interior of union of two sets I have been reading about some properties of interior and closure operator.
I came across the fact that For any topological space $X$ and $A$ and $B$ $\subseteq X$.It is not true in general $i(A \cup B )=i(A) \cup i(B)$
.We have the counter example as the set $A=[0,1] \cup (1,2) \subseteq \mathbb{R}$.
But I would be interseted in knowing that can we impose any condition on $A$ and $B$ in general topological space so that equality holds.Will it hold if the two sets are disjoint??

Let it be that $\overline{A}\cap B=\emptyset=A\cap\overline{B}$.
If $x\in\text{int}\left(A\cup B\right)$
and $x\in A$ then also $x\in\text{int}\left(A\cup B\right)\cap\overline{B}^{c}\subseteq A$.
Note that $\text{int}\left(A\cup B\right)\cap\overline{B}^{c}$ is
open so we are allowed to conclude that $x\in\text{int}\left(A\right)$.
If $x\in\text{int}\left(A\cup B\right)$
and $x\in B$ then likewise we find $x\in\text{int}\left(B\right)$
Proved is now that $\text{int}\left(A\cup B\right)\subseteq\text{int}A\cup\text{int}B$
and it is obvious that also $\text{int}A\cup\text{int}B\subseteq\text{int}\left(A\cup B\right)$.
