Preliminars and notation:
Let $M$ be an $n$-dimensional compact manifold, $T\colon M\rightarrow M$ a diffeomorphism and $( x_n)_{n\in\mathbb{Z}}$ a dense orbit under $T$, ($x_n = T^n(x_0)$). Let $p\in M$ be another point and define, for $\delta>0$, $B_n(\delta) = B(p, e^{-n\delta})$.
Question: Is it true that for every $\delta>0$ the set
$A = $( $n\in\mathbb{N} | x_n \in B_n(\delta)$ )
has finite cardinality? How can it be proven?
Thank you for the answers!
This is called the shrinking target problem, and there is a reasonably large literature on it. For hyperbolic dynamical systems we can usually find quite a few pairs $x$, $p$ such that $A$ is infinite for all $\delta$. Indeed, I believe that there are results showing that in certain cases, for any point $z$ and positive real number $\delta>0$, the set of all $x$ such that $d(T^nx,z)<\exp(-n\delta)$ for infinitely many $n \geq 1$ has positive Hausdorff dimension. A good place to start would be the articles "Ergodic theory of shrinking targets" and "The shrinking target problem for matrix transformations of tori", both by Hill and Velani, but there are many results beyond this.
For illustration, here is a nice example in the case where $T$ is a smooth map of the circle which is not a diffeomorphism. I realise that this falls slightly outside the purview of your question, but it is possible to extend this argument to the case of toral diffemorphisms using the technical device of a Markov partition. (I will not attempt this here because it is very fiddly.) Let $X=\mathbb{R}/\mathbb{Z}$ be the circle, let $T \colon X \to X$ be given by $Tx = 2x \mod 1$, and let $d$ be a metric on $X$ which locally agrees with the standard metric on $\mathbb{R}$. Take $p=0 \in X$ and fix any $\delta>0$. Now, the orbit of $x$ is dense if and only if it enters every interval of the form $(k/2^n,(k+1)/2^n)$, if and only if every possible finite string of 0's and 1's occurs somewhere in the tail of its binary expansion. On the other hand, we have $d(T^nx,0)<2^{-\delta n}$ as long as the binary expansion of $x$ contains a string of zeroes starting at position $n$ and having length $\lceil \delta n \rceil$. I think that it is not difficult to see that we can construct an infinite binary expansion, and hence a point $x$, such that this condition is met for infinitely many $n$, whilst simultaneously meeting the condition that the orbit of $x$ is dense. In particular we can construct a point $x$ for which $A$ is infinite, even for all $\delta$ simultaneously if you like.
$x_n \in B(p,e^{-n\delta})$ iff $p \in B(x_n,e^{-n\delta})$. Thus we ask whether $$ \bigcap_{k=1}^\infty \bigcup_{n=k}^\infty B(x_n,e^{-n\delta}) = \varnothing $$ But that is a countable intersection of dense open sets, so (by Baire category) is NOT empty.
(I hope my quantifiers are right...)
Take $M=\mathbb{R} / \mathbb{Z}$, $T(x)=x+\alpha$ for some $\alpha$. If $\alpha$ is irrational, all orbits will be dense. Set $p=0$, then the set $A$ can be made to be infinite by choosing an $\alpha$ that can be approximated well: choose $\alpha$ from $B_1(\delta-2\cdot 10^{-k_1})$ for some $k_1$, then modify it at most by $10^{-k_1}$to make $10^{k_1} \alpha -[10^{k_1} \alpha] \in B_{10^{k_1}}(\delta-2 \cdot 10^{-k_2})$ and repeat ad infinitum.
