0

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.

Sorey
  • 1,048
  • 6
  • 18
  • In the earlier post you have $2\pi k $ (instead of $k$ )with $k$ irrational. That makes a big difference. Please tell us if you want to stick to the statement. Also, it is obvious that there is no homeomorphism of the real line onto the torus. Are you thinking of a homeomorphism INTO the torus? – Kavi Rama Murthy Feb 24 '19 at 23:33
  • Maybe my language is bad, as I’m not sure which one I mean. I’m trying to show $f$ is not a homeomorphism. – Sorey Feb 24 '19 at 23:44
  • OK. If you want to stick to the current statement then the answer below should clear your doubts. – Kavi Rama Murthy Feb 24 '19 at 23:47

1 Answers1

2

$S^1\times S^1$ 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.

Tsemo Aristide
  • 87,475