# Valuation topology vs modified valuation topology

Let $K$ be a field with valuation $v:K\to G\cup\{\infty\}$ where $G$ is an ordered abelian group. In section 7.62 of the book "Foundations of analysis over surreal number fields." Vol. 141. Elsevier, 1987, by Alling, Norman L., the valuation topology on $K$ is defined as the one with base $\{B(x_0,>g):g\in G,x_0\in K\}$ where $$B(x_0,>g)=\{x\in K:v(x-x_0)>g\}.$$

He shows that $K$ equipped with this topology is a Hausdorff topological field, and if $K$ is an ordered field then the order topology coincides with the valuation topology.

Until here everything seems normal, but then in another section (7.64) he defines the modified valuation topology on $K$ as the one with base $\{B(x_0,\geq g):g\in G,x_0\in K\}$ where $$B(x_0,\geq g)=\{x\in K:v(x-x_0)\geq g\}.$$

He shows that $K$ equipped with this topology is a Hausdorff topological field, and if $K$ is an ordered field then the order topology coincides with the modified valuation topology.

If I am not mistaken, the valuation topology and the modified valuation topology are always identical (when $G$ is not trivial, independently if $K$ is ordered or not) because of the inclusions: $$B(x_0,\geq 2|g|)\subset B(x_0,> g)\subset B(x_0,\geq g),$$ where $|g|=\max\{g,-g\}$.

Questions: Am I mistaken? Do these topologies always coincide? Why is Norman L. Alling treating these topologies as different ones? Am I missing something?

• Neighborhoods are not necessarily centered. – Fan Zheng Dec 3 '16 at 3:50
• Here every element of a ball is a center of the ball, so that shouldn't be an issue. I have updated the post: actually the families of balls defined there form basis for their topologies. – Chilote Dec 3 '16 at 4:18
• Now I see, but there is a bug in your inequality when $g=0$. – Fan Zheng Dec 3 '16 at 4:51
• I believe you are right. See for example, Efrat, Valuations, Orderings, and Milnor K-Theory, AMS 2006, Section 8.1, p. 75, which gives both bases for the topology. The question whether the field can be ordered or not does not seem to be relevant, as far as I can see. – user05811 Dec 3 '16 at 5:53
• But they differ for $G$ trivial (discrete vs coarse), which may be why Alling considers them both. Maybe this unifies some results. – ACL Dec 3 '16 at 11:20