Duality between Ideal and Filters on a poset

Question

I am studying Ideals and Filters on a poset, and it is clear to me that this are dual notions in the sense that reversing inclusions in the definition we can get the one from the other, or in the sense that taking complements of subsets belomging to a structure we obtain the associated dual one. I was wondering if there is more about this duality concept, particularly in relation to general topology: Filters are used directly to assign neighboroods, do Ideal directly play a role as well?

Answer 1

There is a nice duality between filters and ideals in topology, but then we have to go via Boolean algebras and Stone spaces:

If we have a Boolean algebra (BA) $B$ (a bounded distributed and complemented lattice essentially, so with operations $\land, \lor, \lnot$ and a $0$ and $1$.) on the set $S(B)$ of its ultrafilters we can put a natural topology that makes it into a compact Hausdorff space with a base of clopen (closed-and-open) sets.

On the other hand, if we have a compact Hausdorff space with a base of clopen sets (a co-called Stone space) $X$ then $CO(X)$, the set of clopen subsets of $X$, is a natural Boolean algebra (with the standard set operations of union, intersection and complementation and $\emptyset$, $X$ as top and bottom.

It turns out these are "natural" inverses: the Stone space of the BA $CO(X)$ is homeomorphic to $X$ again and $CO(S(B))$ is isomorphic to the BA $B$ for all Boolean algebras.

Now a filter $\mathcal{F}$ in a BA $B$ corresponds to a closed set $C \subseteq S(B)$ : define $C$ to be the set of all ultrafilters on $B$ (i.e. points of $S(B)$) that contain $\mathcal{F}$. And vice versa, if $C$ is a closed subset of $S(B)$, then the corresponding filter in $B$ is the intersection of all $\mathcal{U}_b$ with $\mathcal{U}_b \in C$, where $\mathcal{U}_b=\{: F \subseteq B: b \in F\}$ is the ultrafilter on $B$ determined by $b$.

Dually an open subset $O$ of $S(B)$ corresponds to an ideal of $B$ and vice versa.

So in Stone spaces we see that the duality open vs closed is the mirrror image of the filter and ideal duality.

For more info on this duality and proofs, see these notes by KP Hart.

Answer 2

Yes there is a topological duality.
For a space S and a subset A, $S - A^o = \overline{S - A}$.
It is a self dual statement to $S - \overline A = (S - A)^o$.

One is proved by the other by taking complements and interchanging A and S - A.

An example of its usefullness:
$(A \cap B)^o = A^o \cap B^o$
is an easy theorem to prove.
The dual statement $\overline{A \cup B} = \overline A \cup \overline B$,
is harder to prove directly but is a quick result of the former by the above described process, a bit of DeMorgan juggling. and the self dual theorem.

To apply this to ideals, the dual notion of neighborhood is needed.
A nhood of x is any set K with x in the interior of K.
A surrounding of x is any set K with x not in the closure of K.
Ideals can can be assigned to surroundings, an unused concept.