Do we have $A=\bigcap_{n\in\mathbb{N}}\overline{f^n(V)}$? Let $X$ be a (compact) topological space and $f\colon X\to X$ continuous. Let $A\subset X$ be compact and $V$ a neighborhood of $A$ with $f(V)\subset V$ and $A=\bigcap_{n\in\mathbb{N}}f^n(V)$.


Is then $A=\overline{A}$ and
    $$
A=\bigcap_{n\in\mathbb{N}}\overline{f^n(V)}?
$$



If $X$ is Hausdorff, A is closed since it is compact, hence $A=\overline{A}$.
 A: An example where $\overline A \neq A$: Let $X = (-1,1) \cup \{i\}$ and $V\subset X$ is open if and only if (1) $V = \emptyset$, (2) $V\subset(-1, 1)$ and is open, or (3) $V = \{i\} \cup W$, where $W \subset (-1, 1)$ is open and containing $0$. Then define $f:X\to X$ by 
$$f(x) = \begin{cases} \frac x2 & \text{if } x\in (-1, 1) \\ i &\text{if } x = i. \end{cases}$$
Then $f$ is continuous. Let $V = (-.5, .5)$. Then $V$ is open $f(V) \subset V$. But $A= \{0\}$ is not closed. Indeed $\overline A = \{0, i\}$. 
An example that the equality is not satisfied: Let $X$ be formed by two copies of $[0,1]$ with $(0,1]$ identified. That extra "$0$" I call it $0'$. Let $f: X\to X$ be defined by $f(x) = x^2$ and $f(0') = 0'$. Let $V = [0, \frac 12)$. Then $V$ is open and $f(V)\subset V$. Also
$$\bigcap_{n} f^n(V) = \{0\}=A.$$
However, $\overline{f^n(V)} = \left[0, \frac 1{2^{n+1}}\ \right] \cup \{0'\}$ and so 
$$\bigcap_n \overline{f^n(V)} \neq \bar A = A.$$
(It seems that we can glue this two spaces to construct an example so that both of your assertions are not true).
