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Please let me refer you to:

Example 4.18. The skew line $f: \mathbb R \to S^1 \times S^1$ $$ f(t) = (e^{it}, e^{i\alpha t}). $$ If $\alpha$ is irrational then the image of $f$ is dense in $S^1 \times S^1$, so if $V$ is an open neighborhood of $f(t)$ in $S^1 \times S^1$, then $$ \overline{V \cap f(\mathbb R)} = V $$ so $V \cap f(\mathbb R) \neq f(U)$.

(Source: http://www.math.toronto.edu/mat1300/smoothmaps.4.pdf, Page $16$, example $4.18$.)

Why is the image of the skew line dense (assuming $\alpha$ is not rational)?

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This is called Kronecker winding on the torus. Have a look here: en.wikipedia.org/wiki/Irrational_winding_of_a_torus –  Giuseppe Negro Oct 5 '11 at 22:37

2 Answers 2

up vote 1 down vote accepted

Essentially because, in $S^1$, $e^{it}=e^{i(t+2\pi n)}$ for integer $n$, and the fractional part of $\alpha n$ is dense in $[0,1)$.

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Can you please explain more? –  user17182 Oct 5 '11 at 23:03
    
@user17182: which bit? –  Henry Oct 5 '11 at 23:34
    
why the result follows from what you wrote? I don't see it –  user17182 Oct 6 '11 at 2:26
    
@user17182: If you are looking close to $\left(e^{i2\pi x},e^{i2\pi y}\right)$, consider $\left(e^{i2\pi (x+n)},e^{i2\pi (\alpha x+\alpha n)}\right)$ for integer $n$. –  Henry Oct 6 '11 at 6:15

Hint: It is enough to show that the intersection with $\{1\}\times S^1$ is dense in $S^1$. Because $\alpha$ is irrational, this intersection is infinite, so by compactness of $S^1$ it contains points arbitrarily close together. Take two such points and subtract their parameter values ...

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