6
$\begingroup$

I'm struggling to prove or disprove that the continuity of $f$ implies $f^{-1}(\bar A)\subset\overline{f^{-1}(A)}$.

$f:X\to Y$ is a map between metric spaces $(X,d),(Y,d')$ while $\bar M$ denotes the closure of $M$.

The definition of continuity I'm supposed to use for this exercise is:

$f$ is continuous$\ \Leftrightarrow\ f^{-1}(M)$ is open if $M\subset Y$ is open$\ \Leftrightarrow\ f^{-1}(M)$ is closed if $M\subset Y$ is closed.

$\endgroup$
  • $\begingroup$ This is true if $f$ is an open map. Proof: Let $x\in f^{-1}(\overline{A})$ and take $U$ an arbitrary neighborhood of $x$. Then $f(x)\in f(U)\cap \overline{A}$ so $f(U)\cap A\neq \emptyset$ and then $U\cap f^{-1}(A)\neq \emptyset$. As $U$ was arbitrary we have $x\in \overline{f^{-1}(A)}$. $\endgroup$ – yamete kudasai Oct 31 '18 at 15:24
5
$\begingroup$

Let $Y = \mathbb{R}$ with usual metric and $X = (0,1) \cup \{2\}$ with metric inherited from the standard metric on $\mathbb{R}$. let $f(x) = x$ on $(0,1)$ and $f(2) = 1$ and take $A = (0,1)$. it is then easy to see that for this particular case your statement doesn't hold

$\endgroup$
  • $\begingroup$ Nice and easy counterexample. Thanks a lot! $\endgroup$ – Sora. Jul 29 '14 at 17:35
3
$\begingroup$

Since $A\subseteq\bar{A}$ you know that $f^{-1}(A)\subseteq f^{-1}(\bar{A})$, so $$ \overline{f^{-1}(A)}\subseteq f^{-1}(\bar{A}) $$ because $f^{-1}(\bar{A})$ is closed by continuity of $f$.

Now, if your claim is true, you'd conclude that $\overline{f^{-1}(A)}= f^{-1}(\bar{A})$ for every $A\subseteq Y$. Can you?

$\endgroup$
1
$\begingroup$

Let $X=\mathbb{R}$ with the discrete metric. Then any function is continuous. Let $Y=\mathbb{R}$ with the standard metric. Define $f:X\rightarrow Y$ to be $f(1)=1$ and $f(x)=\sqrt{2}$ for all $x\neq 1$. Finally set $A=\mathbb{Q}$. Then $\overline{\mathbb{Q}}=\mathbb{R}$ and $f^{-1}(\mathbb{R})=\mathbb{R}$. But, $f^{-1}(\mathbb{Q})=\{1\}$.

$\endgroup$

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.