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Isometry group of euclidean space $\Bbb{R}^n$ is displayed by E(n). We say that a subgroup G of E(n) is discrete if and only if the subspace topology (from E(n)) on G is discrete.

If X and Y are Hausdorff spaces and Y is locally compact, then continuous map $f:X\longrightarrow Y$ is called proper if and only if for each compact subset $K\subseteq Y$, the pre-image i.e $f^{-1}(K)$ is compact.

An continuous action of an arbitrary topological group G on the space X is called proper if the associated map $ G \times X\longrightarrow X\times X $ that $ (g,x)\mapsto (gx, x)$ is proper.

My question:

Let G to be discrete subgroup of E(n). Is any (continuous)action of G on $\Bbb{R}^n$ proper?

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As a counterexample, you may consider an irrational rotation on $\mathbb{R}^2$.

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  • $\begingroup$ Rotation group on $\Bbb{R}^2$ is SO(2) that is not discrete. $\endgroup$ – bigli Nov 29 '13 at 8:12
  • $\begingroup$ Just take the subgroup generated by a single irrational rotation. $\endgroup$ – Seirios Nov 29 '13 at 8:13
  • $\begingroup$ Why is action of the subgroup generated by a single irrational rotation on $\Bbb{R}^2$, proper? $\endgroup$ – bigli Nov 29 '13 at 10:57
  • $\begingroup$ It is a counterexample, so the action is not proper. $\endgroup$ – Seirios Nov 29 '13 at 12:19

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