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Jun 17, 2020 at 16:06 comment added tomasz Really? If you just put $d=0$, then this is clearly true, no matter how many points you have. But I guess assuming $d$ has non-negative values (in an ordered abelian group) and that $d(a,a)=0$ for all $a$, $d=0$ is the only solution.
May 10, 2010 at 12:29 vote accept Unknown
Aug 5, 2010 at 10:03
Apr 30, 2010 at 21:21 comment added LSpice I'm pretty sure that, in one of the trivia anthologised in “Lion-hunting and other mathematical pursuits” (amazon.com/Lion-Hunting-Other-Mathematical-Pursuits/dp/…), Boas assigns a name to the hapless lecturer. (I don't have my copy of the book to hand, and so can't check.)
Apr 30, 2010 at 18:53 comment added Andrew Stacey I wonder how old that story is. I certainly heard it when a graduate student in Warwick 'round about 2000 (and the person telling it claimed that the researcher was a former graduate student that they knew).
Apr 30, 2010 at 16:12 comment added Qiaochu Yuan This sounds a lot like the story of the grad student who wrote his thesis on the properties of Holder-continuous functions with $\alpha > 1$. (The only such functions are the constant functions!)
Apr 30, 2010 at 15:08 comment added Mariano Suárez-Álvarez That serves as a great cautionary tale appropriate to tell my students after I suggest that they consider two examples of everything!
Apr 30, 2010 at 14:56 history answered Tom Smith CC BY-SA 2.5