@Sniffnoy You could say X, Y are adjacent iff ∃z∈(X∪Y) such that every open set containing z intersects both X and Y.
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Replying to @InstanceOfClass
@InstanceOfClass That's the same as "cl(X) overlaps Y or cl(Y) overlaps X". It only uses topology. I'm confused about the motivation here.1 reply 0 retweets 1 like -
Replying to @InstanceOfClass
@InstanceOfClass By avoiding making any of them adjacent! Really dont think a topology-only, or even metric-only, approach makes sense here.1 reply 0 retweets 1 like -
Replying to @InstanceOfClass
@InstanceOfClass Sure, but then you're using more info than just the sets; you've picked edges. At which point, why bother with constraints?1 reply 0 retweets 1 like -
Replying to @InstanceOfClass
@InstanceOfClass I know, but while it's not extra info beyond the set, it is extra info in the choice of how to express timezone as a set.1 reply 0 retweets 1 like -
Replying to @InstanceOfClass
@InstanceOfClass Poles are still a problem though. Assigning the north pole to any time zone means they'll all be adjacent.1 reply 0 retweets 1 like
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Replying to @InstanceOfClass
@InstanceOfClass Oh, true. Still unwanted behavior however. I still think this problem is silly. You're on a manifold; make use of that.0 replies 0 retweets 1 likeThanks. Twitter will use this to make your timeline better. UndoUndo
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