a notion of dual for ideals Consider an ideal $A$ on $X$. Define the dual of $A$ as the set of all sets that intersect all sets in $A$ finitely often. What is the name of this dual? Where can I read more about this?
In particular, the dual I'm interested in is defined like so:
consider ideal $A$ over a set $X$, then the dual $A'$ is
$A' = \{x\subseteq X:\forall y\in A,|x\cap y|<\infty\}$
In case this is helpful in identifying this dual, here are some easily proven properties:
For any ideal $A$ we have $A'=A'''$ and $A\subseteq A''$
where $A'$ denotes the dual of $A$
Also if $A\subseteq B$ for ideals $A$ and $B$ then $B'\subseteq A'$
 A: The following is rather a contribution (or partial answer) to your question. It is based on a series of elementary facts:


*

*Let $f:B \to C$ be a surjective morphism of Boolean algebras. If $I$ is an ideal of $B$, then $f(I)$ is an ideal of $C$; if $J$ is an ideal of $C$, then $f^{-1}(J)$ is an ideal of $B$.

*Let $X$ be a set. The relation $\sim$ on ${\cal P}(X)$ defined by 
$$
a \sim b \text{ if and only if the symmetric difference of $a$ and $b$ is finite}
$$
is a congruence of Boolean algebra, that is, $a \sim b$ implies $a \cup c \sim b \cup c$, $a \cap c \sim a \cap c$ and $\bar a \sim \bar b$ (here $\bar a$ denotes the complement of $a$). It follows that the quotient map $p: {\cal P}(X) \to {\cal P}(X)/{\sim}$ is a surjective morphism of Boolean algebras.

*Let $C$ be a Boolean algebra and $I$ be an ideal if $C$. Then the set
$$
I^{\bot} = \{c \in C \mid \text{ for all $a \in I$, $c \wedge a = 0$} \}
$$
is also an ideal.


Now, coming back to your definition, one has $I' = p^{-1}((p(I)^{\bot})$. In other words, $I'$ is obtained from the simpler $I^{\bot}$ construction through the "up-to-finite" congruence.
Notation and terminology. The operator $I^{\bot}$ is likely to be well known
(although a precise reference would be welcome), but I don't know of a specific name for it.
The notation $I^{\bot}$ advocates for the term orthogonal of $I$, but this might suggest that not only $I \wedge I' = 0$, which is true, but also that $I \vee I' = 1$, which is not true in general. 
For instance, in the Boolean algebra ${\cal P}({\Bbb N})$, if $I$ is the ideal of  finite subsets, then $I^\bot = \{\emptyset\}$. 
