Let $G$ and $H$ be locally compact totally disconnected abelian groups, and $f:G\rightarrow H$ a surjective open map. Let $Y\subseteq G$ be a discrete subgroup in the subspace topology. Is it true that the image $f(Y)$ is also discrete in the subspace topology? If so, how can one prove it?



Is the image of discrete set under an open map discrete?
No, that's false. You didn't say that $f$ is a homomorphism, but the answer is still no if we require this.


Let $G = {\bf Z} \times C\_2 \times C\_3 \times C\_5 \times \cdots$ be the product of the infinite cyclic group and the cyclic groups of all prime orders. Let $H = C\_2 \times C\_3 \times C\_5 \times \cdots$ and let $f: G \to H$ be the obvious homomorphism with kernel ${\bf Z}$. Let $Y$ be the subgroup of $G$ generated by the element $(1, 1, 1, \ldots)$. It is a discrete infinite cyclic subgroup, but its image in $H$ is not discrete (in fact, it is dense in $H$).


