TITLE: Why is $f:\mathbb{R} \to S^1, f(t)=(\cos(2\pi t), \sin(2\pi t))$ not closed?
QUESTION [9 upvotes]: I'm studying intro. to topology.
I have the following function: we have to topological spaces, $\mathbb{R}$ with the standard topology and $S^1$ with the subspace topology from $\mathbb{R}^2$.
I am asked to show that the map $f:\mathbb{R} \to S^1, f(t)=(\cos(2\pi t), \sin(2\pi t))$, is not closed.
It actually seems to me that is closed, I think that each closed set in $\mathbb{R}$ is mapped either to a part of the circle which is a closed line or to the whole circle.
Where am I wrong?

REPLY [4 votes]: Here is a very badly behaved counterexample, although P..'one is certainly the best to answer the question. 
Take $\alpha $ irrational. Then $\alpha\mathbb{Z}$ is closed and discrete, but $\alpha\mathbb{Z}+\mathbb{Z}$ is dense in $\mathbb{R}$. Now
$$
f(\alpha\mathbb{Z})=f(\alpha\mathbb{Z}+\mathbb{Z})
$$
is dense in $S^1$, but certainly not the whole of $S^1$.
Note: whatever example you choose, this shows that a quotient map needs not be closed. Here $q:\mathbb{R}\longrightarrow\mathbb{R}/\mathbb{Z}$.