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)?

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    $\begingroup$ Sure; A can be identified with the set of functions from X to the Sierpinski space, and you can give this the compact-open 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
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    $\begingroup$ 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
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    $\begingroup$ 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 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 $X$:

  1. $X$ is an exponentiable space in the category of topological spaces ($Y^X$ exists for all $Y$).
  2. The exponential $\Sigma^X$ exists.
  3. The topology of $X$ is a continuous lattice.
  4. 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.

  • $\begingroup$ 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
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    $\begingroup$ 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 Alexandrov topology

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.

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    $\begingroup$ This is only an actual metric on the non-empty compact subsets. $\endgroup$ – HJRW Apr 8 '10 at 17:20
  • $\begingroup$ For sure, you must restrict attention to non-empty sets. If you allow non-compact 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 non-compact sets may not be interesting, but that is another issue.) $\endgroup$ – Harald Hanche-Olsen 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
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    $\begingroup$ 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

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