The following question arises from trying to understand Lemma 1.3(ii) of [arXiv:math/0405063](https://arxiv.org/pdf/math/0405063.pdf). I believe a particular case of the proof (and in fact I think the proof is essentially equivalent to this claim) is:



> 
> Let $G$ be a locally compact group. Let $C,D$ be cosets (*not assumed* open, closed etc.) each of which has empty interior. Then $C\cup D$ also has empty interior.
> 
> 
> 


This is not try in general topology, of course: let $C,D$ be the rational, respectively, irrationals, in $\mathbb R$. However, I cannot decide if being a coset rules out this sort of example. Is the claim true, and if so, what is a proof?



Empty interior of union of cosets?
This is false. Take the (compact abelian) group $G=(\mathbf{Z}/2\mathbf{Z})^\mathbf{N}$ and let $H$ be a dense subgroup of index 2 (there are many, since $G$ has only countably many closed subgroups of index 2 but has $2^c$ subgroups of index $2$, and clearly a subgroup of index 2 is either closed or dense). Then $G=H\cup (G\smallsetminus H)$ and both $H$ and its coset $G\smallsetminus H$ have empty interior.


