Given $f:X \to Y$, prove that the following sentences are equivalent: (Continuity, topology.) I'm having a hard time trying to prove the following theorem:
Given $f:X \to Y$, prove that the following sentences are equivalent:
1) $f$ is continuous
2)$\overline {f^{-1}(B)}\subset f^{-1}\overline{(B)} $ for all $B\subset Y$
3)$f^{-1}(\operatorname{Int}(B))\subset\operatorname{Int}(f^{-1}(B))$ for all $B\subset Y$
4)$\partial(f^{-1}(B))\subset f^{-1}(\partial B) $
I was able to do 1$\implies$2 and 2 $\implies$3, but I'm having trouble with the remaining ones, since for 3$\implies$4 only thing I get is:
$\partial(f^{-1}(B))\subset  \overline{f^{-1}(B)}- f^{-1}(\operatorname{Int}(B))$
 A: $(1)\Rightarrow(2)$: Suppose that $f$ is continuous. Let $B\subseteq Y$.
Since $B\subseteq\bar{B}$, we have $f^{-1}(B)\subseteq f^{-1}(\bar{B})$.
As $f$ is continuous and $\bar{B}$ is closed, $f^{-1}(\bar{B})$
is also closed. Finally, recall $\overline{f^{-1}(B)}$ is the smallest
closed set containing $f^{-1}(B)$, so $\overline{f^{-1}(B)}\subseteq f^{-1}(\bar{B})$.
$(2)\Rightarrow(3)$: Given $B\subseteq Y$, we denote its complement
and its interior by $B^{c}$ and $B^{o}$ respectively. Recall that
closure and interior has duality: $B^{o}=B^{c-c}$ and $B^{-}=B^{coc}$.
Let $B\subseteq Y$ be given. Then 
\begin{eqnarray*}
\left[f^{-1}(B)\right]^{o} & = & \left[f^{-1}(B)\right]^{c-c}\\
 & = & \left[f^{-1}(B^{c})\right]^{-c}\\
 & \supseteq & \left[f^{-1}(B^{c-})\right]^{c}\\
 & = & f^{-1}(B^{c-c})\\
 & = & f^{-1}(B^{o}).
\end{eqnarray*}
$(3)\Rightarrow(2)$: Let $B\subseteq Y$, then 
\begin{eqnarray*}
\left[f^{-1}(B)\right]^{-} & = & \left[f^{-1}(B)\right]^{coc}\\
 & = & \left[f^{-1}(B^{c})\right]^{oc}\\
 & \subseteq & \left[f^{-1}(B^{co})\right]^{c}\\
 & = & f^{-1}(B^{coc)}\\
 & = & f^{-1}(B^{-}).
\end{eqnarray*}
It follows that $(2)\Longleftrightarrow(3)$.
$(3)\Rightarrow(4)$: Recall that if $A\subseteq X$, $\partial A=A^{-}\setminus A^{o}$.
Let $B\subseteq Y$. By $(2)$ and $(3)$, we have $\left[f^{-1}(B)\right]^{-}\subseteq f^{-1}(B^{-})$
and $\left[f^{-1}(B)\right]^{o}\supseteq f^{-1}(B^{o})$. Therefore,
\begin{eqnarray*}
\partial\left[f^{-1}(B)\right] & = & \left[f^{-1}(B)\right]^{-}\setminus\left[f^{-1}(B)\right]^{o}\\
 & \subseteq & f^{-1}(B^{-})\setminus f^{-1}(B^{o})\\
 & = & f^{-1}(B^{-}\setminus B^{o})\\
 & = & f^{-1}(\partial B).
\end{eqnarray*}
$(4)\Rightarrow(1)$: Given $A\subseteq X$. If $A$ is open, then
$\partial A=A^{-}\setminus A^{o}=A^{-}\setminus A$. Conversely, if
$\partial A=A^{-}\setminus A$, then $A=A^{-}\setminus\partial A=A^{-}\setminus\left(A^{-}\setminus A^{o}\right)=A^{o}$
and hence $A$ is open. That is, $A$ is open iff $\partial A=A^{-}\setminus A$.
Let $B\subseteq Y$ be open, then $\partial B=B^{-}\setminus B.$
Now, we have:
\begin{eqnarray*}
f^{-1}(B^{-})\setminus f^{-1}(B) & = & f^{-1}(B^{-}\setminus B)\\
 & = & f^{-1}(\partial B)\\
 & \supseteq & \partial\left[f^{-1}(B)\right]\\
 & = & \left[f^{-1}(B)\right]^{-}\setminus\left[f^{-1}(B)\right]^{o}.
\end{eqnarray*}
We go to show that $f^{-1}(B)\subseteq\left[f^{-1}(B)\right]^{o}$.
Prove by contradiction. Suppose the contrary that there exists $x\in f^{-1}(B)\setminus\left[f^{-1}(B)\right]^{o}$.
Since $f^{-1}(B)\subseteq\left[f^{-1}(B)\right]^-$, we have $x\in\left[f^{-1}(B)\right]^{-}$.
It follows that $x\in\left[f^{-1}(B)\right]^{-}\setminus\left[f^{-1}(B)\right]^{o}\Rightarrow x\in f^{-1}(B^{-})\setminus f^{-1}(B)$,
which is a contradiction because $x\in f^{-1}(B)$. The reverse inclusion
$f^{-1}(B)\supseteq\left[f^{-1}(B)\right]^{o}$ is automatic. Therefore
$f^{-1}(B)=\left[f^{-1}(B)\right]^{o}$. That is $f^{-1}(B)$ is open
and hence $f$ is continuous.
A: $3\Longrightarrow 4$ follows from
$$
\partial f^{-1}(B)\cup Int(f^{-1}(B))=f^{-1}(B)=f^{-1}(\partial B\cup Int(B))=f^{-1}(\partial B)\cup f^{-1}(Int(B))
$$
and the fact that $\partial f^{-1}(B)\cap Int(f^{-1}(B))=f^{-1}(\partial B)\cap f^{-1}(Int(B))=\emptyset.$
For $4\Longrightarrow 1$, let $U\subseteq Y$ be open and note that $U\cap \partial U=\emptyset,$ implying that for any $x\in f^{-1}(U)$, $x\not\in f^{-1}(\partial U)$. However, then $x\not \in \partial f^{-1}(U)$, and thus, $x\in Int(f^{-1}(U))$. We conclude that $f^{-1}(U)=Int(f^{-1}(U))$, implying that $f$ is continuous.
