I am trying to prove the following exercise from my Topology text
To follow the translation and a possible solution.
A subset $E$ of a topological space $X$ is the difference $A\setminus B$ of two closed sets $A$ and $B$ if and only if $\operatorname{cl}E\setminus E$ is closed. I point out that the closed sets $A$ and $B$ are not generally uniquely determined!
So first of all we remember that if
$U$, $V$ and $W$
are sets then
$$
U\setminus(V\setminus W)=(U\setminus V)\cup(U\cap W)
$$
so that
$$
X\setminus(\operatorname{cl} E\setminus E)=(X\setminus\operatorname{cl}{E}\,) \cup E
$$
Now I tried to prove the statement showing that
$X\setminus(\operatorname{cl}E\setminus E)$
is open so that for any
$x\in X\setminus(\operatorname{cl}E\setminus E)$
I tired to find a neighborhood
$I_x$
contained in
$X\setminus(\operatorname{cl}E\setminus E)$.
So since
$X\setminus\operatorname{cl}E\,$
is open then for any
$x\in X\setminus\operatorname{cl}E$
the set
$X\setminus\operatorname{cl}E\,$
is a neighborhood of
$x$
obviously contained in
$X\setminus(\operatorname{cl}E\setminus E)$
and analogously for any
$x\in\operatorname{int}E$
the set
$\operatorname{int}E$
is an open neighborhood of $x$ contained in
$X\setminus(\operatorname{cl}E\setminus E)$
so that the statement follows finding for any
$x\in E\cap\operatorname{bd}E$
a neighborhood $I_x$ contained in
$X\setminus(\operatorname{cl}E\setminus E)$
and so we let to do it.
So we remember now that
$$
\operatorname{bd}(U\cap V)\subseteq(\operatorname{cl}U\cap\operatorname{bd}V)\cup(\operatorname{bd}U\cap\operatorname{cl}V)
$$
so that
$$
\operatorname{bd}E=\operatorname{bd}\big(A\cap(X\setminus B)\big)\subseteq\big(\operatorname{cl}A\cap\operatorname{bd}(X\setminus B)\big)\cup\big(\operatorname{bd}A\cap\operatorname{cl}(X\setminus B)\big)
$$
Now since $A$ and $B$ are closed we observe that
$$
\big(\operatorname{cl}A\cap\operatorname{bd}(X\setminus B)\big)\cup\big(\operatorname{bd}A\cap\operatorname{cl}(X\setminus B)\big)=\\
\big(A\cap B\cap\operatorname{cl}(X\setminus B)\big)\cup\big(A\cap\operatorname{cl}(X\setminus A)\cap\operatorname{cl}(X\setminus B)\big)
$$
but
$$
E\cap\big(A\cap B\cap\operatorname{cl}(X\setminus B)\big)=\big(A\cap(X\setminus B)\big)\cap\big(A\cap B\cap\operatorname{cl}(X\setminus B)\big)=\emptyset
$$
so that we conclude that
$$
E\cap\operatorname{bd} E\subseteq E\cap\big(A\cap\operatorname{cl}(X\setminus A)\cap\operatorname{cl}(X\setminus B)\big)=\\
\big(A\cap(X\setminus B)\big)\cap\big(A\cap\operatorname{cl}(X\setminus A)\cap\operatorname{cl}(X\setminus B)\big)=\\
\big(A\cap(X\setminus B)\big)\cap\operatorname{cl}(X\setminus A)=E\cap\operatorname{cl}(X\setminus A)
$$
but unfortunately, this does not seem help so that I ask how find for any
$x\in E\cap\operatorname{bd} E$
a neighborhood
$I_x$
that is contained in
$X\setminus(\operatorname{cl}E\setminus E)$. Anyway if $\operatorname{cl}E\setminus E$ is closed then we observe that
$$
E=\operatorname{cl}E\cap E=\operatorname{cl}E\setminus(\operatorname{cl}E\setminus E)
$$
so that the reverse implication effectively holds. So how prove the direct implication? Could someone help me, please?
Best Answer
So we know that $$ E\subseteq A $$ so that if the clousure operator $ \operatorname{cl} $ preserve the inclusion then $$ \operatorname{cl}E\subseteq\operatorname{cl}A $$ but $A$ is closed and thus $$ \operatorname{cl}E\subseteq A $$ and so we can conclude that $$ X\setminus A\subseteq X\setminus\operatorname{cl}E\subseteq (X\setminus\operatorname{cl}E)\cup E=X\setminus\big(\operatorname{cl}E\setminus E) $$ Moreover, we know that $$ E=A\setminus B=A\cap(X\setminus B) $$ so that any element of $ E\cap\operatorname{bd}E $ is an element of $ X\setminus B $ which is open. So we let to prove that $$ (X\setminus A)\cup(X\setminus B) $$ is an open neighrbohood of any $ x\in E\cap\operatorname{bd} E $ that is contained in $ X\setminus\big(\operatorname{cl}E\setminus E) $.
So first of all we observe that if $A$ and $B$ are closed then $A\cap B$ is closed and so by the first De Morgan law it is not hard to conclude that $ (X\setminus A)\cup(X\setminus B) $ is open and moreover it obviously contains any element of $ E\cap\operatorname{bd}E $. So if $(X\setminus A)$ is contained in $ X\setminus(\operatorname{cl}E\setminus E) $ then the statement follows proving that $X\setminus B$ is contained in $ X\setminus(\operatorname{cl}E\setminus E) $ too. So to this purpose we observe that $$ X\setminus B=(X\setminus B)\cap X=(X\setminus B)\cap\big((X\setminus A)\cup A\big)=\big((X\setminus B)\cap(X\setminus A)\big)\cup\big((X\setminus B)\cap A))= \\ \big((X\setminus B)\cap(X\setminus A)\big)\cup E\subseteq(X\setminus A)\cup E\subseteq(X\setminus\operatorname{cl}E)\cup E=X\setminus(\operatorname{cl}E\setminus E) $$ and thus we finally conclude that $$ (X\setminus A)\cup(X\setminus B)\subseteq X\setminus(\operatorname{cl}E\setminus E) $$ as we desidred.