$\mathcal{B}:=(B,\leq,\lor,\land,^c,0,1)$ is a Boolean algebra. Show that the smallest filter in $\mathcal{B}$ containing $a\in B$ is given by $$F(a) = \{x\in B:a\leq x\}$$
Here's my work:
First, I think it's important to verify that $F(a)$ is a filter. Clearly, $a \in F(a)$ so $F(a)\neq \phi$. $F(a)$ is also upwards closed, i.e. if $x \in F(a), x\leq y$ then $y \in F(a)$. This follows from the definition of $F(a)$ and transitivity of $\leq$, the partial order on $B$. Lastly, we must show that if $x,y\in F(a)$ then $x\land y \in F(a)$. $x\land y$ is the greatest lower bound of $x$ and $y$. Since $x,y\in F(a)$, we know that $a\leq x, a\leq y$. From here, we see that $a\land x =x$ and $a\land y=y$. So $a \land (x\land y) = (a\land x)\land y = a\land y = a$, implying $a \leq x\land y$, so $x\land y\in F(a)$. So F(a) is indeed a filter!
Why is $F(a)$ the smallest filter containing $a$?
P.S. For reference, $F \subseteq B$ is a filter if:
- $F \neq \phi$
- If $x,y \in F$ then $x\land y\in F$
- If $x\in F$ and $x\leq y$ then $y\in F$
Hint: any filter to which $a$ belongs is “upwards closed”, so it has to include $F(a)$