{T}}}}}\subset{{\mathcal {G}}}_n$ be the set of marked groups $(\Gamma,S)$ having a free action on a Bruhat–Tits tree.

Then ${{{{\mathcal {B}}}{{\mathcal {T}}}}}$ is closed in ${{\mathcal {G}}}_n$, and ${{{{\mathcal {B}}}{{\mathcal {T}}}}}$ consists in groups acting freely on $\Lambda$–trees where $\Lambda{\otimes}{{\mathbb{Q}}}$ has dimension at most $3n+1$ over ${{\mathbb{Q}}}$. In particular, ${{{{\mathcal {B}}}{{\mathcal {T}}}}}$ consists in groups acting freely on ${{\mathbb{R}}}^{3n+1}$–trees where ${{\mathbb{R}}}^{3n+1}$ has the lexicographic ordering.

[Remeslennikov]{}

[[@Remeslennikov_exist_ukrainian]]{}A limit group has a free action on an ${{\mathbb{R}}}^n$–tree.

This follows from the theorem above since a free group acts freely on a Bruhat–Tits tree.

We first prove that ${{{{\mathcal {B}}}{{\mathcal {T}}}}}$ is closed. Let $(\Gamma_i,S_i)\in{{{{\mathcal {B}}}{{\mathcal {T}}}}}$ be a sequence of marked groups converging to $(\Gamma,S)$. For each index $i$, consider a field $K_i$ and a valuation $v_i\co K_i{\rightarrow}\Lambda_i\cup\{\infty\}$ and an embedding $j_i\co \Gamma_i{\rightarrow}SL_2(K_i)$ such that $j_i(\Gamma_i)$ acts freely without inversions on the corresponding Bruhat–Tits tree $BT_i$.

Consider $\omega$ an ultrafilter on ${{\mathbb{N}}}$, [i.e. ]{} a finitely additive measure of total mass $1$ (a *mean*), defined on all subsets of ${{\mathbb{N}}}$, and with values in $\{0,1\}$, and assume that this ultrafilter is non-principal, [i.e. ]{} that the mass of finite subsets is zero. Say that a property $P(k)$ depending on $k\in{{\mathbb{N}}}$ is true 