# [Math] $f: \mathbb{R} \to S^1$ where $f(x) = (\cos x, \sin x)$ open and closed mapping

general-topology

Show that $f: \mathbb{R} \to S^1$ where $f(x) = (\cos x, \sin x)$ is both an open and closed mapping, or provide counter-examples if one or both are not true.

Well, my hypothesis is that they are both true, that open sets in $\mathbb{R}$ map to open sets, and closed sets in $\mathbb{R}$ map to closed sets. I could not think of any counter-examples since yesterday, unless I am missing something.

Anyway, it is a trivial case if $f(X) = S^1$ since $S^1$ is both and open and closed in itself.

I feel intuitively that an open or closed interval will map to an open or closed arc segment on the given unit circle, respectively. But I don't know how to write this in mathematical form. Or this not true in general?

This map is not closed. Consider for example $F = \{2n\pi + \frac 1n: n = 1,2,\ldots\}$. It is open though. To prove it note that every open subset of $\mathbb{R}$ is a union of open intervals. Since an open interval is mapped onto an open arc (which is open if we think about $S_1$ with arc metric) or the whole circle, the image is a union of open sets which is open.