$\mathbf{Problem}$: Define diffeomorphism $\Phi: \mathbb{R}^2 - \{0 \} \to \mathbb{R}^2 - \{0 \}$, where $\Phi(x,y) = (2x, \frac{1}{2}y)$. Define $\sim$ as $p \sim \Phi(p)$ (i.e. $p \sim q \iff p = \Phi^n(q)$ or $q = \Phi^n(p)$). Show that there exists two inequivalent points $p_1, p_2 \in \mathbb{R}^2 - \{0 \}$ such that for every open set $U, V \subset \mathbb{R}^2 - \{0 \}$ where $p_1 \in U, p_2 \in V$, there is some $n$ such that $\Phi^n(U) \cap V \neq \emptyset$.
$\mathbf{Attempt}$: Denote $p_1 = (x_1, y_1)$, and $p_2 = (x_2, y_2)$. Since any open set contains an open rectangle, I will simply use rectangles of width $dx$ and height $dy$ to represent all the open sets $U$ around $p_1$. In general, we need to have that, given $dx, dy$, my choice of $p_1$ and $p_2$ must be such that there exist some $n$ where $$(2^n x_1 - 2^{n-1} dx < x_2 < 2^n x_1 + 2^{n-1} dx) \land ((\frac{1}{2})^n y_1 - (\frac{1}{2})^{n+1} dy < y_2 < (\frac{1}{2})^n y_1 + (\frac{1}{2})^{n+1} dy)$$ Another observation is that $\Phi^n(x)$ never escapes the quadrant which $x$ is in.
However, my attempts in placing $p_1, p_2$ in the plane failed in satisfying the required conditions. It seems that no matter where I place them, I can always find a rectangle small enough around $p_1$ such that it never gets deformed enough by $\Phi^n$ to intersect $V$. Any help is appreciated!
Here is how I approached it: first I wondered a little since $\Phi$ seems to send a point to infinity if you apply it repeatedly. But of course there is an exception to this: when $p$ lies on the $y$-axis then $\Phi$ sends $p$ closer and closer to zero. So we have to take $p_1=(0,y)$ for some $y\neq 0$.
Now what about $p_2$? It probably has to lie on the $x$-axis. Why? Well look at $\Phi^{-1}=(\frac{1}{2}x,2y)$. If you apply $\Phi^{-1}$ repeatedly to such a $p_2$ this $p_2$ will get send closer and closer to zero.
Why is this is important? Well to prove that $\Phi^n(U)\cap V\neq \emptyset$ it suffices to show that there are $k,l$ such that $\Phi^k(U)\cap (\Phi^{-1})^l(V)\neq \emptyset$. Just apply $\Phi^l$ to the lefthandside to get $\Phi^{l+k}(U)\cap V\neq \emptyset$. But $\Phi$ send $p_1$ closer and closer to zero as does $\Phi^{-1}$ with $p_2$. So $U$ and $V$ have to meet eventually.
This is just an idea and you still need to fill in some details: you need to observe that $\Phi$ "stretches" a $\varepsilon$-ball in $x$ direction as does $\Phi^{-1}$ in $y$ direction.
I hope it helps anyway.