Prove $W$ is closed iff for all $A\subset X$, $int(K\cup int(A))=int(K\cup A)$ Let $(X,\tau)$ be a topological space and $W\subset X$. Prove that if $W$ is closed, then for all $A\subset X$, $(W\cup A^\circ)^\circ=(W\cup A)^\circ$.
Here's what I have:
$A^\circ \subset A$, so $A^\circ\cup W\subset A\cup W$. Then $(A^\circ \cup W)^\circ \subset (A\cup W)^\circ$.
Now, let $p\in (A\cup W)^\circ$. Then there is $U\in \tau$ such that $p\in U\subset A\cup W$.
I want to find $V\in \tau$ such that $p\in V\subset W\cap A^\circ$.
First I noticed that $U=(U\cap W)\cup (U\cap (X\setminus W))$ and $(U\cap (X\setminus W))\in \tau$. 
I still can't manage to find the adequate $V$ to conclude that $p\in(W\cap A^\circ)^\circ$.
Any help would be appreciated.
 A: It suffices to take $V:=U$. As you commented $U=(U-W)\cup(W\cap U)$ (Note that I am writing $U-Q=U\cap (X-W)$). Since $U\subset A\cup W$, then necessarily $U-W\subset A$. Besides, since $W$ is closed, then $U-W$ is open. Therefore $U-W\subset A^o$. 
Consequently, since $U-W\subset A^o$ and $W\cap U\subset W$, we finally obtain  $U=(U-W)\cup(W\cap U)\subset A^o\cup W$.
A: Let $X,\tau$ be a topological space and let $A\subset X$.
First, if $O\subset A$ and $O$ is open, then $O\subset A^\circ$. 
Proof:
Let $O\subset A \implies O^\circ \subset A^\circ$, but since $O$ is open, then $O=O^\circ$. Then $O\subset A^\circ$.
Now, take $p\in (W\cup A)^\circ$, from the definition of the interior of a set, we know that there is $U\in \tau$ such that $p\in U\subset A\cup W$.
We are looking for $V\in \tau$ such that $p\in V\subset W\cup A^\circ$.
Take $V=U$. 
Notice that $U=(U\cap W)\cup (U\cap (X-W))=(U\cap W)\cup (U-W)$ and $(U\cap (X-W))\in \tau$. 
Now, $U-W\subset A$. Then $U-W\subset A^\circ$.
Then $U\subset (U\cap W)\cup A^\circ\subset W\cup A^\circ$.
So $p\in U\subset W\cup A^\circ\implies p\in (W\cup A^\circ)^\circ$.
