The meet of two filters

Question

Given two filters $\mathcal{F}$ and $\mathcal{G}$ on a set $X$, there's a smallest filter containing $\{F \cup G : F \in \mathcal{F},G \in \mathcal{G}\}.$

(Actually, I can't quite tell if this is a filter in its own right, or whether we have to look at the smallest filter that contains it.)

In any event, let's write $\mathcal{F} \wedge \mathcal{G}$ for the smallest filter containing the aforementioned set. It seems to be the order-theoretic meet of $\mathcal{F}$ and $\mathcal{G}$ (ordering filters by inclusion).

Question. Is this correct? (Proof request.)

Answer 1

The order-theoretic meet of $\mathcal{F}$ and $\mathcal{G}$ is just $\mathcal{F}\cap \mathcal{G}$, since this is the order-theoretic meet on all of $\mathcal{P}(\mathcal{P}(X))$, and it is easily checked to be a filter. Let $\mathcal{H} = \{F\cup G\mid F\in \mathcal{F}, G\in \mathcal{G}\}$. I claim that $\mathcal{H} = \mathcal{F}\cap \mathcal{G}$, so it is the order-theoretic meet (and a filter), as you suspected.

• For any $F\cup G\in \mathcal{H}$, $F\subseteq F\cup G$, so $F\cup G\in \mathcal{F}$, and $G\subseteq F\cup G$, so $F\cup G\in \mathcal{G}$. So $F\cup G\in \mathcal{F}\cap \mathcal{G}$.
• For any $H\in \mathcal{F}\cap \mathcal{G}$, $H = H\cup H$, so $H\in \mathcal{H}$.

Answer 2

Yes, it is a filter.
Let F,G be the filters and S the aforementioned set.
The empty set is not in S because it is not in F nor G.

If K = A $\cup$ B, L = C $\cup$ D, A,C in F, B,D in G, then
(A $\cap$ C) $\cup$ (B $\cap$ D) in S is a lower bound of K,L.
Thus S is lower directed.

Assume K = A $\cup$ B, A in F, B in G and K subset L.
As F and G are filters, K is in both F,G.
Thus L in F,G and L = L $\cup$ L in S.
Consequently S is an upper set.

So S is a lower bound of F,G. Is it actually the meet?
M = F $\cap$ G is a filter.
Empty set not in M since it isn't in F nor G.

If A,B in M, then some K in F, L in G with K,L subset A,B.
Thus U = K $\cup$ L subset A,B and U in F,G.
Whence M is lower directed.

If A in M, A subset K, then K in F,G. Thus M is upper set.

Since M is the greatest lower bound of F,G, S subset M.
Does S = M?

These results for both S and M generalize from filters of a powerset to filters of an upper semi-lattice.