The closure of all periodic homeomorphisms of circle

Question

Let $X$ be the space of all continuous maps from $S^{1}$ to $S^{1}$. $X$ is equipped with $C^{0}$ topology. Let $Y\subset X$ be the union of all finite order homeomorphisms i.e: all homeomorphisms $\phi$ with $\phi^{n}=Id$ for some natural number $n$. Assume that $g\in X$ belongs to $\bar{Y}-Y$. Some questions about such $g$:

1.Is $g$ necessarilly a homeomorphism?If yes, is it necessarilly order preserving?

2. Is it true to say that $g$ satisfies at least one of the following conditions:

I: The periodic points of $g$ is dense

II: $g$ has a dense orbit

More generally, to what extent, the elements of $\bar{Y}-Y$ are known, dynamically?

What about if we replace $S^{1}$ with another manifold and study the elements of $\bar{Y}-Y$? What about if we consider $C^{k}$ topology for $k>0$?

Note: A possible operator theoretical version for question $1$ could be the following:Let $A$ be a $C^{*}$ algebra. Let $X$ be the space of all unital morphisms and $Y\subset X$ be the union of all finite order automorphisms. Assume that $T\in \bar{Y}-Y$. Is $T$ necessarilly an automorphisms.(Actually, is it surjective?)

Answer 1

The answer to both questions 1 and 2 is false, due to the following example. Consider $(n+1)$ consecutive intervals $I_0,...,I_n$ of lengths $1,2,4,...,2^{n-1}, 2^n$. Let the map $T_n$ cyclically permute them in an affine way. In fact, if you choose $I_j=[2^j,2^{j+1}]$, then $$ T_n(x)=\begin{cases} 2x, & x \not\in I_{{n}}\\ 2^{-n}x, & x\in I_{n}. \end{cases} $$

Now, make an affine change of the coordinates so that the union $I_0\cup\dots\cup I_n=[1,2^{n+1}]$ becomes (always the same) circle $[0,1]/(0\sim 1)$. Then, the constructed maps on this circle $C^0$-converge, as $n\to\infty$, to $$ T(x)=\begin{cases} 2x, & x \in [0,1/2]\\ 0, & x\in [1/2,1]. \end{cases} $$

This map is not a homeomorphism (it collapses the interval $[1/2,1]$ to a single point 0). Also, the orbit of any point falls to the fixed point 0 in a finite number of steps, so there is only one periodic point, and there is no dense orbit.

Answer 2

The answer to Question 2 is no. In fact, there is an orientation-preserving homeomorphism $g\colon \mathbb{R}/\mathbb{Z}\to\mathbb{R}/\mathbb{Z}$ such that $g\in\overline{Y}\setminus Y$ and such that it has exactly one fixed point (consequently, it has no dense orbit and it is not periodic):

For each $t\in\mathbb{R}$, let $\tilde g_t\colon\mathbb{R}\to\mathbb{R}$ be given by $\tilde g_t(x):= x + 1/8(1-\cos(2\pi x))+t$. Notice $\tilde g_t$ is the lift of a circle homeomorphism $g_t$. Let $g=g_0$ and observe $0\in\mathbb{R}/\mathbb{Z}$ is the only fixed point of $g$.

We want to show that $g\in\overline{Y}\setminus Y$. To do this, let $\rho_t=\rho(\tilde g_t)$ denote the rotation (also called translation) number of $\tilde g_t$. Observe that $\rho_t$ depends continuously on $t$ and $$ \rho_t > \rho_0=0, \quad\forall t>0.$$

In particular, this implies that there exists a strictly decreasing sequence $(t_n)$ with $t_n\to 0$ such that $\rho(t_n)$ is a Diophantine (irrational) number. Hence, by Herman-Yoccoz linearization theorem (see for instance https://eudml.org/doc/82144), for each $n$ there exists a $C^\infty$-diffeomorphism $h_n\colon\mathbb{R}/\mathbb{Z}\to\mathbb{R}/\mathbb{Z}$ such that $h_n\circ g_{t_n}\circ h_n^{-1}=R_{\rho_{t_n}}$, the rigid rotation of angle $\rho_{t_n}$. Then one can choose a sequence of sufficiently small positive real numbers $(\epsilon_n)$ such that $t_n+\epsilon_n\in\mathbb{Q}$ and $$h_n^{-1}\circ R_{t_n+\epsilon_n}\circ h_n\to g, \quad \text{in the } C^\infty\text{-topology}.$$

Finally, since $t_n+\epsilon_n\in\mathbb{Q}$, we have that $h_n^{-1}\circ R_{t_n+\epsilon_n}\circ h_n\in Y$ and consequently, $g\in\overline{Y}$.