# Why is $f:\mathbb{R} \to S^1, f(t)=(\cos(2\pi t), \sin(2\pi t))$ not closed?

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?

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I suppose you mean $(\cos(2\pi t),\sin(2\pi t))$. – P.. Mar 18 '13 at 18:58

Consider the set $A=\left\{n+\frac1n:n\in\mathbb N, n\neq1\right\}\subseteq \mathbb R$.

It is a closed subset of $\mathbb R$ but $f(A)$ is not closed in $S^1$.

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Couple of minutes after I have posted the question a friend of mine gave me a solution, which made me think exactly this solution. thanks :) – Shay Ben Moshe Mar 18 '13 at 21:20

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 $q:\mathbb{R}\longrightarrow\mathbb{R}/\mathbb{Z}$.

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