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Nov 3, 2010 at 16:24 answer added Andrej Bauer timeline score: 3
Nov 3, 2010 at 16:09 comment added Andrej Bauer You should change "which is not compact" to "which is not necessarily compact", just to emphasize you don't assume compactness. The condition "which is not compact" just makes your question ugly.
Nov 3, 2010 at 15:40 answer added HenrikRüping timeline score: 3
Oct 28, 2010 at 21:55 comment added François G. Dorais The lattice of compact-open sets is also distributive.
Oct 28, 2010 at 20:09 comment added HenrikRüping to make this work one hass of course to remove the smallest element, i.e. the emptyset.
Oct 28, 2010 at 19:49 comment added HenrikRüping One idea might be: Call a decreasing sequence $K_0\supset K_1\supset\ldots$ less or equal to another decreasing sequence $L_0\supset L_1\supset \ldots$ , iff $\forall n\exists i(n) : L_{i(n)}\subset K_n$. Call two sequences equivalent, iff the are less or equal to each other. Then the equivalence classes are ordered and I hope, that the minimal classes correspond to the points of $X$.
Oct 28, 2010 at 17:35 history edited Martin Brandenburg CC BY-SA 2.5
edited body
Oct 28, 2010 at 17:18 comment added Mariano Suárez-Álvarez I don't understand the definition of $P$: what is $U$?
Oct 28, 2010 at 17:15 history edited Martin Brandenburg CC BY-SA 2.5
added 10 characters in body; edited title
Oct 28, 2010 at 16:55 history asked Martin Brandenburg CC BY-SA 2.5