
> 
> Let $G$ be a finitely generated residually free group.
> 
> 
> 


(i.e. for each $1 \neq g \in G$ there exists a homomorphism $\tau \colon G \to F$ such that $F$ is a free group, and $\tau(g) \neq 1$.)



> 
> Let $H \lneq G$ be a proper finitely generated subgroup. Must there be
>  a finite index proper subgroup $U$ of $G$ containing $H$?
> 
> 
> 


If $G$ must be LERF then the answer is positive. Henry Wilton, in a beautiful work, has proved that limit groups (fully residually free groups) are LERF.



Is a finitely generated residually free group "almost LERF"?
*(converted from the comments)* No, $F\_2\times F\_2$ is a counterexample, where $F\_2$ is free on 2 generators.


Recall that a group is LPF if the profinite closure of every f.g. subgroup of infinite index has infinite index. This fails if there is a profinitely dense f.g. subgroup.


[Remark: your property appeared in the 3-manifold literature as "$G$ has the engulfing property". It is an elementary remark (see [page 10-11 here](https://arxiv.org/abs/math/0509090), where Property LPF is introduced) that a group has Property LPF iff each of its finite index subgroups has the engulfing property.] 


Now a way to get profinitely dense f.g. subgroups in $F\_2\times F\_2$ is as follows. Consider an aperiodic (= with no nontrivial finite quotient) infinite finitely presented group $P$ on 2 generators (or with $n$ generators, but then work with $F\_n\times F\_n$), and fix an epimorphism $f:F\_2\to P$. Then the fibre product $$F\_2\times\_P F\_2=\{(x,y)\in F\_2\times F\_2:f(x)=f(y)\}$$ is profinitely dense in $F\_2\times F\_2$ (because $P$ is aperiodic) and finitely presented (because $P$ is finitely presented) and has infinite index (because $P$ is infinite).


