What is the closure of $\mathbb{Q}$ in Homeo$(\mathbb{R})$? I read in a paper that if I consider the group of rationals, $\mathbb{Q}$ acting on the real line, $\mathbb{R}$ by addition then taking the closure of $\mathbb{Q}$ in Homeo$(\mathbb{R})$  gives a copy of $\mathbb{R}$ acting on itself by translation. Here Homeo$(\mathbb{R})$ is the group of all homeomorphisms of $\mathbb{R}$ with itself and is given the compact open topology. I have two questions about this -
1. How do I prove closure of $\mathbb{Q}$ in Homeo$(\mathbb{R})$ is $\mathbb{R}$?
My guess was to prove that the subspace topology on $\mathbb{R} \subseteq$  Homeo$(\mathbb{R})$ is the same as the usual topology, in which case $\overline{\mathbb{Q}} = \mathbb{R}$. But I couldn't do so. Is this guess wrong? If not how do I prove it?
2. I want to know if there is a way to characterize all homeomorphisms of $\mathbb{R}$ with itself.
If it is just continuous linear maps of $\mathbb{R}$ with itself then they are of the form $r \mapsto \lambda r$ for some $\lambda \in \mathbb{R}$ and if I need linear homeomorphisms then I should just take $\lambda \neq 0$. Is that right? 
But here I have all homeomorphisms. So how should I proceed?
Thanks in advance!
 A: For 1, as you guessed, you just need to show that $(\mathbb{R},+)$ included in $(Homeo(\mathbb{R}),\circ)$ is a closed subgroup. The inclusion being given by $x_0\mapsto (x\mapsto x+x_0)$.
To do so, you can show that $\mathbb{R}$ is closed using sequences.
Now the topology (I assume it is uniform convergence on compacts) on $Homeo(\mathbb{R})$ imposes that if a sequence of homeomorphism  $f_n$ converges toward $f$ then for all $x$, $f_n(x)$ converges toward $f(x)$. 
Assume that a sequence $(f_n)\subseteq \mathbb{R}$ converges toward $f\in Homeo(\mathbb{R})$. Let $f_n$ be associated to $x_n$, it means that for all $x$, $f_n(x)$ converges toward $f(x)$ hence $x+x_n\rightarrow f(x)$. In particular $(x_n)$ converges toward $f(x)-x$ which is then independant of $x$, we can call it $x_{\infty}$. One then sees that $f(x)=x+x_{\infty}$ for all $x$ so that $f\in\mathbb{R}$ is a translation. 
Finally $\mathbb{R}$ is closed in $Homeo(\mathbb{R})$ since it is obvious that $\bar{\mathbb{Q}}\supseteq\mathbb{R}$ we finally get that $\bar{\mathbb{Q}}=\mathbb{R}$.
For 2, apart from the fact that they should be strictly increasing or decreasing going to infinity on both side, I don't think one can say something else. Considering that any strictly increasing pieceweise affine function from $\mathbb{R}$ to $\mathbb{R}$ is an homeomorphism, we already see that they can be very wild.
Edit : following Eric Wofsey's remark.
The compact open topology comes from $C(\mathbb{R},\mathbb{R})$ the continuous functions. A base of neighborhood for this :
$$V_{K,U}:=\{f\in C(\mathbb{R},\mathbb{R})\mid f(x)\in U\text{ whenever }  x\in K\}$$
Now we see that $V_{K,U}\subseteq V_{K',U'}$ if $K'\subseteq K$ and $U\subseteq U'$. Now if :
$$(U_n)\text{ is a coutable neighborhood base for } \mathbb{R}$$
$$K_m:=[-m,m]$$
THen we see that $(V_{K_m,U_n})$ is a countable neighborhood base for $C(\mathbb{R},\mathbb{R})$. 
Now it follows that $Homeo(\mathbb{R})$ has a countable neighborhood base, whence it is sequential.
