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Feb 13, 2019 at 4:46 comment added G. P thanks for help me :)
Feb 12, 2019 at 23:42 vote accept G. P
Jan 29, 2019 at 8:59 comment added Vincent So in order to have $diam(B) = r_x(B)$ we need that $diam(B) = \sqrt{2}$ as well and not 1 as the OP states. Now Nick Weavers example in the comments to the OP is interesting in the context, because by exhibiting two elements $x, y$ that are $\sqrt{2}$ apart he shows that indeed we have $diam(B) \geq \sqrt{2}$.
Jan 5, 2019 at 21:49 history edited Pietro Majer CC BY-SA 4.0
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Jan 5, 2019 at 18:51 history answered Pietro Majer CC BY-SA 4.0