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A rigid analytic space $Y$ over a complete non-archimedean valued field $k$ is said to be of countable type if it has a countable (possibly finite) admissible covering by affinoids over $k$.

Suppose that $X$ is an affinoid space over such a $k$ and $U$ is an admissible open subset of $X$ with respect to the strong $G$-topology on $X$ (see Non-Archimedean Analysis by Bosch, Guntzer, Remmert section 9.1.4).

In section 4 of 'On one-dimensional separated rigid spaces'. Indag. Math. (N.S.) 6 (1995), no. 4, 439–451 by Q. Liu and M. van der Put there is given an example of such a pair $X$ and $U$ with $U$ not of countable type. However, I believe that example only works if the residue field is uncountable. My question is whether there is such an example if the residue field is itself countable. I suspect that there are examples but would be pleased if not.

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If the field admits a countable dense subfield is countable, the corresponding Berkovich space will be a metrizable topological space. Therefore, every open subset is metrizable, hence paracompact, hence each connected component will be countable at infinity.

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  • $\begingroup$ I'm struggling to process this slightly. Do you have a reference for the first sentence? $\endgroup$
    – Simon Wadsley
    Commented Jan 24, 2014 at 9:37
  • $\begingroup$ @SimonWadsley: See, eg, the beginning of Section 2 of my paper « Mesures et équidistribution sur les espaces de Berkovich », Crelle (2006), DOI:10.1515/CRELLE.2006.049 $\endgroup$
    – ACL
    Commented Jan 24, 2014 at 14:56
  • $\begingroup$ Something here is slightly misleading: an admissible open does not necessarily correspond to an open subspace of the Berkovich space. Nevertheless, it does correspond to some subspace, so the reasoning is still sound. $\endgroup$
    – Andrew Dudzik
    Commented Jan 27, 2014 at 6:15
  • $\begingroup$ @AndrewDudzik: That's true, admissible open correspond to (locally closed) analytic domains. $\endgroup$
    – ACL
    Commented Jan 27, 2014 at 15:38

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