Let $W$ be any non-empty set. We call a subset $A$ of $\mathscr{P}(W)$ a 3-fold filter on $W$ if the following hold:


(i) $W\in A$ and $\varnothing\notin A$;


(ii) If $a\in A$ and $a\subset b$ then $b\in A$;


(iii) If $a,b,c\in A$ then $(a\cap b)\cup(b\cap c)\cup(c\cap a)\in A$.


Now let $A,B$ be two 3-fold filters on $W$. Let $X=\{a\times W\mid a\in A\}$ and $Y=\{W\times b\mid b\in B\}$. Prove that $X\cup Y$ can be extended to a 3-fold filter on $W\times W$, i.e., $\varnothing$ is not contained in the closure of $X\cup Y$ under the operation $(a\cap b)\cup(b\cap c)\cup(c\cap a)$.



A question about the generalization of filters
First, a bit of notation: if $a\_1,\dots,a\_n\subseteq W$, let $\chi\_{a\_1,\dots,a\_n}\colon W\to\{0,1\}^n$ denote the “$n$-fold characteristic function”
$$\chi\_{a\_1,\dots,a\_n}(u)=([u\in a\_1],\dots,[u\in a\_n]),\qquad u\in W,$$
where $[\ ]$ is the Iverson bracket. For a Boolean function $f\colon\{0,1\}^n\to\{0,1\}$, let $S\_f\colon\mathcal P(W)^n\to\mathcal P(W)$ be the corresponding set operator
$$S\_f(a\_1,\dots,a\_n)=\{u\in W:f(\chi\_{a\_1,\dots,a\_n}(u))=1\}.$$


In this notation, a 3-fold filter is a nonempty upper subset $A\subseteq\mathcal P(W)$ closed under $S\_m$, where $m(x,y,z)$ is the ternary majority function. (I am allowing $\varnothing\in A$ here; I will call a 3-fold filter *proper* if $\varnothing\notin A$.)


I will also need a little information on Boolean clones, see [Post’s lattice](https://en.wikipedia.org/wiki/Post's_lattice). The majority function $m(x,y,z)$ generates the clone of all monotone self-dual Boolean functions. It follows that for a nonempty set $X\subseteq\mathcal P(W)$, the 3-fold filter generated by $X$ consists of sets $a$ such that
$$a\supseteq S\_f(a\_1,\dots,a\_n)$$
for some $a\_1,\dots,a\_n\in X$, and a monotone self-dual function $f\colon\{0,1\}^n\to\{0,1\}$.


It will be actually more convenient to use a slightly different characterization. Since 3-fold filters are closed upwards, they are closed under $S\_f$ not only for monotone self-dual $f$, but also for Boolean functions $f$ such that $f(x\_1,\dots,x\_n)\ge g(x\_1,\dots,x\_n)$ for some monotone self-dual function $g$. Now, the collection of all such functions $f$ forms the clone denoted $T\_1^2$ on the linked page; it can be described intrinsically as the set of all functions $f\colon\{0,1\}^n\to\{0,1\}$ such that
$$\alpha\lor\beta=\mathbf1\implies f(\alpha)\lor f(\beta)=1$$
for all $\alpha,\beta\in\{0,1\}^n$, where $\alpha\lor\beta$ is defined coordinatewise, and $\mathbf1=(1,\dots,1)$. (That is, $f$ is polymorphism of the relation $\{(1,1),(1,0),(0,1)\}$.)


So, the 3-fold filter generated by $X\subseteq\mathcal P(W)$ is
$$\langle X\rangle=\{a\subseteq W:\exists a\_1,\dots,a\_n\in X\,\exists f\in T\_1^2\,a\supseteq S\_f(a\_1,\dots,a\_n)\}.$$



> 
> **Lemma:** If $A\subseteq\mathcal P(W)$ is a proper 3-fold filter, and $a\_1,\dots,a\_n\in A$, there exist $u,u'\in W$ such that $\chi\_{a\_1,\dots,a\_n}(u)\lor\chi\_{a\_1,\dots,a\_n}(u')=\mathbf1$.
> 
> 
> 


**Proof:** Define a function $f\colon\{0,1\}^n\to\{0,1\}$ by
$$f(\alpha)=0\iff\exists u\in W\,\chi\_{a\_1,\dots,a\_n}(u)=\alpha.$$
By definition, $S\_f(a\_1,\dots,a\_n)=\varnothing\notin A$, hence $f\notin T\_1^2$. Thus, there are $\alpha,\beta$ such that $\alpha\lor\beta=\mathbf1$, and $f(\alpha)=f(\beta)=0$. The latter means there are $u,u'\in W$ such that $\alpha=\chi\_{a\_1,\dots,a\_n}(u)$ and $\beta=\chi\_{a\_1,\dots,a\_n}(u')$. QED


Now, let $A$, $B$, $X$, and $Y$ be as in the question, and assume for contradiction that $\varnothing\in\langle X\cup Y\rangle$. Thus, there are $a\_1,\dots,a\_n\in A$, $b\_1,\dots,b\_m\in B$, and a function $f\in T\_1^2$ such that
$$S\_f(a\_1\times W,\dots,a\_n\times W,W\times b\_1,\dots,W\times b\_m)=\varnothing.$$
Using the lemma, there are $u,u',v,v'\in W$ such that
$$\begin{align}
\chi\_{a\_1,\dots,a\_n}(u)\lor\chi\_{a\_1,\dots,a\_n}(u')&=\mathbf1,\\
\chi\_{b\_1,\dots,b\_m}(v)\lor\chi\_{b\_1,\dots,b\_m}(v')&=\mathbf1.
\end{align}$$
Put $w=(u,v)$ and $w'=(u',v')$. Then $w,w'\in W\times W$ satisfy
$$\chi\_{a\_1\times W,\dots,a\_n\times W,W\times b\_1,\dots,W\times b\_m}(w)\lor\chi\_{a\_1\times W,\dots,a\_n\times W,W\times b\_1,\dots,W\times b\_m}(w')=\mathbf1,$$
thus
$$f(\chi\_{a\_1\times W,\dots,a\_n\times W,W\times b\_1,\dots,W\times b\_m}(w))\lor f(\chi\_{a\_1\times W,\dots,a\_n\times W,W\times b\_1,\dots,W\times b\_m}(w'))=1$$
as $f\in T\_1^2$. However, this means that $w$ or $w'$ is an element of $S\_f(a\_1\times W,\dots,a\_n\times W,W\times b\_1,\dots,W\times b\_m)$, a contradiction.


