In a number of texts (including Cassels' "Local Fields" and Artin's "Algebraic Numbers and Algebraic Functions") I've met the definition of an absolute value on a field that's almost the same as the usual, except that instead of the triangle inequality it satisfies $$|x+y|\leq C\max(|x|,|y|)$$ for some real $C$ (which is then necessarily $\geq1$).
According to these texts such values induce a topology on the field with the open balls $U_r(x)=\{y:|x-y|<r\}$ as basis sets. Now in order for this family of sets to be a basis for the topology they generate the intersection of any two should contain a third (around any point of the intersection) and this can be reduced to showing $|x|<r\implies\exists s$ such that $|x-y|<s\implies|y|<r$. How do I show this? I tried playing with stuff like $s=\min(\frac1Cr,|x|)$, but it's not working.
I am aware that any such function is a power of a regular absolute value, so one could just decree the topology of this to be the topology of any so associated absolute value (since powers don't affect the topology), however I'm looking for an elementary proof of this proposition.
The usual proof goes like this:
$$|y| = |(y-x) + x| \leq |y-x| + |x|$$
Now if we let $s= r-|x|>0$ then if $y\in U_s(x)$ the above simplifies to
$$|y| < r$$
Which completes the proof.
The only thing that changes is that now we can no longer conclude
$$|(y-x)+x| \leq |y-x| + |x|$$
But since we have
$$|(y-x)+x|\leq C\max(|y-x|,|x|)$$
We can conclude
$$|y|=|(y-x)+x| \leq \frac{C}{2}(|y-x|+|x|) $$
And now note that if $\frac{C}{2}>1$ then this bound is greater than $|x|$, and may thus be greater than $r$. At this point I started suspecting that what we have here is simply not enough, and to make this precise, I present a counterexample of an absolute value function on $\mathbb{R}$ that satisfies the conditions in question, but does not satisfy the property you are trying to prove, define:
\begin{align} &|\cdot|: \mathbb{R}\to \mathbb{R} \\ &|x| =\begin{cases} \operatorname{sgn}(x)x & \mbox{ if } x\leq 4 \\ x+2&\mbox{ otherwise} \end{cases} \end{align}
Which you can verify satisfies
$$ |x+y| \leq 3\max(|x|,|y|) $$
Then consider $r=5$ and $x=4\in U_5(0)$
But now any open ball around $4$ contains an element of the form $4+\varepsilon$ where $\varepsilon>0$, but then $$|(4+\varepsilon)-0| > 6 \geq 5$$
So $4+\varepsilon\notin U_5(0)$, thus there is no $s$ for which the inclusion
$$ U_s(x)\subseteq U_5(0)$$
Holds. I am not sure if I missed something, but it seems like we need to add some additional constraints for the theorem to be true