In their book Topological Vector Spaces (2nd ed.) Lawrence Narici and Edward Beckenstein generalise convex sets for TVS over ultravalued field $K$ as $K$-convex sets. The definition goes as following:


Ultravalue $|\bullet|:K \to \mathbb{R}\_{+}$ can be viewed as a valuation on the field which defines an ultrametric $\rho(x,y) = |x-y|$. On the other hand there are can be ultravalues taking values in arbitrary ordered group with a minimal element $0$, but I think this approach is too general for the book of Narici. I think he assumes all valuations to be real-valued; The key properties are that $|\alpha \beta| = |\alpha||\beta|,|\alpha+\beta| \le \max( |\alpha|, |\beta|)$ and $|\alpha| = 0 \iff \alpha = 0$.


Let $V$ be TVS over ultravalued field $K$. Then the set $A \subset V$ is called absolutely $K$-convex or $K$-disc iff $\mathbb{D}\_K(0,1)A + \mathbb{D}\_K(0,1)A \subset A$, there by $\mathbb{D}\_K(0,1)$ I denote a closed unit ball centered at $0$ with respect to the ultravalue. A set $C \subset V$ is $K$-convex iff it is a translate of an absolutely K-convex set.


Then, there is an exercise 4.202.c : If $\mathrm{char}\; K \neq 2$ then a set $C \subset V$ is $K$-convex iff $\alpha x + (1 - \alpha) y \in C$ for any $x,y \in C$ and any $\alpha \in \mathbb{D}\_K(0,1)$, call this property an A-convexity for further discussion.


Personally, I treated characteristic of the field as hind and used $1/2 \in K$ in the $(\Leftarrow)$ direction of the proof. So, the $K$-convex set is always A-convex even in char 2. But now I'm wondering if the condition on characteristic of the field is just to make the exercise doable by the book's audience, as having $1/2 \in K$ felt very handy. Or is there actually an example of the A-convex set which is not $K$-convex?


I don't have much experience with ultravalued fields of char 2. So, can you suggest an example? I'm not sure where to start.


P.S.


I found a preprint ["Combinatorial properties of non-archimedean convex sets"](https://arxiv.org/abs/2109.04591) by A. Chernikov and A. Mennin. It uses slightly different system of definitions and covers group-valued ultravalues. But the definitions of convex sets are equivalent: see the triple convex combination condition. the correct statement of the theorem is indeed:



> 
> If $\mathrm{res} \; K = \frac{\{\alpha \in K : |\alpha|\le 1\}}{\{\alpha \in K : |\alpha|< 1\}} \not \cong \mathbb{F}\_2$ then a set $C \subset V$ is $K$-convex iff $\alpha x + (1 - \alpha) y \in C$ for any $x,y \in C$ and any $\alpha \in \mathbb{D}\_K(0,1)$.
> 
> 
> 


This article also provides nice counter example (see proposition 2.7), a set $\Big \{ (\alpha\_1,\alpha\_2,\alpha\_3) \in \mathbb{D}^3(0,1) : \exists i \in \{1,2,3\} \; . \; |\alpha\_i| < 1 \Big\} \subset K^3$. What is important here is the fact that the equation $1=x+y$ has no nontrivial solutions in a residue field. Actually Will Sawin also uses this in his answer: "In this field, since we must have $α=0$ or $α=1\ldots$". This actually implies that residue field is $\mathbb{F}\_2$.


1. *Narici, Lawrence; Beckenstein, Edward*, Topological vector spaces, Pure and Applied Mathematics (Boca Raton) 296. Boca Raton, FL: CRC Press (ISBN 978-1-58488-866-6/hbk). xvii, 610 p. (2011). [ZBL1219.46001](https://zbmath.org/?q=an:1219.46001).
2. *Chernikov, Artem; Mennin, Alex*, Combinatorial properties of non-archimedean convex sets (<https://arxiv.org/abs/2109.04591>).



Are there "pathological convex sets" over ultravalued fields of char 2?
You can see this already with the ultravalued field $\mathbb F\_2$, with $|1|= 1$, $|0|=0$.


In this field, since we must have $\alpha=0$ or $\alpha=1$, the $\alpha x + (1-\alpha) y$ condition is trivial — every set satisfies it.


But not every set is convex. Since every element of the field is at most $1$, a set is absolutely $K$-convex if and only if it is a vector space, so $K$-convex if and only if it is an affine space.


So there are many sets which satisfy the weaker condition but are not convex, the simplest consisting of three points in the plane.


For an example over a meatier field like $\mathbb F\_2((t))$, simply take a pathological example over $\mathbb F\_2$ and look at its inverse image under the reduction mod $t$ map $\mathbb F\_2[[t]]^n \to \mathbb F\_2^n$.


