Condition for a function to be open and continuous I'm proving but get many problems
$f$ is continuous and open mapping if and only if $\overline{f^{-1}(B)}=f^{-1}(\overline{B})$
 A: In fact, we have the conditions
$$f^{-1}(\overline{B}) \supset \overline{f^{-1}(B)} \iff f\text{ continuous}$$
$$f^{-1}(\overline{B}) \subset \overline{f^{-1}(B)} \iff f\text{ open}.$$
The first equivalence is less tricky and I will leave it to you. For the second, my suggestion is to first show that 
$$f^{-1}(\overline{B}) \subset \overline{f^{-1}(B)} \quad \text{ for all }B$$ 
is equivalent to 
$$f^{-1}(T^\circ) \supset f^{-1}(T)^\circ \quad \text{ for all } T.$$
(try $T=B^c$). Then I would note that $A \subset f^{-1}(B) \iff f(A) \subset B$ for any sets $A$ and $B$. Now you can translate the condition $f^{-1}(T^\circ) \supset f^{-1}(T)^\circ$ into $T^\circ \supset f(f^{-1}(T)^\circ)$. This isn't too far from $f$ being open.
For more general versions of these, see this question I asked once.
A: $f^{-1}[\overline{B}] \supseteq \overline{f^{-1}[B]}$ is equivalent to $f$ being continuous. 
For supppose $f: X \rightarrow Y$ is continuous and $B \subseteq Y$. Then as $B \subseteq \overline{B}$ we also have $f^{-1}[B] \subseteq f^{-1}[\overline{B}]$ and the latter set is closed, as $f$ is continuous and $\overline{B}$ is closed, so $\overline{f^{-1}[B]} \subseteq \overline{f^{-1}[\overline{B}]} = f^{-1}[\overline{B}]$, and this shows one implication.
Suppose the inclusion holds for all $B$ and let $C$ be a closed subset of $Y$. Then to show continuity of $f$ we need $f^{-1}[C]$ to be closed. But by the inclusion applied to $C$, $\overline{f^{-1}[C]} \subseteq f^{-1}[\overline{C}] = f^{-1}[C] \subseteq \overline{f^{-1}[C]}$, using $C$ is closed, so $\overline{C} = C$. So we have that $f^{-1}[C]$ equals its closure, hence it's closed.
Also, $f^{-1}[\overline{B}] \subseteq \overline{f^{-1}[B]}$ is equivalent to $f$ being open. 
For suppose $f$ is open. Pick $x \in f^{-1}[\overline{B}]$, so that $f(x) \in \overline{B}$. Let $O$ be any open neighbourhood of $x$. Then $f[O]$ is an open neighbourhood of $f(x)$ so $f[O]$ intersects $B$, as $f(x) \in \overline{B}$. So there exists some $x' \in O$ such that $f(x) \in B$, so $O$ intersects $f^{-1}[B]$. As $O$ was an arbitrary neighbourhood of $x$, $x \in \overline{f^{-1}[B]}$, proving the inclusion.
Suppose the inclusion holds for all $B \subseteq Y$. We want to show $f$ is open so let $O \subseteq X$ be open. We want $f[O]$ to be open, so pick any $y \in f[O]$. Then there is some $x \in O$ with $f(x) = y$. Suppose $y$ is not an interior point of $O$, then every neighbourhood of $y$ intersects $B = Y\setminus f[O]$, so $y \in \overline{B}$. So applying the inclusion to $x \in f^{-1}[\overline{B}]$ we get that $x \in \overline{f^{-1}[B]}$, which implies that $O$ (as a neighbourhood of $x$) intersects $f^{-1}[B]$, so some $x' \in O$ exists with $f(x') \in B = Y \setminus f[O]$, but this is clearly absurd, as $f(x')$ by definition lies in $f[O]$. This contradiction shows that $y$ is an interior point of $f[O]$, so $f[O]$ is open, as required.     
There are similar characterisation for $f$ being continuous and closed as well. These use forward images: $f[\overline{A}] = \overline{f[A]}$ for all $A \subseteq X$ characterises closed and continuous maps, and again, one inclusion is equivalent to continuity and the other to closedness. I'll leave this as an extra exercise.
