$(0,+\infty)/\Bbb{Z}$ homeomorphic to $S^1$ 
Let $a\in (0,1)$ and $Z$ acting on $(0,+\infty)$ by $n\cdot x=a^nx$ for all $n,x.$
I would like to prove that $(0,+\infty)/\Bbb{Z}$  is homeomorphic to $S^1.$

Let $x\mapsto \exp(2i\pi\log_a(x))$ be $h.$ It's continuous. We have $$h(n\cdot x)=\exp(2i\pi\log_a(a^nx))$$
$$=\exp\bigr(2i\pi(\log_a(a^n)+\log_a(x))\bigl)=\exp(2i\pi\log_a(x))\exp(2i\pi n)=\exp(2i\pi\log_a(x)).$$ So that $h$ is equivariant, then $h$ can be factorized to $h=\overline{h}\circ p$ where $p$ is the "canonical projection" and $\overline{h}:(0,+\infty)/\Bbb{Z}\to S^1$ is continuous.
I need to prove that $\overline{h}$ is bijective.
If $\overline{h}(\overline{x})=\overline{h}(\overline{y})$ then $h(x)=h(y)$ then $\log_a(x)-\log_a(y)\in\Bbb{Z}$ so that there exist $m\in \Bbb{Z}$ such that $\log(x/y)=m\log(a).$ Finally as $x=a^ny$ we have $\overline{x}=\overline{y.}$
Now as $exp: i\Bbb{R}\to S^1$ is surjective and $p$ as well it follow that $\overline{h}$ is surjective.
To conclude as $S^1$ is $T_1$ I just need to prove that $(0,+\infty)/\Bbb{Z}$ is compact, not sure how can I do that.
 A: The action is clearly free (and in fact, it is a covering space action), so each orbit is in bijection with $\mathbb{Z}$ (indeed any orbit looks like $\{\ldots,a^{-2}x,a^{-1}x,x,ax,a^2x,\ldots\}$).
The key point is that each orbit intersects the half-open interval $(a,1]$ exactly once. Indeed, note that any point in this interval leaves the interval instantly: any $x \in (a,1]$ satisfies $ax \leq a$ and $a^{-1}x > 1$. Moreover, any point not in this interval eventually reaches it. This can be seen by contradiction: if that were not the case, then some $x > 1$ would satisfy $ax \leq a$. (Because $a^nx \xrightarrow{n \to \infty} 0$.) But this implies $x \leq 1$, a contradiction.
Now the inclusion $[a,1] \to (0,\infty)$, which is clearly continuous, composes with the quotient map to yield a map $[a,1] \to (0,\infty)/\mathbb{Z}$. As this $\mathbb{Z}$-action, restricted to $[a,1]$, is simply the identification of $a$ with $1$, we get an induced map $[a,1]/\{a,1\} \to (0,\infty)/\mathbb{Z}$ which is a bijection by the observation above. The former is compact, and the latter is Hausdorff (two distinct points in $(a,1]$ may be separated by open subsets of this interval, and the orbits of such neighborhoods provide neighborhoods for the orbits), so we are done. (Assuming that you are comfortable with the fact that $S^1 \cong [a,1]/\{a,1\}$.)

This idea can be used to talk about $S^1 \subset \mathbb{C}$ directly too: by the discussion above, the map $S^1 \to (0,\infty)/\mathbb{Z}$ that sends $e^{2i\pi \theta}$ to the orbit of $1+\theta(a-1)$ is a continuous bijection from a compact space to a Hausdorff space.
