# Open sets having an empty intersection but the intersection of their closure is not empty

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 $$V_{n}=(0,1/n)$$)

• I think when $\bigcap_n V_n =\emptyset$ that $\bigcap_n \overline{V_n}$ will always have empty interior, so the supposition is superfluous. Jul 13, 2019 at 22:19
• You are right. I fixed it. Jul 14, 2019 at 2:26

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 $$n\geq m$$.
• $V_n$ is not a single interval, it is a union of intervals, one for each $m$. Jul 13, 2019 at 22:28