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!

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

                    / \
                   x   y
                    \ /

(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 $x\land y$.

  • $\begingroup$ 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? $\endgroup$ – strawberry-sunshine Nov 8 '20 at 3:51
  • $\begingroup$ @strawberry-sunshine: Looks good to me! $\endgroup$ – Brian M. Scott Nov 8 '20 at 4:19

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