# Is every filter generated by a (non-trivial) filter base?

Suppose $F$ is any filter defined on a set $X$ (that is, with the partial order given by $\subset$). I know that every filter base generates a filter and that a filter is always a filter base itself (a base of itself, actually). But the question that intrigues me is: Is it always possible to find some filter base of $F$ that is not $F$ itself?

I ask this because my lecturer defined the supremum of a family of filters as the filter generated by the set of all finite intersections of elements of each base of each filter. One could simply take these bases as the filters themselves, but were my first question true, would the supremum change if one takes non-trivial bases for some (or all) of the filters?

The first question interests me most...

• Isn't $F\setminus\{X\}$ a filter base? – bof Nov 14 '17 at 4:41
• Such a supremum of even two filters might not exist, we can easily have two filters with a set $A$ in one of them, and the set $X\setminus A$ in the other. But if it exists it can be described as the filter generated by the intersections which must then always be non-empty.. – Henno Brandsma Nov 14 '17 at 22:01
• @bof. Yes, unless X has just one member. (Annoying trivial case.) – DanielWainfleet Nov 19 '17 at 7:34

No. If $F = \{X\}$, the only filter basis generating $F$ is $F$ itself.
• Isn't $\emptyset$ a base for the filter $\{X\}$? Okay, okay, i looked it up in Wikipedia and see a definition that requires a filter base to be nonempty. But that seems pretty arbitrary to me. That's not the way I would define it. – bof Nov 19 '17 at 8:02
• A family $\mathcal B$ of subsets of $X$ with the finite intersection property. If $\mathcal B=\emptyset$ then $\mathcal B$ has the finite intersection property, and the smallest filter containing $\mathcal B$ is $\{X\}.$ – bof Nov 19 '17 at 9:20