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Let be $T_{\beta}:[0,1]\to [0,1]$ defined by $T_{\beta}(x)=\beta x \bmod 1$ where $\beta \in (1,2).$

Questions:

  1. $T_{\beta}$ is topologically transitive?

  2. What about the periodic points?

  3. $T_{\beta}$ is topologically mixing ?

The answers to this question when beta is equal to 2 comes from the fact that $ T_{\beta}$ is conjugated to two side shift.

However in this case $ \beta \in (1,2).$ So my attempt is in brute force, i.e, I ventured to say that the points $x\in \mathbb{R}\setminus \mathbb{Q}$ has dense orbit, since the orbit of $x $ is the set

$$\operatorname{Orb}(x)=\{T^n_{\beta}(x), ~~n\in\mathbb{N}\}\;,$$ and

$$T^n(x)=\beta^n(x)\bmod 1$$

that is, $$T^n(x)=\beta^nx+l,~~~l\in\mathbb{Z},~~n\in \mathbb{N}.$$ think the above set is dense, for sets of the form $$A=\{nx+m,~~ ~~~l\in\mathbb{Z},~~n\in \mathbb{N}\}$$ when $x$ is irrational. this is not quite a proof ... I'm only showing where my intuition is guiding me.

I wonder if anyone can do it more elegantly.

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Enven for the doubling map, not all irrational points have a dense orbit (actually, uncountably many points have a non-dense orbit). –  D. Thomine Jun 7 '12 at 16:11

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I think you mean $T_\beta\colon [0,1)\to [0,1)$ where $T_\beta(x) = \beta x$ mod $1$, as otherwise the map is not well-defined.

Unless I misunderstand, topologically mixing is a stronger condition than topological transitivity:

  • $T_\beta$ is topologically transitive if for every two nonempty open sets $U$ and $V$ there is some $n$ for which $f^n(U)\cap V \neq \emptyset$.
  • $T_\beta$ is topologically mixing if for every two nonempty open sets $U$ and $V$ one has $f^n(U)\cap V\neq\emptyset$ for all sufficiently large $n$.

For the maps $T_\beta$ something stronger is true: for every nonempty open set $U$, one has $T_\beta^n(U) = [0,1)$ for sufficiently large $n$. Essentially this is because if $U$ is an interval of length $L$, then $\beta^nU$ is an interval of length $\beta^nL$, which for large $n$ will always be $>1$.

For periodic points: Suppose $x\in [0,1)$ is $n$-periodic, so $T_\beta^n(x) = x$. This is equivalent to saying that $\beta^nx = x + m$ for some integer $m$, which is in turn equivalent to $$x = \frac{m}{\beta^n-1}$$ for some integer $m$. You therefore get quite a lot of periodic points.

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