Let $f:X \to Y$ be a continuous function between topological spaces then $f(\bar A) \subseteq \overline{f(A)}$. I know that for every $A \subset X$ the subset $f^{-1}( \overline{f(A)})$ is a closed one that containing $A$. Now how can I deduce that $\bar A \subseteq f^{-1}( \overline{f(A)})$?
Characterization of continuity in terms of closure
general-topology
Best Answer
Indeed $A \subseteq f^{-1}[f[A]] \subseteq f^{-1}[\overline{f[A]}]$, the first by definition of $f^{-1}$ and $f[A]$ the second as $f[A] \subseteq \overline{f[A]}$. And by continuity of $f$, the right hand set is closed, as the inverse image of a closed set.
Now, two ways to proceed: always $B \subseteq C$ implies $\overline{B} \subseteq \overline{C}$ and so $\overline{A} \subseteq \overline{f^{-1}[\overline{f[A]}]} = f^{-1}[\overline{f[A]}]$ (as $\overline{B}=B$ iff $B$ is closed). Or just use that the closure of $A$ is the smallest closed set that contains $A$ and $f^{-1}[\overline{f[A]}]$ is a closed subset that contains $A$, so the minimality of closure can be invoked.
Either way $\overline{A} \subseteq f^{-1}[\overline{f[A]}]$ which implies that if $x \in \overline{A}$, $f(x) \in \overline{f[A]}$ so that $f[\overline{A}] \subseteq \overline{f[A]}$, for all $A \subseteq X$ and $f$ continuous.
For the reverse see this question, which shows that this property could be used as an alternative definition of continuity.
A "pointwise" way to see the inclusion: let $x \in \overline{A}$ and let $V$ be any neighbourhood of $f(x)$, as $f$ is continuous at $x$, there is some neighbourhood $U$ of $x$ so that $f[U] \subseteq V$. Because $x \in \overline{A}$, there is some $a \in A \cap U$ and then $f(a) \in f[A] \cap f[U] \subseteq f[A] \cap V$. So $V$ intersects $f[A]$ and as $V$ was arbitrary, we conclude that indeed $f(x) \in \overline{f[A]}$.