How to find all invariant sets under decimal expansion Transformation?

Question

Assume that $$T:[0,1]\to[0,1]$$

$$T(x) = 10x \quad\text{mod 1} \\[2ex] $$

How to find all set $A$ such that $T^{-1}A=A$

this question refer to ergodicity of decimal transformation

Answer 1

HINT: Suppose that $x,y\in[0,1]$, and $x$ and $y$ have decimal expansions $x=0.d_1d_2d_3\ldots$ and $y=0.e_1e_2e_3\ldots$ such that there is an $m\in\Bbb Z^+$ such that $d_n=e_n$ whenever $n\ge m$. (In other words, $x$ and $y$ have decimal expansions that are identical from some point on.) Show that if $T^{-1}[A]=A$, then $x\in A$ if and only if $y\in A$. Then define an equivalence relation $\sim$ on $[0,1]$ in such a way that $T^{-1}[A]=A$ if and only if $A$ is a union of $\sim$-equivalence classes.

Answer 2

Let $[0,1]=:I$, and call two numbers $x$, $y\in I$ equivalent, if there exist $m$, $n\in {\mathbb N}_{\geq0}$ such that $T^m x=T^n y$. (It's easy to check that this is indeed an equivalence relation.)

I claim that any $T$-invariant set $A\subset I$ is a union of equivalence classes.

Proof. Let $A$ be a nonempty $T$-invariant set, and assume $x\in A$. If $y\sim x$ then $T^mx=T^n y$ for suitable $m$ and $n$. From $x\in A=T^{-m}(A)$ it follows that $T^n y=T^mx\in A$, hence $y\in T^{-n}A= A$.$\qquad\square$

It remains to describe the equivalence relation in an intuitive way. Two numbers $x$, $y\in I$ with decimal expansions $$x=0.x_1x_2x_3\ldots,\quad y=0.y_1y_2y_3\ldots$$ are equivalent iff there exist $m$, $n\in{\mathbb N}_{\geq0}$ such that $$x_{m+k}=y_{n+k}\quad\forall k\geq0\ ,$$ i.e., iff their expansions have the same ends, up to an arbitrary shift.