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?

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.

• Why is the topology on $\operatorname{Homeo}(\mathbb{R})$ sequential? – Eric Wofsey Oct 14 '15 at 16:09
• @EricWofsey, because the topology is Haussdorf and every point has a countable neighborhood base. But I might be wrong on this last affirmation, I should check this. – Clément Guérin Oct 14 '15 at 16:11
• Why is there a countable neighborhood base? (In any case, this doesn't matter, because your argument works equally well with nets.) – Eric Wofsey Oct 14 '15 at 16:13
• Your countable basis doesn't work: a set of the form $V_{K,U}$ need not be a union of sets of the form $V_{K_m,U_n}$ (it contains some such sets, but they will probably not cover it; think about the case where $K$ is a point). But, as I said, you don't need to prove this at all: you can just do your argument with nets instead of sequences. – Eric Wofsey Oct 14 '15 at 16:43
• @EricWofsey: as far as I know, since $\mathbb{R}$ is a hemicompact metric space, $C(\mathbb{R}, \mathbb{R})$ is in fact metrizable. – Niels J. Diepeveen Oct 15 '15 at 15:56