Is the action of $\mathbb Z$ on $\mathbb R$ by translation the only such action? It is well known that $\mathbb Z$ acts on $\mathbb R$ by translation. That is by $n\cdot r=n+r$. The quotient space of this action is $S^1$. 
Could someone give me an example where $\mathbb Z$ acts on $\mathbb R$ in some other (non trivial) way to give (possibly) some other quotient? I am only interested in continuous actions. 
Specifically, is there an action such that the quotient is a closed interval? 
Thank you. 

Recall that a continuous action of $\Bbb{Z}$ on $\Bbb{R}$ consists of a group morphism $\Bbb{Z} \to \operatorname{Aut}(\Bbb{R})$, where $\operatorname{Aut}(\Bbb{R})$ denotes the group of homeomorphisms of $\Bbb{R}$ into itself.
Now, $\Bbb{Z}$ is cyclic, hence any morphism $\mu:\Bbb{Z} \to \operatorname{Aut}(\Bbb{R})$ is uniquely determined by $\mu(1)$. So, if you fix any homeomorphism $f \in \operatorname{Aut}(\Bbb{R})$, you will get the action
$$n \cdot x = f^n(x)$$
where $f^n$ is the $n$-iterated map $f(f( \dots))$ (for $n < 0$ it's $(f^{-1})^{-n}$).
Now, let's see some examples:
$f(x)=x+1$ leads to the action in your question.
$f(x)=x$ leads to the trivial action.
$f(x)=-x$ leads to the action $n \cdot r = (-1)^nr$, whose quotient is $\Bbb{R}_{\ge 0}$.
