I am reading the book Introduction to Dynamical Systems by Brin and Stuck. Here is the definition of attracting fixed points.
Let $X$ be a locally compact metric space and $f: X \to X$ a continuous map. A fixed point $p$ of $f$ is attracting if it has a neighborhood $U$ such that $\overline{U}$ is compact, $f(\overline{U}) \subset U$, and $\cap_{n \ge 0} f^n(U) = \{p\}$.
My intuition for this definition is when we act the map $f$ on $U$ infinitely many times, it will make $U$ contract to the fixed point $p$. My question is "Is there any related to the limit $\lim_{n \to \infty}f^n(x) = p$, for any $x \in U$?" Or, is $\cap_{n \ge 0} f^n(U) = \{p\}$ equivalent to $\lim_{n \to \infty}f^n(x) = p$?
If yes, how can we show that? For example, in the case when $X = \mathbb{R}$?
We can show that $\cap_{n \ge 0} f^n(U) = \{p\}$ implies $\lim_{n \to \infty}f^n(x) = p$, for all $x \in U$. Indeed, fix $x \in U$ and let $V$ be an open neighborhood of $p$. Then, $\cap_{n \ge 0} f^n(U) \subset V$ since the intersection equals to $\{p\}$. Hence,
$$f^n(x) \in f^n(U) \subset V, \text{ for all } n \in \mathbb{N}.$$
Therefore, $\{f^n(x)\}_n$ converges to $p$.
The converse is not true according to Amadeus Maldonado's answer. We need some extra condition. For example, in the case $f: \mathbb{R} \to \mathbb{R}$, $f \in C^1$ and $p$ is a fixed point. If $|f'(p)| < 1$, then we can show that $\lim_{n \to \infty}f^n(x) = p$ and then $\cap_{n \ge 0} f^n(U) = \{p\}$.
Best Answer
It does imply that all points in $U$ converge to $p$: Given $x \in U$ and $V \subset X$ neighborhood of $p$ that we may assume is contained in $U$, we wish to find $n_0$ such that $n > n_0 \implies f^n(x) \in V$. I claim that there exists $n_0$ such that $f^{n_0}(\bar{U}) \subset V$, otherwise for all $n \in \mathbb{N}$,the compact set $ f^n(\bar{U}) \cap \bar{U}\setminus V$ is nonempty. Since $f^{n+1}(\bar{U}) \subset f^n(U) \subset f^n(\bar{U})$, for all $M>0$ and $i_1,...,i_M \in \mathbb{N}$, $$ \cap_{k=1}^M f^{i_k}(\bar{U}) \cap \bar{U}\setminus V = f^{\max\{i_1,...,i_M\}}(\bar{U}) \cap \bar{U}\setminus V \neq \emptyset $$ so these compact sets have the finite intersections property, which implies that their intersection is nonempty, but $\cap_{n \geq 1} f^n(\bar{U}) \subset \cap_{n \geq 1} f^{n-1}(U) = \{p\} \not\subset \bar{U}\setminus V$, which is a contradiction. Therefore such $n_0$ does exists and it satisfies what we wanted. Notice also that we have uniform convergence to $p$, that is, the $n_0$ does not depend on the starting $x \in U$.
The converse is not true: if you only know that there exists a set $U$ such that $\lim f^n(x) = p$ for all $x \in U$ that is not enough to conclude that $\cap_{n\geq 0} f^n(U) = \{p\}$. Take for instance $f(x) = x^3$ on the real line, $p=0$ and $U = (-1,1)$, then $f(U) = U$ but for any $x \in U$, $\lim f^n(x) = 0$.
Of course in this case you can decrease $U$ a little to obtain the desired $U$ for the definition, but this is not always the case: Take the circle $S^1 = \mathbb{R} \cup {\infty}$ and the map $f$ that keeps infinity fixed and for $x \in \mathbb{R}$, $f(x) = x+1$. Then all points converge to the fixed point $\infty$, but every neighborhood of $\infty$ has points that take a long while until they start converging.