Stone duality for ideals and filters (exercise) 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.
 A: $\newcommand{\Lra}{\Leftrightarrow}$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 $N_a \subseteq I^*$.
A: 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.
