3
$\begingroup$

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!

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ Why is the topology on $\operatorname{Homeo}(\mathbb{R})$ sequential? $\endgroup$ – Eric Wofsey Oct 14 '15 at 16:09
  • $\begingroup$ @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. $\endgroup$ – Clément Guérin Oct 14 '15 at 16:11
  • $\begingroup$ Why is there a countable neighborhood base? (In any case, this doesn't matter, because your argument works equally well with nets.) $\endgroup$ – Eric Wofsey Oct 14 '15 at 16:13
  • 1
    $\begingroup$ 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. $\endgroup$ – Eric Wofsey Oct 14 '15 at 16:43
  • 1
    $\begingroup$ @EricWofsey: as far as I know, since $\mathbb{R}$ is a hemicompact metric space, $C(\mathbb{R}, \mathbb{R})$ is in fact metrizable. $\endgroup$ – Niels J. Diepeveen Oct 15 '15 at 15:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.