for any $r\in {\mathcal{L}}_R\setminus \set{R} $; all such relations are closed.

A projective sequence $(A_n, {\varphi}_n^m)$ of finite ${\mathcal{L}}_{R}$-structures and epimorphisms is *fine* whenever its projective limit is a prespace. If $(A_{n}, {\varphi}_{n}^m)$ is a fine projective sequence in $ {\mathcal{L}}_R$ with projective limit $ \mathbb A $ and $X$ is a compact metrizable space homeomorphic to $ {{\raisebox{.1em}{$ \mathbb A \!$}\left/\raisebox{-.1em}{$R^{ \mathbb A }$}\right.}} $, we say that $(A_{n}, {\varphi}_{n}^m)$ *approximates* $X$.

Given a reflexive graph (that is, a reflexive and symmetric relation) $R$ on some set, denote by $d_R$ the distance on the graph, where $d_R(a, b)=\infty $ if $a, b$ belong to distinct connected components of the graph. Note that if $R, S$ are reflexive graphs and $\varphi $ is a function between them such that $x{\mathbin{R}}y\Rightarrow x\mathbin{S}y$ for all $x, y$, then the inequality $d_S(\varphi (x), \varphi (y))\le d_R(x, y)$ holds for every $x, y$.

We can determine whether a sequence is fine by checking that the $R$-distance of points which are not $R$-related tends to infinity. More precisely:

\[lemmafour\] Let $(A_{n}, {\varphi}_{n}^m)$ be a projective sequence of finite $ \mathcal L_R$-structures, with projective limit $ \mathbb A $. Assume that $R^{A_n}$ is reflexive and symmetric 