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In A Course in Universal Algebra (Burris, Sankapannavar), the exercise 4.4.7-8, p.158, says:

Let $A$ be a Boolean algebra. Denote $A^\ast:=\{\text{ultrafilters of }A\}$, and give $A^\ast$ the topology, defined by the basis of open sets $\{N_a; a\!\in\!A\}$, where $N_a\!:=\!\{U\!\in\!A^\ast; a\!\in\!U\}$.

(a) The map $(\{\text{ideals of }A\},\subseteq)\!\rightarrow\!(\{\text{open subsets of }A^\ast\},\subseteq),\, I\!\mapsto\!I^\ast\!:= \bigcup_{a\in I}\!N_a$ is a lattice isomorphism, with $a\!\in\!I \Leftrightarrow N_a\!\subseteq\!I^\ast$.

(b) The map $(\{\text{filters of }A\},\subseteq)\!\rightarrow\!(\{\text{closed subsets of }A^\ast\},\subseteq),\, F\!\mapsto\!F^\ast\!:= \bigcap_{a\in F}\!N_a$ is a lattice isomorphism, with $a\!\in\!F \Leftrightarrow N_a\!\supseteq\!F^\ast$.

For any $S\!\subseteq\!A$, let $\mathfrak{I}(S)$ denote the ideal generated by $S$, and $\mathfrak{F}(S)$ the filter generated by $S$. Then $$\bigcup_{a\in S}\!N_a=\!\bigcup_{a\in \mathfrak{I}(S)}\!N_a~~~\text{ and }~~~\bigcap_{a\in S}\!N_a=\!\bigcap_{a\in \mathfrak{F}(S)}\!N_a.$$ In $(\{\text{ideals of }A\},\subseteq)$, the supremum is described as $I\!\vee\!I'\!=\!\{x\!\in\!A; \exists a\!\in\!I\,\exists a'\!\in\!I'\!: x\!\leq\!a\!\vee\!a'\}$, and in $(\{\text{filters of }A\},\subseteq)$, the supremum is described as $F\!\vee\!F'\!=\!\{x\!\in\!A; \exists a\!\in\!F\,\exists a'\!\in\!F'\!: x\!\geq\!a\!\wedge\!a'\}$. Moreover, $N_a\!\cup\!N_b\!=\!N_{a\vee b}$; $N_{a}\!\cap\!N_{b}\!=\!N_{a\wedge b}$; $(N_a)^c=\!N_{a^c}$. In Boolean algebras, an ideal $I$ of $A$ is maximal (i.e. maximal w.r.t. $\subseteq$ among all ideals $I'$ with $1\!\notin\!I'$) iff it is prime (i.e. $1\!\notin\!I$ and $\forall x,y\!\in\!A\!: x\!\wedge\!y\!\in\!I \Leftrightarrow (x\text{ or }y\!\in\!I)$). In Boolean algebras, a filter $F$ of $A$ is maximal (or an ultrafilter, i.e. maximal w.r.t. $\subseteq$ among all filters $F'$ with $0\!\notin\!F'$) iff it is prime (i.e. $0\!\notin\!F$ and $\forall x,y\!\in\!A\!: x\!\vee\!y\!\in\!F \Leftrightarrow (x\text{ or }y\!\in\!F)$). (Stone) If $I$ is an ideal of $A$ and $a\!\in\!A\!\setminus\!I$, then there is a maximal ideal $M$ with $F\!\subseteq\!M\!\subseteq\!A\!\setminus\!\{a\}$. (Stone) If $F$ is a filter of $A$ and $a\!\in\!A\!\setminus\!F$, then there is an ultrafilter $U$ with $F\!\subseteq\!U \!\subseteq\! A\!\setminus\!\{a\}$.

Questions: Here are the things that I didn't yet manage to prove and am having problems with.

(1) We have $F^\ast\cap F'^\ast=(F\!\cap\!F')^\ast$ iff $(\bigcap_{a\in F}\!N_a)\cap(\bigcap_{a'\in F'}\!N_{a'}) = \bigcap_{x\in F\cup F'}\!N_x = \bigcap_{y\in F\cap F'}\!N_y$ iff for each ultrafilter $U$, we have $F\!\cap\!F'\!\subseteq U \Rightarrow F\!\cup\!F'\!\subseteq U$, but I don't see why this would be true.

(2) Proving $F^\ast\!\cup F'^\ast=(F \vee\!F')^\ast$ boils down to showing that the following inclusion holds: $\{U\!\in\!A^\ast; \forall a\!\in\!F\,\forall a'\!\in\!F'\!: a\!\vee\!a'\!\in\!U\}\subseteq\{U\!\in\!A^\ast; \forall b\!\in\!F\,\forall b'\!\in\!F'\, \forall x\!\geq\!b\!\wedge\!b'\!: x\!\in\!U\}$. Now for $b,b'$, we have $b\!\vee\!b'\!\in\!U$, and from primality of $U$, we have w.l.o.g. $b\!\in\!U$. But how do we show $b\!\wedge\!b'\!\in\!U$?

(3) Injectivity: We have $I^\ast\!=\!I'^\ast$ iff $\forall U\!\in\!A^\ast\!: (\exists a\!\in\!I\!: a\!\in\!U)\Leftrightarrow(\exists a'\!\in\!I'\!: a'\!\in\!U)$ iff $\forall U\!\in\!A^\ast\!: U\!\cap\!I\!=\!\emptyset \Leftrightarrow U\!\cap\!I'\!=\!\emptyset$. I've proved the injectivity of $F^\ast\!=\!F'^\ast$ by using Stone's theorem above, but for $I^\ast\!=\!I'^\ast$, I must produce an ultrafilter by using ideals, so I'm not sure what to do.

(4) We have $a\!\in\!I\Leftarrow N_a\!\subseteq\!I^\ast$ iff $\{U\!\in\!A^\ast;a\!\in\!U\}\!\subseteq\!\{U\!\in\!A^\ast\!; I\!\cap\!U\!\neq\!\emptyset\}\Rightarrow a\!\in\!I$. I don't know where to go from here. I've proved $a\!\in\!F\Leftarrow N_a\!\supseteq\!F^\ast$, by using Stone's theorem above, but here, we must find an ultrafilter by using ideals, so I'm out of good ideas.

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I think that (1) and (2) could be easier if you use the observation that $F^*=\{U\in A^*; F\subseteq U\}$. – Martin Sleziak Jun 6 '12 at 13:58
There is an error in part (b) of the exercise. The map $F\mapsto F^*$ is not a lattice isomorphism but an antiisomorphism, as can be quickly checked on, say, the Boolean algebra 4. – Miha Habič Jun 6 '12 at 14:08
Also, since in (3) and (4) you seem to have shown exactly half of what you need, duality between ideals and filters springs to mind. Concretely, try using the fact that, for an ideal $I$, we have $I^*=A^*\setminus I'^*$ where $I'$ is the dual filter to $I$. – Miha Habič Jun 6 '12 at 14:15
Aha, I thought that $F\!\mapsto\!F^\ast$ being a homomorphism was fishy. Also, your equality $I^\ast\!=\!A^\ast\!\setminus\!{I^c}^\ast$ and $F^\ast\!=\!A^\ast\!\setminus\!{F^c}^\ast$ proved most useful, and shortened my proof considerably. Thanks! @MihaHabič – Leon Jun 7 '12 at 1:23
up vote 2 down vote accepted

You have $F^*=\bigcap\{N_a:a\in F\}=\{U\in A^*:F\subseteq U\}$. This clearly means that if $F_0\subseteq F_1$, then $F_0^*\supseteq F_1^*$: the bigger the filter $F$, the more sets $N_a$ you’re intersecting to form $F^*$, so the smaller $F^*$ must be. In fact, if $F$ is an ultrafilter, then $F^*=\{U\in A^*:F\subseteq U\}=\{F\}$: the only ultrafilter that contains $F$ is $F$ itself. Thus, if $\mathscr{F}$ is the set of filters on $A$, and $\mathscr{C}$ is the family of closed subsets of $A^*$, the map $\mathscr{F}\to\mathscr{C}:F\mapsto F^*$ is order-reversing. In particular, you can’t hope to prove that $\langle\mathscr{F},\subseteq\rangle$ and $\langle\mathscr{C},\subseteq\rangle$ are isomorphic lattices: if the map is a lattice isomorphism, it must be an isomorphism between $\langle\mathscr{F},\subseteq\rangle$ and $\langle\mathscr{C},\supseteq\rangle$.

In particular, this means that you should be trying to prove that if $F_0,F_1\in\mathscr{F}$, then $$(F_0\cap F_1)^*={F_0}^*\cup{F_1}^*$$ and $$(F_0\lor F_1)^*={F_0}^*\cap{F_1}^*\;.$$ This should dispose of most of your difficulties with (1) and (2).

For (3) and (4), note that (maximal) ideals and (ultra)filters are complementary to each other: a set $S\subseteq A$ is a (maximal) ideal iff $\{\lnot a:a\in A\}$ is an (ultra)filter. Thus, by taking complements you can work with filters or with ideals, as you choose.

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Your first paragraph is very clear on why we actually have an order reversing map. I should have come to that conclusion myself, but I was too preoccupied with proving what was asked of me. Thank you! – Leon Jun 7 '12 at 1:26

$\newcommand{\Lra}{\Leftrightarrow}$I think that (1) and (2) could be easier if you use the observation that $F^*=\{U\in A^*; F\subseteq U\}$.

Similarly you have $I^*=\{U\in A^*; I\cap U\ne\emptyset\}$.

(3) Let $I$ be any ideal. Then $F=\{a'; a\in I\}$ is a filter and $$U\in I^* \Lra (\exists a\in I) a\in U \Lra (\exists a\in I) a'\notin U \Lra U\notin F^*.$$ Hence $I^*=A^* \setminus F^*$.

(4) Suppose that $a\notin I$. Then $F=\{b\in A; b\ge a\}$ is a filter such that $F\cap I=\emptyset$. Then you can use Boolean prime ideal theorem to get an ultrafilter $U$ such that $U\cap I=\emptyset$ and $F\subseteq U$. In particular, we have $a\in U$, and this contradicts the assumption $N_a \subseteq I^*$.

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