*$\omega$–almost everywhere* if $\omega(\{k\in{{\mathbb{N}}}|P(k)\})=1$. Note that a property which is not true almost everywhere is false almost everywhere. Given a sequence of sets $(E_i)_{i\in{{\mathbb{N}}}}$, the ultraproduct $E^*$ of $(E_i)$ is the quotient $(\prod_{i\in {{\mathbb{N}}}} E_i)/{{\sim}}_\omega$ where ${{\sim}}_\omega$ is the natural equivalence relation on $\prod_{i\in {{\mathbb{N}}}} E_i$ defined by equality $\omega$–almost everywhere.

Consider $K^*$ the ultraproduct of the fields $K_i$, $\Gamma^*$ the ultraproduct of the groups $\Gamma_i$, and $\Lambda^*$ the ultraproduct of the totally ordered abelian groups $\Lambda_i$. As a warmup, we prove the easy fact that the natural ring structure on $K^*$ makes it a field: if $k^*=(k_i)_{i\in{{\mathbb{N}}}}\neq 0$ in $K^*$, then for almost all $i\in{{\mathbb{N}}}$, $k_i\neq 0$, and $1/k_i$ is defined for almost every index $i$, and defines an inverse $(1/k_i)_{i\in{{\mathbb{N}}}}$ for $k^*$ in $K^*$.

Similarly, $\Gamma^*$ is a group, and $\Lambda^*$ a totally ordered abelian group (for the total order $(x_i)_{i\in{{\mathbb{N}}}}\leq (y_i)_{i\in{{\mathbb{N}}}}$ if and only if $x_i\leq y_i$ almost everywhere). Now consider the map $v^*\co K^*{\rightarrow}\Lambda^*\cup\infty$ defined by $v^*((k_i)_{i\in{{\mathbb{N}}}})=(v_i(k_i))_{i\in{{\mathbb{N}}}}$, and the map $j^*\co \Gamma^*{\rightarrow}SL_2(K^*)$ defined by $j^*((g_i)_{i\in{{\mathbb{N}}}})=(j_i(g_i))_{i\in{{\mathbb{N}}}}$. Then $v^*$ is a valuation on $K^*$, and $j^*$ a monomorphism of groups. We denote by $BT^*$ the Bruhat–Tits tree of $SL_2(K^*)$.([^2])

Now, given a field $K$ with a valuation $v\co 