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

Question

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?

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

P.. · Accepted Answer · 2013-03-18T20:58:28.097

10

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 $S^1$.

edited Mar 18 '13 at 20:58

answered Mar 18 '13 at 19:02

P..

14,929

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

score 4 · Answer 2 · answered Mar 18 '13 at 19:21

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

Why is $\alpha\mathbb{Z}+\mathbb{Z}$ dense in $\mathbb{R}$? – JKnecht Feb 27 '16 at 21:25 — JKnecht, Feb 27 '16 at 21:25

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

2 Answers2