TITLE: Does the set of open sets in a topological space have a topology itself?
QUESTION [8 upvotes]: 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)?

REPLY [15 votes]: 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.