Consider an action of $\mathbb{R}$ on the torus $S^1\times S^1$. $$ h_r:\mathbb{R}\times \left( S^1\times S^1 \right) \longrightarrow S^1\times S^1 $$ $$ \left( r,\left( e^{2\pi ix},e^{2\pi iy} \right) \right) \longmapsto \left( e^{2\pi i\left( x+r \right)},e^{2\pi i\left( y+\sqrt{2}r \right)} \right) $$ For a certain point on $S^1\times S^1$, prove:The orbit formed by this action is a dense＆proper subset on $S^1\times S^1$.
Ps:I know the result is right. Here is a mapping: $$ \pi:E^2\longrightarrow S^1\times S^1 $$ $$ \left( x,y \right) \longmapsto \left( e^{2\pi ix},e^{2\pi iy} \right) $$ Obviously $\pi$ is a quotient map, it maps the points on the plane to the torus.
Take $x=y=0$, the problem turns out that under map $\pi$, the image of $\pi$ can be dense in square $I\times I$.And it equivalent to that this path doesn't go through point $(1,1)$. The question I want to ask is how to prove that $\pi \left( \text{0,}0 \right) $ does not pass through the point $(1,1)$.
We may only consider the orbit of $(1,1)$ since one obit may be obtained from another orbit by applying a diffeomorphism.
Consider a point $(1,e^{2i\pi x})$ in the orbit of $(1,1)$, we deduce that $(e^{2i\pi r},e^{2i\pi\sqrt2 r})=(1,e^{2i\pi x})$ and $r=k\in\mathbb{Z}$, $2i\pi x=2i\pi\sqrt2 k$, and $x=\sqrt2 k,k\in\mathbb{Z}$ if $x$ is $\sqrt2 y$ where $y$ is irrational, $(1,e^{2i\pi x})$ is not in the image.
Remark that the subgroup $H=\{\sqrt 2+k,k\in\mathbb{Z}\}$ of $\mathbb{R}$ is dense. we have $(e^{2i\pi (x+k)},e^{2i\sqrt2(x+k)})$ $=(e^{2i\pi x},e^{2i\pi\sqrt2 (x+k)})$, for every neighborhood $U_y$ of any $y\in\mathbb{R}$, there exists $n\sqrt2+m$ in that neighborhood, $e^{2i\pi n\sqrt2}=e^{2i\pi(n\sqrt 2+m)}$. This implies that the orbit of $(1,1)$ is dense.