Let $\mathfrak{B}=(B,\bot,\top,\lnot,\wedge,\vee)$ be a boolean algebra. $B_F$ be the set of all filters on $\mathfrak B$. And for all filter $F$, $G$, $F \wedge_{B_F} G \colon= \mathbf C(F \cup G)$ in which $\mathbf C$ denotes the filter closure operator; $F \vee_{B_F} G \colon= F \cap G$; $0 \colon= B$, $1 \colon= \{\top\}$. Then $(B_F,0,1,\wedge_{B_F},\vee_{B_F})$ makes a complete lattice.

My question is can we add a negation in order to make it be a Boolean algebra?


It is relatively easy to see that it does not matter whether we work with filters or with ideals.

The following is taken verbatim from Steven R. Givant,Paul Richard Halmos: Introduction to Boolean Algebras p.167:

The ideals of a Boolean algebra form a complete, distributive lattice, but they do not, in general, form a Boolean algebra. To give an example, it is helpful to introduce some terminology. An ideal is maximal if it is a proper ideal that is not properly included in any other proper ideal. We shall see in the next chapter that an infinite Boolean algebra $B$ always has at least one maximal ideal that is not principal. Assume this result for the moment. A "complement" of such an ideal $M$ in the lattice of ideals of $B$ would be an ideal $N$ with the property that $$M\wedge N=\{0\} \qquad\text{and}\qquad M\vee N=B.$$ Suppose the first equality holds. If $q$ is any element in $N$, then $p \wedge q = 0$, and therefore $p \le q'$, for every element $p$ in $M$, by Lemma 1. In other words, the ideal $M$ is included in the principal ideal $(q')$. The two ideals must be distinct, since $M$ is not principal. This forces $(q')$ to equal $B$, by the maximality of $M$. In other words, $q' = 1$, and therefore $q = 0$. What has been shown is that the meet $M\wedge N$ can be the trivial ideal only if $N$ itself is trivial. In this case, of course, $M \vee N$ is $M$, not $B$. Conclusion: a maximal, non-principal ideal does not have a complement in the lattice of ideals.

The existence of maximal ideals, which was used in the above excerpt, is guaranteed by Boolean prime ideal theorem.

  • $\begingroup$ I understand. Thank you very much. $\endgroup$ – Popopo Jun 20 '12 at 14:01

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