PIT implies: In a boolean lattice, every filter can be enlarged to a maximal one

Question

I am working through this proof of Herrlich's Axiom of Choice:

$(1)\Rightarrow(2)$: How do you define the quotient lattice $B$ modulo a Filter? And why is the preimage of the maximal filter $\mathcal{G}$ maximal in $B$?

Answer 1

Boolean lattice has a complement operation. So you can define what would be a symmetric difference, $a\oplus b=(a\lor b)-(a\land b)$. Now $a\equiv_\cal F b$ if and only if $1-(a\oplus b)\in\cal F$.

This is just like the usual quotient by a filter, or usually an ultrafilter. Or an ideal in ring theory. You want two things to be identified if their difference is negligible.

Verifying that the preimage of a maximal filter in $B/\cal F$ is an extension of $\cal F$ to a maximal filter is just verifying the definitions, and I will leave this part to you. Unless you have a specific point you wish clarified.