## Tag Archives: backward transitive

### topological transitivity

There are various types of transitivity. Let $X$ be a compact metric space and $f:X\to X$ a continuous map on $X$. There are several definitions of transitivity.

(Topological Transitivity). For every pair of nonempty open sets $U$ and $V$ in $X$, there is a positive integer $n$ such that $f^n(U)\cap V\neq\emptyset$,

(Point Transitivity). There is a point $x\in X$ such that the orbit of $x$ is dense in $X$.

Since we do not assume that $f$ is invertible, above only make sense for positive orbits. However if $f$ is a homeomorphism, the second can be subdivided into three cases:

(W-Transitivity) The whole orbit $\mathcal{O}_{\mathbb{Z}}(x)$ is dense in $X$.

(F-Transitivity) The forward orbit $\mathcal{O}_{\mathbb{N}_0}(x)$ is dense in $X$.

(B-Transitivity) The backward orbit $\mathcal{O}_{-\mathbb{N}_0}(x)$ is dense in $X$.

Example 1. $X=\{a,b\}$ with discrete topology, $f(a)=f(b)=b$. Then $(X,f)$ is PT but not TT.

Example 2. $f:\mathbb{T}\to\mathbb{T},x\mapsto 2x$. Let $X$ be the set of periodic points of $f$ with induced topology. Then $(X,f)$ is TT, but not PT.

Example 3. $X=\mathbb{Z}\cup\{\infty\}$ be the one-point compactification of $\mathbb{Z}$. Let $f(n)=n+1$ and $f(\infty)=\infty$. Then $(X,f)$ is WT, but not FT or BT.

Proposition 1. If $X$ has isolated points and the homeomorphism $f:X\to X$ is transitive, then $X$ is of a single orbit of $f$.

Proposition 2. For continuous maps

‘PT implies TT’ if $X$ has no isolated point.
‘TT implies PT’ if $X$ is separable and second category.

For good spaces (for example, a connected closed manifold) the two definitions PT and TT are equivalent. For now on we focus on the good spaces and try to distinguish the three subcategories of PT: WT, FT and BT.

Let $W^s(x)=\{y\in X: d(f^nx,f^ny)\to 0, n\to+\infty\}$ and $W^u(x)=\{y\in X: d(f^nx,f^ny)\to 0, n\to-\infty\}$. Let $WT$ be the set of points whose whole orbit is dense. Similarly we define $FT$ and $~BT$.

Proposition 3. $FT$ is $W^s$-saturated and $~BT$ is $W^u$-saturated. Note that $WT$ may not be bi-saturated.

In general all these are totally different. Take the full shift $(\Sigma_2,\sigma)$ for example (hence for all horseshoes and hyperbolic systems). If we arrange the whole word $\mathcal{W}=\bigcup_{n\ge1}\{0,1\}^n$ one by one to generate a $(i_n)_{n\ge0}$ and assign arbitrary values for $(i_n)_{n<0}$ (for example $i_n=0$ for all $n<0$), then the resulting point is a FT but not BT. Similarly there is some point that is BT but not FT.

Question 1: Is $WT=FT\cup BT?$

Answer: Yes if $X$ is a compact Baire space.

Proof. Let $f$ be a homeomorphism on $X$ and $x\in WT\neq\emptyset$. Then every open invariant set is dense in $X$ since it contains $\mathcal{O}_{\mathbb{Z}}(x)$. On the other hand we have $X=\omega(x)\cup\alpha(x)$. By closeness, one of them, say $\omega(x)$, has nonempty interior. Then $\omega(x)$ contains an open dense subset of $X$ and hence $\omega(x)=X$ by the closeness of $\omega(x)$. That is $x\in FT$.

Question 2: what about the set $FT\cap BT$?

Let $X$ be a compact Baire space. Assume that $WT\neq\emptyset$. Then a more involve argument shows that both $FT, BT$ are nonempty and automatically dense $G_\delta$ subset. In particular, $FT\cap BT$ is also dense $G_\delta$.
Question 3: what is the difference of $WT\backslash FT$?
Answer: It is of zero probability. That is, $\mu(WT\backslash FT)=0$ for each $\mu\in\mathcal{M}(f)$.
Let $\mu\in\mathcal{M}(f)$. Then for each open set $U$, $\mu(\bigcup_{k\in\mathbb{Z}}f^kU\backslash \bigcup_{k\ge0}f^kU)=0$. Let $\mathcal{U}=\{U_n\}$ be a subbasis of the topology on $X$. Then $\mu(WT\backslash FT)\le\sum_{n}\mu(\bigcup_{k\in\mathbb{Z}}f^kU_n\backslash \bigcup_{k\ge0}f^kU_n)=0$.