Let $R$ be a valuation ring, with fraction field $K$ and residue field $k$. Denote by $\Gamma=K^{\times}/R^{\times}$ the valuation group (assumed nontrivial).  

The valuation $v:K^{\times}\to\Gamma$ decomposes $K^{\times}$ as a disjoint union of nonempty open subsets, indexed by $\Gamma$. Each of these is homeomorphic to $R^{\times}$, which is in turn (using the reduction map to $k$) a disjoint union of nonempty open subsets, indexed by $k^{\times}$.  

 We conclude that any basis for the topology of $K$ must have cardinality at least $\kappa:=\max(\mathrm{Card}\,\Gamma, \mathrm{Card}\,k)$. 


Question: does there exist a basis of open subets of $K$ with cardinality $\kappa$? 


Remarks:  

(1) It is true if $v$ is discrete, i.e. $\Gamma\cong\mathbb{Z}$. Proof: take a set $S\subset R$ of representatives of $k$, and a uniformizing parameter $\pi$. Let $X\subset K$ be the set of finite sums $\sum\_i s\_i\pi^{n\_i}$ ($s\_i\in S$, $n\_i\in\mathbb{Z}$). Then the balls centered on $X$ form a basis.  

(2) I am especially intereseted in the case $\kappa=\omega$. Explicitly: if $\Gamma$ and $k$ are countable, does it follow that $K$ is second-countable (or, equivalently, separable)? 



When is a valued field second-countable?
I now think that the answer is no. Let $K$ be the field of Hahn series for the group $\mathbb{Q}$ over a countable field $k$, in the variable $t$ (so $\kappa=\omega$). For any sequence $a=(a\_i)$ of elements of $k$, and any increasing sequence $\gamma=(\gamma\_i)$ of rational numbers between $0$ and $1$, we have an element $x\_{a,\gamma}=\sum\_{i\in\omega}a\_it^{\gamma\_i}$ in $K$. The distance between different such elements is less than $1$ (in the valuative sense), so if we take an open ball of radius $2$ around each we get a disjoint union of uncountably many balls.


