is the function from $\mathbb{R}$ to $S^1$ open or closed?

Question

I am new to topology...so the question might be very trivial and stupid.

Is the function $f:\mathbb{R}\to S^1 $(unit circle) such that $x\mapsto (sin(x),cox(x))$ open or closed?

So first I assume that $S^1$ is equipped with the relative topology. Thus any open set in this topology is an intersection of open balls in $\mathbb{R}^2$ with $S^1$.

Then I think...if we take $(0,\pi)$ in $\mathbb{R}$, then the image of $(0,\pi)$ under $f$ should be $(0,1]\times (-1,1)$ in $S^1$, which is not open since $(0,1]$ is not?

But then my confusion is that I actually only know that $(0,1]$ is not open in $\mathbb{R}$, how can I connect this to the relative topology of $S^1$ in $\mathbb{R}^2$?

Did I make some trivial mistakes? Could you please help me? Also to prove if this function is closed or not...I have on clue...

Thanks!

Answer 1

An open subset of $\Bbb R$ is a union of open intervals and an open interval is a union of open intervals of length less than $2\pi.$ If $a<b<a+2\pi$ then the image of $(a,b)$ is $\{(\sin (a+x), \cos (a+x)): 0<x<b-a\},$ which is open in $S^1.$