If X is a topological space, and A consists of all of X's open sets, can we define a natural topology on A (using the topology of X)?
Of course there are many answers to your question. The interesting thing to ask is if there is a "best" or "right" answer. In many respects the "correct" topology for the lattice of open sets is the Scott topology. In case $X$ is locally compact, the Scott topology coincides with the compact-open topology of the continuous function space $C(X,\Sigma)$, where $\Sigma$ is the Sierpinski space (where we identify open sets with their characteristic functions into $\Sigma$).
There are several reasons why the Scott topology is the "right" one. One of them is that the following are equivalent for a space $X$:
- $X$ is an exponentiable space in the category of topological spaces ($Y^X$ exists for all $Y$).
- The exponential $\Sigma^X$ exists.
- The topology of $X$ is a continuous lattice.
- The lattice of open sets of $X$ equipped with the Scott topology is the exponential $\Sigma^X$.
I recommend the following paper by Martin Escardó and Reinhold Heckmann in which they explain many things related to topology of the lattice of open sets (and function spaces in general):
M.H. Escardo and R. Heckmann. Topologies on spaces of continuous functions. Topology Proceedings, volume 26, number 2, pp. 545-564, 2001-2002.
If $X$ is compact Hausdorff then the Vietoris topology (Wikipedia is lacking here, consult your standard topology textbook) on the compact (i.e. closed) subsets of $X$ implicitly defines a compact Hausdorff topology on the open subsets of $X$ via complements.