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

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

1 Answer 1

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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 $n\geq m$.

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  • $\begingroup$ As the intervals do not overlap, they are not nested as required. $\endgroup$ Jul 13, 2019 at 22:28
  • $\begingroup$ $V_n$ is not a single interval, it is a union of intervals, one for each $m$. $\endgroup$ Jul 13, 2019 at 22:28
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    $\begingroup$ The verbal description fails. $\endgroup$ Jul 13, 2019 at 22:29

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