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Apr 1, 2011 at 16:58 comment added Gerhard Paseman If we view it in terms of raw cardinality of the point set, then I agree. Lately, I have had posters change/generalize their questions on me. So I am half prepared to see cardinality replaced by some measure. I still think the argument works because of the shape of the covering set (it is false for non-convex sets, and I am unsure about general rectangles) for infinitely many rectangles. Gerhard "Get Ready, Get Set, Generalize!" Paseman, 2011.04.01
Apr 1, 2011 at 16:19 comment added Juris Steprans I don't think there is anything to consider for infinitely many rectangles. This question only makes sense for finite S and in this case the set of possible $S_R$ is also finite. Any two rectangles with the same $S_R$ are equivalent for the purposes of this question.
Apr 1, 2011 at 9:08 history answered Gerhard Paseman CC BY-SA 2.5