Topology of the Malcev-Neumann group ring For a ring $R$ and a group $G$ the group ring $R[G]$ consist of maps from $G$ to $R$ with finite support.
It was shown that if the group is fully ordered them this ring can be embedded in a division ring $R[[G]]$ that generalises the previous construction by allowing infinite but well-founded support.
I would like to construct a topology on $R[[G]]$ that is compatible with the ring structure such that $R[G]$ is a dense subring. Intuitively this makes sense as a well-founded infinite set can be approximated by an infinite series of finite sets.
Consider as a fundamental neighborhood of zero the following family of sets:
$$
B_x = \{ r \in R[[G]] \mid \forall y \in \textrm{supp}(r).\, x 
\leq y \} 
$$
i.e. sets where there is a minimum possible element in the support.
Clearly $R[G]$ is a dense subset for this topology as it will intersect each set in the basis. But is this topology compatible with the ring structure?
 A: It is unclear to me whether you are working with commutative or non-commutative objects, but I will take $G$ to be Abelian and linearly ordered, and $R$ to be possibly non-commutative.
The topology you are considering is compatible with the ring structure, and is often called the valuation topology (at least in the case when $R$ is a field, so $R[[G]]$ has a standard valuation $v \ r:= \min \operatorname{supp} r$ for non-zero $r \in R[[G]]$).
But the group ring $R[G]$ is not dense in $R[[G]]$ in general. For instance, take $G$ to be the ordered additive group of rational numbers. Consider the neighborhood $r+B_1$ of the series $r$ which as a function $\mathbb{Q} \rightarrow R$ is the indicator function of the well-ordered set $X:=\{\frac{n}{n+1} \ : \ n \in \mathbb{N}\}$. In other words $r=1+t^{\frac{1}{2}}+t^{\frac{2}{3}}+ \cdot \cdot \cdot$ in the series representation with variable $t$.
Then $r+B_1$ contains no element of $R[G]$. Indeed the support of any element in the intersection would have to contain $X$.
The only group for which this couterexample (or similar ones) fails is that of the integers.
I don't know a natural ring topology which makes the inclusion $R[G] \subseteq R[[G]]$ dense.
