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I'm looking for a hint:

Suppose $T:X→X$ is a dynamical system. Assume that $T$ is both topologically transitive and a contraction, i.e. $d(T(x),T(y))≤d(x,y)$ for all $x,y∈X$. Prove that T is minimal.

What I did: Let us assume that T is not minimal. Then exists T-invariant closed unempty set F which is not equal to $X$ and by topological transitivity for all pairs of open sets $U,V$ exists $n\geq 0$ such that $T^n(U)\cap V\neq\emptyset$. Here, I tried to set $U=V=X\setminus F$, but get nothing. How can I use contraction condition?

Any help is welcome. Thanks in advance.

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1 Answer 1

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If you assume $X$ to be a compact metric space, then transitivity implies the existence (in fact, denseness) of transitive points. (A transitive point for $T$ is a point $x$ whose orbit is dense in $X$.) If $T$ were not minimal, then it would also have non-transitive points. But by the contraction assumption, orbits of close points remain close forever, and this would lead to a contradiction.

Without the compactness assumption, you can proceed as follows. Let $x\in F$ and $y\in X\setminus F$ be arbitrary. Let $\varepsilon>0$ be such that $B_\varepsilon(y)$ (i.e., the $\varepsilon$-ball around $y$) is entirely in $X\setminus F$. By transitivity, there exists a point $z\in B_{\varepsilon/2}(x)$ and a time $n\geq 0$ such that $T^n(z)\in B_{\varepsilon/2}(y)$. By the contraction property of $T$, we have $d\big(T^n(z),T^n(x)\big)\leq d(z,x)<\varepsilon/2$. Using the triangular inequality, this implies $T^n(x)\in B_\varepsilon(y)\subseteq X\setminus F$, contradicting the invariance of $F$.

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    $\begingroup$ Nice answer! A little more is actually true: there is a dense $G_{\delta}$ set of transitive points. $\endgroup$
    – 3-in-441
    Jun 9, 2018 at 21:48

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