TITLE: Open sets having an empty intersection but the intersection of their closure is not empty
QUESTION [7 upvotes]: Suppose $V_{n}$ is a decreasing sequence of (bounded) open sets in $\mathbb{R}^{m}$ with $m\geq1$. Suppose  the intersection of all $V_{n}$ is empty, and let $F$ be the intersection of the closures of $V_{n}$. Can we say that there exists $N$ such that every $x$ in $F$ belongs to the boundary of $V_{n}$, for $n\geq N$? 
(This question is suggested by setting $V_{n}=(0,1/n)$)

REPLY [5 votes]: No.  For instance, let $V_n$ be the union of a small open interval around $1/m$ for each $m>n$ and a small open interval with left endpoint $1/m$ for each $m\leq n$, the intervals being small enough to not overlap and shrinking to $0$ as $n\to\infty$.  Then $F=\{0\}\cup\{1/n:n\in\mathbb{Z}_+\}$ but $1/m$ is only in the boundary of $V_n$ if $n\geq m$.