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Am trying to prove the following :

Let ($X,T$) a uniquely ergodic topological system.We suppose that the uniquely $T-$invariant measure $\mu$ satisfies the property :$\mu(U)>0$ for every non-empty open set $U\subseteq X$.Then show under these conditions that the pair $(X,T)$ is minimal.

Let me clarify some definitions:

1)Topological System is the pair $(X,T)$ where $X $ is a compact metric space and $T:X\to X$ is a continuous transformation.

2)Minimal topological System is the pair $(X,T)$ where there does not exist a closed set $\emptyset \neq F \subseteq X$ such that $F\neq X$ and $T(F)\subseteq F$.

I tried to approach this by showing that every $x \in X$ has dense orbit but i only showed that $$\mu\biggl(\bigl\{x \in X: \overline{\{T^n(x):n \in \mathbb{N}_0\}}=X\bigr\}\biggr) =1$$

and i dont know has to use the assumption that $\mu$ is uniquely determined by being $T-$invariant.

Do you have any ideas? Thanks in advance !

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  • $\begingroup$ Could you recall for me what you mean by the measure being unique? Does it mean, it is really the only measure with those properties or are there some almost surely's floating around? $\endgroup$ Jan 2, 2019 at 14:19
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    $\begingroup$ It's the only probability measure defined on Borel sets of $X$ that is T-invariant , meaning that $\mu(T^{-1}(A))=\mu(A)$ for every Borel set $A$. Such measure must be ergodic. $\endgroup$ Jan 2, 2019 at 14:23

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Ok guys i think proved it, i used the following preposition which is equivalent for a dynamical topological pair $(X,T)$ to be uniquely ergodic :

For every continuous function $f\in C(X)$ $$\frac{1}{n}\sum_{k=0}^{n-1}f(T^k(x)) \to \int fd\mu \tag 1$$ for every $x\in X$ where $\mu$ is the uniquely determined $T-$invariant Borel probability measure.

Now i can prove the the desired result as follows:

Suppose we have a set $F$ which closed, $T-$invariant $\bigl(T(F)\subseteq F\bigr)$ and $F \neq X$ then $X\setminus F$ is non empty and open. Let $x_1 \in X\setminus F$, then there is an $\epsilon>0$ such that $$E=\hat{B}(x_1,\epsilon) = \{x\in X: d(x,x_1)\leq \epsilon\} \subseteq X\setminus F$$ From our assumptions we get that $\mu(E)>0 $ since $\hat{B}(x_1,\epsilon)$ contains the open ball $B(x_1,\epsilon).$

Now we are trying to find the crucial continuous function $f$ in order to use $(1)$.Since we have two closed and disjoint sets $F,E$ we define $$f(x) = \frac{dist(x,F)}{dist(x,F)+dist(x,E)}$$ then we know that

$\alpha)$ $f$ is continuous

$\beta)$ $0\leq f \leq 1$

$\gamma$) $f|F =0$ and $f|E=1$

So if we combine these facts and $(1)$ we get that $$\frac{1}{n}\sum_{k=0}^{n-1}f\bigl(T^k(x)\bigr) \to \int fd\mu=\int_{X\setminus F}f d\mu \geq \int_{E}fd\mu \geq \mu(E)>0 \tag 2$$ for every $x\in X$.

But now (2) tells us that $F$ must be the empty set since if we can find $x^*\in F$ since $T(F) \subseteq F$ we get that $T^k(x^*) \in F$ for every k. In other words $$\frac{1}{n}\sum_{k=0}^{n-1}f\bigl(T^k(x^*)\bigr) =0 $$ for every $n \in \mathbb{N}$ which contradicts $(2)$ since it holds for every $x\in X$.

So, for every $F$ closed which is $T-$invariant $\bigl(T(F)\subseteq F\bigr)$ it must be either $F= \emptyset$ either $F=X$.So, $(X,T)$ must be minimal.

Is this correct? thanks again !

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