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Suppose that $U$ is a compact neigborhood of $0$ in the additive group $\mathbb{R}$. Then it is obvious that we can find a sequence $(U_n)$ of translations of $U$, $U_n= x_n +U$, such that $U_m\cap U_n=\emptyset$ for $m\neq n$.

Is this result true for every commutative, locally compact, non-compact group?

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Let's call the group $G$. We have $(x+U) \cap (y+U) \neq \varnothing \iff x-y \in U - U$. If $U$ is a compact neighbourhood of $0$, then $V := U - U$ is also a compact neighbourhood of $0$. For every finite set $F\subset G$, the set $F + V = \bigcup_{x\in F} x + V$ is then compact, hence $F + V \subsetneqq G$.

Thus we can inductively choose $x_n \in G$, starting with an arbitrary $x_0$ such that $x_{n+1} \notin F_n + V$, where $F_n = \{ x_k : 0 \leqslant k \leqslant n\}$. It follows that then $U_n = x_n + U$ gives a sequence of disjoint translates of $U$.

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