Definition: A proper nonempty open subset of $U$ of a topological space $X$ is said to be a maximal open set if any open set which contains $U$ is $X$ or $U$.
i.e.,
Let $(X,\tau)$ be a topological space and $U\subseteq X$. Then,
$U$ is a maximal open set $\iff$ ($\forall U_{1} \subset \tau$ and $U\subset U_{1}): U_{1}=X$ or $U_{1}=U$
Theorem: Let $(X,\tau)$ be a topological space and U be a maximal open set.
1) Assume that $X-U \subsetneq W$ for all $x\in X-U$ and open neighborhood $W$ of $x$. Then $(X-U)^{\mathrm{o}}=\emptyset$.
2) Assume that there exists an open set $W\in\tau$ such that $X-U=W\neq X$. Then $(X-U)^{\mathrm{o}}=X-U$.
I tried to prove the second part:
Proof: 2) We have there is an open set $W\in\tau$ such that $X-U=W $. Since $W$ is an open set, $X-U$ is also an open set. Then $(X-U)^{\mathrm{o}}=X-U$.
But honestly, I do not have an idea about the first part…
Best Answer
For the first part you can do like this. Assume $(X-U)^\circ\neq\emptyset$. It is an open set by definition. The so is $(X-U)^\circ\cup U$ and this open set is larger than $U$. Since $U$ is maximal the $(X-U)^\circ\cup U=X$. in other words $X-U=(X-U)^\circ$. $X-U$ is open and so a neighborhood for each of its points. Applying the property for $W=X-U$ you get $X-U\subsetneq X-U$ which is obviously a contradiction.