Pointwise recurrent functions on the Cantor set Let $X$ be a compact metric space and $f:X \to X$ be a continuous function. Given $x\in X$, $\mathrm{orb}(x,f)=\{x,f(x),f^2(x),...\}$. We say that $f$ is pointwise recurrent provided that, for each $x\in X$, there is an increasing sequence of natural numbers $\{n_i\}$ such that $x=\lim_{i\to\infty}f^{n_i}(x)$. Also, we say that $f$ is minimal if for any $x\in X$ we have that $\mathrm{Cl}(\mathrm{orb}(x,f))=X$.
My question:
Is there a sequence of pointwise recurrent (minimal) functions defined on the Cantor set that converge to the identity map?
Thank you.
 A: No, a sequence of pointwise recurrent minimal functions defined on the Cantor set cannot converge to the identity map.
Define the Cantor set as $X=\{0,1\}^{\mathbb N}$ with metric $d(x,y)=\sum_{i=1}^\infty 2^{-i}|x_i-y_i|.$ For any minimal $f:X\to X,$ since $(1,1,\dots)$ is in the closure of $\{f^n(0,0,\dots)\mid n\in\mathbb N\},$ there must exist some $n$ such that the first digit of $z=f^n(0,0,\dots)$ is $0$ but the first digit of $f(z)$ is $1.$ This means the distance between $z=\operatorname{id}(z)$ and $f(z)$ is at least $1/2.$
A: Define the Cantor set as $X=\{0,1\}^{\mathbb N}$ with metric $d(x,y)=\sum_{i=1}^\infty 2^{-i}|x_i-y_i|.$
There exists a sequence $\{f_n\}$ of pointwise recurrent functions defined on the Cantor set converging to the identity  map $\operatorname{id}$. A cheap way to construct it is to put $f_n=\operatorname{id}$ for each $n$. If we wish to have all $f_n$ different, we may put $f_n(x)=(x_1,\dots,x_{n-1},1-x_n,x_{n+1},x_{n+2},\dots)$ for each $x=(x_1,x_2,\dots)\in X$. 
If we wish to avoid that $f_n^k=\operatorname{id}$ for some natural $k$ then we can define $f_n$'s as follows. Let $\Bbb N=\{1\}\cup\{2,3\}\cup\{4,5,6\}\cup\{7,8,9,10\}$ be a partition of $\Bbb N$ into parts $A_i$, where $|A_i|=i$ for each $i$. For each $n$ let $\sigma_n$ be a permutation of $\Bbb N$ which is the identity on $A_i$, if $i<n$ and a cyclic shift, if $i\ge n$. For each $n$ definite the $i$-th coordinate  $f_n(x)_i=x_{\sigma_n(i)}$ for each $x=(x_1,x_2,\dots)\in X$ and each $i$. 
The map $f_n$ is poinwise recurrent, becasue $x=\lim_{k\to\infty}f^{k!}(x)$ for each $x\in X$.
