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?
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$\begingroup$ Thanks for your quick replies. But I forgot one more assumption. The space $H$ is non-compact. Also as is pointed out in the previous answer, the map $f$ is a homomorphism. Does anyone know the answer? $\endgroup$– user32758Apr 2, 2013 at 14:22
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$\begingroup$ Can´t you just add an extra $\mathbb{Z}$ factor to $G$ and $H$ and then let $Y$ be generated by the element $(1,0,1,1,1,\dots)$? $\endgroup$– Ramiro de la VegaApr 2, 2013 at 14:37
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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 $H$).
answered Apr 2, 2013 at 7:38
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1$\begingroup$ Here is a simpler one: if $t\in\mathbb{R}$ is irrational, the image of $t\mathbb{Z}$ under the projection $\mathbb{R}\to\mathbb{R}/\mathbb{Z}$ is not discrete. $\endgroup$ Apr 2, 2013 at 7:52
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1$\begingroup$ @Laurent: The question asked for $G$ and $H$ to be totally disconnected. $\endgroup$ Apr 2, 2013 at 8:20
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