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 $A=\overline{A}$.


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).

  • $\begingroup$ What do you mean with "two copies of $[0,1]$"? $\endgroup$ – M. Meyer Sep 24 '15 at 10:19
  • $\begingroup$ $X = [0,1] \times \{0\} \cup [0,1] \times \{1\}/ \sim$ where $(x, 0) \sim (y, 1)$ if and only if $x = y$ and are nonzero. @M.Meyer $\endgroup$ – user99914 Sep 24 '15 at 10:24
  • $\begingroup$ This is not with respect to your example: In the general setting, we have $V\supseteq f(V)\supseteq f^2(V)\ldots$, hence $\overline{V}\supseteq\overline{f(V)}\supseteq\overline{f^2(V)}\ldots$, hence $\overline{V}\cap\overline{f(V)}=\overline{f(V)}=\overline{V\cap f(V)}$. And for any finite intersections this should hold too. Why is it not true for the infinite intersection then? $\endgroup$ – M. Meyer Sep 24 '15 at 10:30
  • $\begingroup$ That's a nice discussion here: math.stackexchange.com/questions/356758/… @M.Meyer $\endgroup$ – user99914 Sep 24 '15 at 10:33
  • $\begingroup$ So in order to get the equality I have to assume that X is a discrete topological space? $\endgroup$ – M. Meyer Sep 24 '15 at 10:57

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