1
$\begingroup$

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 $A=\overline{A}$.

$\endgroup$
0
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.