TITLE: Alexandrov Spaces or Topology on a finite point set - is there a distinction between open and closed sets?
QUESTION [5 upvotes]: A topology can be defined as a family of sets that is closed under finite intersections and unions. We then call these open sets.
But, if the point-set is finite, why should we call these open sets, and not closed sets?
Consider:
The union of any number of open sets is open, and the intersection of a finite number of open sets is open.
The union of a finite number of closed sets is closed, and the intersection of any number of closed sets is closed.
But if our family is finite, there is no distinction because we don't have infinite sets to union or intersect. If it is an Alexandrov space, then infinite unions of open sets are always open, so this also removes that distinction.
So, given just a family of sets which is finite, closed under unions and intersections, is there a reason to declare these as "open" rather than "closed" other than convention (and then declaring complements of members of the family the "closed" and "open" respectively)? 
I cannot talk about limit points, or any such things, so it seems like in these cases I am just calling them open, and their complements closed, as a matter of convention. Without extra information about the topology in this case, can open sets and closed sets be viewed as "opposites, but qualitatively equivalent"?

REPLY [4 votes]: You have made a correct, contentful observation here: given any topology $\tau$ on a finite set $X$, you can take $\tau' = \{ X \setminus U \mid U \in \tau\}$ and that is another topology on $X$, not necessarily homeomorphic to the first, although of course $\tau'' = \tau$.
As you say, this does not generalize to all topological spaces but does generalize to all Alexandrov spaces, which by definition are the topological spaces in which arbitrary intersections of open subsets remain open, hence equivalently arbitrary unions of closed subsets remain closed.  We find in fact that for any topological space $(X,\tau)$, $\tau'$ is also a topology on $X$ iff $\tau$ is an Alexandrov topology.  Thus your procedure gives an involution on the class of Alexandrov spaces.
This can be better understood via the equivalence between Alexandrov spaces and preordered sets.  This equivalence is discussed in the above wikipedia article; for a more succinct version, see 5.16) on this problem set (from a general topology course I am currently teaching).
It is not hard to check that on the preordered set side, this involution takes 
$(X,\prec)$ to $(X,\prec'$) where $x \prec' y \iff y \prec x$.  This operation might be called the "dual preordering".
Using this equivalence we can see for instance that if $\# X \leq 2$ we have $(X,\tau') \cong (X,\tau)$: any preordered set with at most two elements is self-dual.  However, there is a partially ordered set with 3 elements which is not self-dual: take two points which are incomparable to each other and each less than a third point.  The dual has one point which is less than each of two other points which are incomparable to each other.  (We can see the difference between a triangle and an upside down triangle!)  If you want to work with total orders, then any finite total order is self-dual, but for instance the order dual of the positive integers $\mathbb{Z}^+$ is the negative integers $\mathbb{Z}^-$, and these are not order-isomorphic.  Notice that maximal elements on the preorder side correspond to closed points on the Alexandrov space side, so the Alexandrov space associated to $\mathbb{Z}^+$ has no closed points whereas the Alexandrov space associated to $\mathbb{Z}^-$ has one.  
I do not know of any application of this involution on Alexandrov spaces, but it is a nice observation and every nice observation is at least potentially useful...