I have a function $f:\mathbb{R}\to S^1 \times S^1$, defined by $f(t) = (\exp(it),\exp(ikt))$, where $k$ is irrational and I want to show that it is not a homeomorphism.
There is a similar question posted here, where the author says in the comments that he shows his (very similar) function $f$ is not a homeomorphism by showing that $f(0)$ is a limit point of $f(\mathbb{Z})$, and then uses the fact that $\mathbb{Z}$ has no limit in $\mathbb{R}$, and hence $f$ can't be homeomorphic onto its image.
Can I use the same argument here? My function $f$ is slightly different (the author of the linked question has their arguments multiplied by $2\pi$, whereas I do not), and if so, how do I go about showing those limit point properties the author claims? Thank you.
$S^1\times S^1$ is compact, but not $\mathbb{R}$ so $f$ cannot be an homeomorphism.Otherwise if $g$ is the inverse of $f$, $g(S^1\times S^1)=\mathbb{R}$ this is impossible since the image of a compact by a continuous map is compact.