[proofplan] We show $B(x, r) = B(z, r)$ by proving mutual containment. Each direction uses the ultrametric (strong triangle) inequality to replace the center: if $z \in B(x, r)$ and $y \in B(z, r)$, then $|x - y| \leq \max(|x - z|, |z - y|) \leq r$, so $y \in B(x, r)$. The reverse inclusion follows by the same argument with $x$ and $z$ swapped. [/proofplan] [step:Show $B(z, r) \subseteq B(x, r)$ using the ultrametric inequality] Let $y \in B(z, r)$, so $|z - y| \leq r$. Since $z \in B(x, r)$, we have $|x - z| \leq r$. Write $x - y = (x - z) + (z - y)$ and apply the strong triangle inequality (the non-archimedean property of $|\cdot|$): \begin{align*} |x - y| = |(x - z) + (z - y)| \leq \max(|x - z|, |z - y|) \leq \max(r, r) = r. \end{align*} Therefore $y \in B(x, r)$, and since $y$ was arbitrary, $B(z, r) \subseteq B(x, r)$. [guided] The ultrametric inequality states that $|a + b| \leq \max(|a|, |b|)$ for all $a, b \in K$. This is strictly stronger than the usual triangle inequality $|a + b| \leq |a| + |b|$, and it is the defining property that makes the geometry of non-archimedean fields so different from archimedean ones. Let $y \in B(z, r)$, meaning $|z - y| \leq r$. We also know $|x - z| \leq r$ since $z \in B(x, r)$. We want to show $|x - y| \leq r$. Decompose $x - y = (x - z) + (z - y)$ and apply the ultrametric inequality: \begin{align*} |x - y| = |(x - z) + (z - y)| \leq \max(|x - z|, |z - y|) \leq \max(r, r) = r. \end{align*} So $y \in B(x, r)$. Since $y \in B(z, r)$ was arbitrary, $B(z, r) \subseteq B(x, r)$. Notice that this argument would fail with the ordinary triangle inequality: we would only get $|x - y| \leq 2r$, which does not show $y \in B(x, r)$. The ultrametric inequality is essential here. [/guided] [/step] [step:Show $B(x, r) \subseteq B(z, r)$ by the same argument with roles swapped] Since $z \in B(x, r)$, we have $|z - x| = |x - z| \leq r$, so $x \in B(z, r)$. Applying the argument from the previous step with the roles of $x$ and $z$ interchanged (which is valid since $x \in B(z, r)$) gives $B(x, r) \subseteq B(z, r)$. [guided] For the reverse inclusion, we need $B(x, r) \subseteq B(z, r)$, i.e., for any $y \in B(x, r)$ we want $|z - y| \leq r$. Note first that $z \in B(x, r)$ implies $|x - z| \leq r$, and by symmetry of the absolute value $|z - x| = |x - z| \leq r$. So $x \in B(z, r)$. Now let $y \in B(x, r)$, so $|x - y| \leq r$. We also have $|z - x| \leq r$. Write $z - y = (z - x) + (x - y)$ and apply the ultrametric inequality: \begin{align*} |z - y| = |(z - x) + (x - y)| \leq \max(|z - x|, |x - y|) \leq \max(r, r) = r. \end{align*} So $y \in B(z, r)$. Since $y \in B(x, r)$ was arbitrary, $B(x, r) \subseteq B(z, r)$. This is exactly the same ultrametric argument as in the first step, with $x$ and $z$ playing swapped roles. The symmetry of the argument reflects the fact that being "a center" is not a special property of any particular point — it is a universal feature of non-archimedean balls. [/guided] [/step] [step:Conclude equality of the two balls] Combining the two containments $B(z, r) \subseteq B(x, r)$ and $B(x, r) \subseteq B(z, r)$ yields $B(x, r) = B(z, r)$. This means: in a non-archimedean field, every point of a closed ball $B(x, r)$ serves equally well as a center. There is no distinguished "true" center — any point inside the ball can be used to define the same ball. [/step]