Is the closure of the set of all irrational rotation maps on $S^1$ dense in $Homeo(S^1)$?

Question

I study about rotation maps on circle, and I have a question.

Let $Homeo(S^1)$ be the set of all circle homeomorphisms with sup-metric $d(f,g)= \sup \{ d(f(x),g(x)| x \in S^1 \}$, and rotation map $f_\alpha: S^1 \rightarrow S^1$ is given by $f_\alpha (x) = x + \alpha \ mod \ 1$.

For rational rotation map $f_\alpha$, we can construct a sequence of irrational rotation maps $(f_{\alpha_n})$ converging to $f_\alpha$.

Then, at least, all rotation maps is in $\overline{\{\text{rotation maps} \ f_\alpha| \alpha \in \mathbb{Q}^c \}}$.

Is it possible that $\overline{\{\text{rotation maps} \ f_\alpha| \alpha \in \mathbb{Q}^c \}} = Homeo(S^1)$?

If it is true, could you give me a hint for this question?

Thank you in advance!

Answer 1

Let $R=\{f_\alpha:\alpha\in\Bbb R\setminus Q\}$. $\langle\operatorname{Homeo}(S^1),d\rangle$ is a metric space, so $f\in\operatorname{cl}R$ if and only if there is a sequence $\langle\alpha_n:n\in\Bbb N\rangle$ in $\Bbb R\setminus\Bbb Q$ such that $\langle f_{\alpha_n}:n\in\Bbb N\rangle$ converges to $f$. Suppose that $f(0)=\alpha$. Then $\langle f_{\alpha_n}(0):n\in\Bbb N\rangle=\langle\alpha_n:n\in\Bbb N\rangle$ converges to $\alpha$ in $S^1$. But then $\langle f_{\alpha_n}:n\in\Bbb N\rangle\to f_\alpha$, so $f=f_\alpha$. Clearly, however, there are autohomeomorphisms of $S^1$ that are not rotations, so $\operatorname{cl}R\ne\operatorname{Homeo}(S^1)$.

Answer 2

You seem to be thinking of the circle as the interval $[0,1]$ with endpoints identified, but it is more helpful to think of it as the unit circle in $\mathbb{C}$ or equivalently the real line $\mathbb{R}$ with a point at infinity added. In the latter model, the fractional linear transformations also known as Mobius transformations $x\mapsto \frac{ax+b}{cx+d}$ where $ad-bc\not=0$ give many examples of selfmaps of the circle that are not anywhere near the irrational rotations. The Mobius transformations are homeomorphisms of the circle.

Answer 3

To give an explicit counterexample, consider the map $m:[0,1]\rightarrow[0,1]$ given by $m(x)=x^2$, this map satisfies $m(0)=0$, and $m(1)=1$ and is monotone increasing (not strictly so). The map is continuous with continuous inverse, hence it is a homeomorphism between $[0,1]$ and itself (an autohomeomorphism). Applying this map to $S^1$ will move points along the circle in a non-uniform way, for instance $0=1$ is stationary, but $m(\frac{1}{2})=\frac{1}{4}$.