} \text{ and } b_1 \le b_2) \implies b_2 \in \mathcal{F}$.

Note the similarity to the conditions for subsets, which is not coincidental, as the powerset of $S$ is a Boolean algebra (where $\land$ is intersection and $\le$ inclusion).

In either case an ultrafilter is a filter that is maximal w.r.t inclusion; there can be no filter that is a strict superset of it.