$X \subseteq \mathcal{P}(B)$ contains filters on $\mathcal{B}$, are $\bigcap_{F\in X}F$ and $\bigcup_{F\in X}F$ filters too?

Consider a Boolean algebra $$\mathcal{B}:=(B,\leq,\lor,\land,^c,0,1)$$ and $$\phi \neq X \subseteq \mathcal{P}(B)$$ whose elements are filters on $$\mathcal{B}$$. Show that:

1. $$\bigcap_{F\in X}F$$ is also a filter on $$\mathcal{B}$$. However, $$\bigcup_{F\in X}F$$ may not be a filter.
2. If $$X$$ is totally ordered by the inclusion relation $$\subseteq$$, then $$\bigcup_{F\in X}F$$ is a filter on $$\mathcal{B}$$.

I'm working with the definition: $$F \subseteq B$$ 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 $$y\in F$$

To start with, I need to show that $$\bigcap_{F\in X}F \neq \phi$$ - which I'm unable to do. Can we find an element common to all filters on $$\mathcal{B}$$, which may help us conclude that the intersection is not empty? I'm thinking in this direction because $$X$$ might as well be the set of all filters on $$\mathcal{B}$$.

Next, I want to show: if $$x,y \in \bigcap_{F\in X}F$$ then $$x\land y\in \bigcap_{F\in X}F$$. This seems easy, since $$x,y \in \bigcap_{F\in X}F$$ means that $$x$$ and $$y$$ are contained in every filter in $$X\subseteq\mathcal{P}(B)$$, and so is $$x\land y$$ (property of filters). Similarly for the last property, i.e. if $$x \in \bigcap_{F\in X}F$$ then $$x$$ is in every filter in $$X$$, and we know that filters are upwards closed - so if $$x\leq y$$ then $$y$$ is in every filter in $$X$$ (and hence in $$\bigcap_{F\in X}F$$)

Next, I want to show that $$\bigcup_{F\in X}F$$ (non-empty, of course) may not always be a filter - which calls for a counterexample? I'm unable to think of one. So, when will $$\bigcup_{F\in X}F$$ not be a filter? From the 2nd part, it seems that this may have something to do with ordering?

For the last part, since X is totally ordered, we could probably start off with $$X= \{X_1,X_2,...\}$$ (X may not be finite, who knows?), and w.l.o.g assume that $$X_1 \subseteq X_2 \subseteq ...\subseteq X_i\subseteq X_{i+1}...$$ (that's the total ordering defined by inclusion, yes?). How do I take it from here?

TL;DR I have shared my thoughts and work for every part of the question, and it would be a great help if I could get hints or insights that could help me complete my solution (happy to see other solutions also, though)! Thanks!

• $B = 1_{\mathcal P(B)}$ is an element of every filter of $\mathcal P(B)$. – amrsa Nov 1 '20 at 17:16
• I'm not sure what the notation means? – strawberry-sunshine Nov 1 '20 at 17:17
• $B$ is the top element of $\mathcal P(B)$, so it belongs to every filter of $\mathcal P(B)$. – amrsa Nov 1 '20 at 17:18
• Makes sense, since filters are upwards closed. Thanks! – strawberry-sunshine Nov 1 '20 at 17:19
• If $B=\{a,b\}$ and $X=\{ \{\{a\},\{a,b\}\}, \{\{b\},\{a,b\}\} \}$, then $$\bigcup X = \{ \{a\}, \{b\}, \{a,b\} \},$$ which is not a filter. – amrsa Nov 1 '20 at 17:22

For the second question you cannot assume that $$\langle X,\subseteq\rangle$$ is even countable, let alone that it can be ordered like the positive integers: it might be ordered like $$\Bbb R$$, for instance. All that you can assume is that if $$F_1,F_2\in X$$, then either $$F_1\subseteq F_2$$, or $$F_2\subseteq F_1$$.

Let $$G=\bigcup_{F\in X}F$$. It’s clear that $$G\ne\varnothing$$. Suppose that $$x,y\in G$$; then there are $$F_x,F_y\in X$$ such that $$x\in F_x$$ and $$y\in F_y$$. Without loss of generality we may assume that $$F_x\subseteq F_y$$. Can you finish it from there and go on to show that $$G$$ is upward closed?

This proof should suggest how to find a counterexample when $$X$$ is not linearly ordered by inclusion: when you’ve finished it, you’ll see that we used the linear order only to show that $$G$$ was closed under $$\land$$. For a counterexample, then, we probably want an $$X$$ that contains filters $$F_x$$ and $$F_y$$ containing elements $$x$$ and $$y$$, respectively, but no filter containing both $$x$$ and $$y$$. The simplest way to do that is to let $$X=\{F_x,F_y\}$$, where $$x,y\in B$$, $$x\in F_x\setminus F_y$$, and $$y\in F_y\setminus F_x$$, and if we can ensure that $$x\land y=0$$, we’ll make certain that $$F_x\cup F_y$$ is not a filter.

Clearly we need $$B$$ to have at least two elements, and they have to be incomparable. (Otherwise, the larger one will be in the filter containing the smaller one.) If we set $$x\lor y=1$$, $$x\land y=0$$, $$x^c=y$$, and $$y^c=x$$, we have simple Boolean algebra whose partial order has this Hasse diagram:

                     1
/ \
x   y
\ /
0


(It’s really just the power set algebra on a $$2$$-point set, as in amrsa’s comment.) And we can take $$F_x=\{x,1\}$$ and $$F_y=\{y,1\}$$ to get the desired counterexample: $$F_x\cup F_y=\{x,y,1\}$$, which is clearly not a filter, precisely because it doesn’t contain $$x\land y$$.

• Just to complete the proof: (1) $x\in F_x, y\in F_y$ and $F_x\subseteq F_y$ means $x\in F_y$ so that $x\land y \in F_y \implies x\land y\in G$ (2) If $x \in G$ and $x\leq y$, consider the filter $F_x$ such that $x \in F_x$. Since $F_x$ is upwards closed, $y\in F_x$ hence $y\in G$. Sounds good? – strawberry-sunshine Nov 8 '20 at 3:51
• @strawberry-sunshine: Looks good to me! – Brian M. Scott Nov 8 '20 at 4:19