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Behrends pointed out \\cite{Be1} that the converse implication does not hold; that is, the BFPP does not imply completeness, in particular, there is a non-closed subset of $\\RR^2$ possessing the BFPP. He also asked \\cite{Be2} if there is even an open example in $\\RR^n$, and whether there is a 'nice' example in $\\RR$. 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