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Apr 13, 2017 at 12:58 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
Feb 16, 2017 at 20:39 comment added LSpice @YCor, ah, of course that's true; how embarrassing. I think I answered the original question (can the growth function be discontinuous?) positively, and satisfied 'propriety' and non-discreteness, but, as you point out, did not satisfy the additional condition that the closure of an open ball is the corresponding closed ball (although I incorrectly said that I did).
Feb 16, 2017 at 20:28 comment added YCor I actually don't know which question you claim to answer.
Feb 16, 2017 at 20:22 comment added YCor But in the additive group, the open 1-ball is closed, i.e. equal to its closure, but not equal to the "closed 1-ball".
Feb 16, 2017 at 20:16 history edited LSpice CC BY-SA 3.0
Wrong claim modified to be less ambitious, and hopefully right
Feb 16, 2017 at 20:14 comment added LSpice @YCor, oh, right, I ignored the crucial 'left-invariant' part. I guess the additive group (with its usual metric) should still work, though.
Feb 16, 2017 at 19:56 comment added YCor And clearly $\mathbf{Q}_p^*$ admits no proper compatible left-invariant ultrametric.
Feb 16, 2017 at 19:54 comment added YCor You need to specify the metric on the multiplicative group.
Feb 16, 2017 at 19:46 history answered LSpice CC BY-SA 3.0