Let $F: X \rightarrow Y$ be a map between topological spaces. Prove that the following property is equivalent to continuity of $F$:
For every subset $A \subseteq X$, $F(\overline A) \subseteq \overline {F( A)}$.
The definition of continuity is that the preimage of an open set is open.
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For one direction: Assume continuity. $\Gamma_X$ is the set of all open sets in $X$.
$F(\overline A) \subseteq \overline {F( A)}$
$\iff$ $\overline A \subseteq F^{-1} (\overline {F(A)})$
$\iff$ $X-\overline A \supseteq F^{-1}(Y-\overline{F(A)})$
$\iff$ $\cup \{Q | Q \in \Gamma_X, A \cap Q = \emptyset\} \supseteq F^{-1}(Y-\overline{F(A)}) $.
Since $F$ is continuous and $Y-\overline{F(A)} \in \Gamma_Y$, $$ F^{-1}(Y-\overline{F(A)}) \in \Gamma_X$$.
It can be proved that $$ F^{-1}(Y-\overline{F(A)})\cap Q = \emptyset$$.
Thus the last statement is true.
QUESTION:
I don't know how to prove the other direction, i.e.
If $\forall A$, $\cup \{Q | Q \in \Gamma_X, A \cap Q = \emptyset\} \supseteq F^{-1}(Y-\overline{F(A)}) $,
then $F^{-1}(Y-\overline{F(A)}) \in \Gamma_X$.
If you can solve it, please tell me how you approach the problem instead of directly giving a solution.
