Is the interior of this closure in the union of the closures? 2 Hi everyone: Suppose $V$ is an open set, and $(F_{n})$ is an increasing sequence (so $F_{n}\subset F_{n+1}$) of closed sets, all of them in $\mathbb{R}^{k}$, $k\geq2$. If 
$V$ is included into the union of the interiors of the $F_{n}$'s,   can we say that the interior of the closure of $V$ is included into the union of the $F_{n}$'s?
The closure of $V$ is just $V$ union boundary. Boundary points are not interior. Also a set is open iff it is equal to its interior. So i think this should make the closure of the interior of $V$ just $V$. As $V$ is in the union of the sets $F_n$, then the interior of the closure should be in the union of $F_n$.
The closure of $V$ is just $V$ union boundary. Boundary points are not interior. Also a set is open iff it is equal to its interior. So i think this should make the closure of the interior of $V$ just $V$. As $V$ is in the union of the sets $F_n$, then the interior of the closure should be in the union of $F_n$.