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)?

7$\begingroup$ Sure; A can be identified with the set of functions from X to the Sierpinski space, and you can give this the compactopen topology. But any topology you might want to put on A depends on what you want to use it for; what application do you have in mind? $\endgroup$ – Qiaochu Yuan Apr 8 '10 at 15:08

3$\begingroup$ You might want to give some more thought to your titles. It's recommended that your title actually be a question, so people know what you're asking. $\endgroup$ – Ben Webster♦ Apr 8 '10 at 15:24

1$\begingroup$ I've changed your title to a more descriptive one. Obviously, you can still edit your question and pick a different title if you prefer. $\endgroup$ – Ben Webster♦ Apr 8 '10 at 15:26

$\begingroup$ Perhaps looking at the order topology on the lattice Powerset(X) will induce an interesting topology on the subspace of open sets? At the moment, I can't tell if this is the same as other topologies being suggested. Gerhard "Ask Me About System Design" Paseman, 2010.04.08 $\endgroup$ – Gerhard Paseman Apr 8 '10 at 15:50

$\begingroup$ Why/how is this interesting? (I'm not saying it isn't interesting; I've just never seen such considerations before.) $\endgroup$ – Kevin H. Lin Apr 9 '10 at 6:48
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 compactopen 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. 545564, 20012002.
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.

$\begingroup$ This question has a connection with my first question,so X is in general not compact Housdorff. $\endgroup$ – cao Apr 8 '10 at 15:26

1$\begingroup$ It doesn't have to be, but you get a weaker conclusion. The Fell topology is an alternative. Qiaochu's construction is completely general. $\endgroup$ – François G. Dorais♦ Apr 8 '10 at 15:31
The topology is a preorder/post/lattice (amongst other things), and there are various topologies one can put on lattices:
the Scott topology
the Lawson topology
In general domain theory brings up lots of things along this line
If $X$ is a metric space, you can use Hausdorff distance to get a metric on the closed sets.

4$\begingroup$ This is only an actual metric on the nonempty compact subsets. $\endgroup$ – HJRW Apr 8 '10 at 17:20

$\begingroup$ For sure, you must restrict attention to nonempty sets. If you allow noncompact closed sets, the Hausdorff distance between two sets may be infinite, but that ought to be easily remedied, for example by replacing the original metric by a bounded one. Does anything else go wrong? (The extension to noncompact sets may not be interesting, but that is another issue.) $\endgroup$ – Harald HancheOlsen Apr 8 '10 at 19:26

$\begingroup$ Harald, one could also simply allow distances to be infinite. In many situations this is, I think, the right thing to do. $\endgroup$ – Tom Leinster Apr 8 '10 at 19:50

1$\begingroup$ On compact sets you get the Vietoris topology, what do you get on all closed sets? Is there a topological characterization? $\endgroup$ – François G. Dorais♦ Apr 8 '10 at 22:38