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Let $g:\mathbb{Z}\to [0,1)$ be given by $$g(z)=rz - [rz]$$ where $[y]$ is the greatest integer that is smaller or equal to $y$ and $0<r<1$ is irrational.

Is the image of $g$ dense in $[0,1)$ and is it an embedding? I think so, but I fail to formally proof it. Thanks for your help!

Jhgd
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    The image is dense if and only if $r$ is irrational. It’s easy to show that the image is finite if $r$ is rational, and there is a proof that it’s dense for irrational $r$ here. In what sense are you using the term embedding? – Brian M. Scott Nov 13 '16 at 17:38
  • I forgot to mention that $r$ should be irrational. Thanks for pointing that out and the reference. $g$ is an embedding iff $g:\mathbb{Z}\to g(\mathbb{Z})$ is a homeomorphism, i. e. it's homeomorphic on it's image. – Jhgd Nov 13 '16 at 17:54
  • Assuming that you’re talking about the usual topologies on $\Bbb Z$ and $[0,1)$, that last part is easy: $\Bbb Z$ is discrete, and a dense subset of $[0,1)$ clearly isn’t. – Brian M. Scott Nov 13 '16 at 17:56
  • Yes, it should be discrete topology on $\mathbb{Z}$ and subspace topology on $[0,1)$. So it's not an embedding? – Jhgd Nov 13 '16 at 19:42
  • That’s right: it’s easy to verify that no dense subset of $[0,1)$ has the discrete topology, so $\Bbb Z$ and $g[\Bbb Z]$ cannot be homeomorphic. (As a matter of fact, $g[\Bbb Z]$ is a countable metric space without isolated points, so it’s homeomorphic to $\Bbb Q$.) – Brian M. Scott Nov 13 '16 at 19:49
  • All right, thanks a lot for your help! – Jhgd Nov 13 '16 at 21:56

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