Open sets having an empty intersection but the intersection of their closure is not empty 2 Consider a decreasing sequence of bounded open sets $V_{n}$ in $\mathbb{R}^{m}$ and $m\geq1$. Suppose $\cap V_{n}=\emptyset$ and $F:=\cap \overline{V_{n}}$ is connected. Can we say there is $N$ such that each $x\in F$ belongs to the boundary of $V_{n}$, for all $n\geq N$?
 A: Again (see previous question) the answer is no, though my example uses $m=2$. It is easily generalizable for higher dimensions, but maybe the answer is different for $m=1$.
Consider the plane with polar coordinates $(r,\phi)$. Define $\forall n \ge 1$
$$V_n=\{(r,\phi) \in \mathbb R_{\ge 0}\times [0,2\pi): 1-\frac1n < r < 1 + \frac1n\} - \{1\}\times[0,2\pi-\frac1n] .$$

$V_n$ is an open annulus, where in the middle a closed circle arc has been removed (sorry for the mspaint-art). That a point $(r,\phi) \in V_n$ is in the interior of $V_n$ is trivial for $r \neq 1$, as the removed arc is 'far away'. $(1,\phi) \in V_n$ means $\phi \in (2\pi-\frac1n,2\pi)$, so again a small open circle around $(1,\phi)$ can be found that doesn't contain the removed arc. That means $V_n$ is open.
We have $V_{n+1} \subsetneq V_n$, because when incrementing $n$ the open annulus shrinks and the removed arc increases. 
What is $\cap V_n$? Because $V_n \subset \{(r,\phi) \in \mathbb R_{\ge 0}\times [0,2\pi): 1-\frac1n < r < 1 + \frac1n\}$ we get 
$$\cap V_n \subseteq \cap \{(r,\phi) \in \mathbb R_{\ge 0}\times [0,2\pi): 1-\frac1n < r < 1 + \frac1n\}  = \{1\}\times[0,2\pi).$$
OTOH, each $(1, \phi) \in \{1\}\times[0,2\pi)$ is in the removed arc for all high enough $n$, so we actually get that
$$\cap V_n = \emptyset,$$
as required.
The closure $\overline{V_n}$ is the corresponding closed annulus
$$\overline{V_n} = \{(r,\phi) \in \mathbb R_{\ge 0}\times [0,2\pi): 1-\frac1n \le r \le 1 + \frac1n\},$$
the removed arc gets 'added back' by the closure operation. We get
$$F=\cap\overline{V_n} = \cap \{(r,\phi) \in \mathbb R_{\ge 0}\times [0,2\pi): 1-\frac1n < r < 1 + \frac1n\}  = \{1\}\times[0,2\pi),$$
which is the unit circle and hence connected (even path-connected).
Now that we've checked all the conditions imposed by the problem, let's see how the hoped for conclusion fares. The boundary of $V_n$ are the bounding circles of the annulus and the removed arc:
$$\partial{V_n} = \{1-\frac1n, 1+\frac1n\}\times [0,2\pi) \cup \{1\}\times[0,2\pi-\frac1n].$$
We have $\forall n \ge 1: F \cap \{1-\frac1n, 1+\frac1n\}\times [0,2\pi) = \emptyset$ anyway and if we choose any positive $N \in \mathbb N$, then we can find $(1, 2\pi-\frac1{2N}) \in \{1\}\times[0,2\pi) = F$ and $(1, 2\pi-\frac1{2N}) \notin \partial{V_N}$.
That means there is no such $N$ where $\partial{V_N} \supseteq F$, let alone $\partial{V_n} \supseteq F\; \forall n \ge N$.
A: A weaker version of the claim is true: For each $x\in F$ there is $N$ such that $x\in \partial V_n$ for all $n>N$.
Take some $x\in F$. Suppose $x$ does not fulfill the claim. Then there are infinitely many indices $(n_k)$ such that $x$ does not belong to the boundary of $V_n$. Since $x\in F=\cap \overline{V_n}$, $x$ is in the closure of $V_n$ for all $n$. Hence, $x$ is an interior point of $V_{n_k}$ implying $x\in V_{n_k}$. Since the $(V_n)$ are decreasing,
it follows $x\in V_n$ for all $n\le n_k$. This implies $x\in V_n$ for all $n$. Hence $x\in \cap V_n$, which is a contradiction.
