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 $$n\geq N$$?

• By decreasing sequence of sets what exactly do you mean?
– irh
Jul 14 '19 at 1:49
• $V_{n+1}\subset V_{n}$ for all $n$. Jul 14 '19 at 2:21
• Can you show us an example of such a sequence, please? Jul 14 '19 at 5:36
• Yes. Set $V_{n}=(0,1/n)$. Jul 14 '19 at 5:40
• Yes. What did you try to solve this? Jul 14 '19 at 7:38

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 $$\partial{V_n} \supseteq F\; \forall n \ge N$$.

• Exellent! But steel I don't see why $V_{n+1}\subset V_{n}$? As I understand the $V_{n}$ are nither increasing nor decreasing. Isn't it that $V_{n+1}$ gets closer to the $x-$axis than $V_{n}$? Jul 14 '19 at 19:10
• All of these are polar coordinates, not cartesian ones. For the monotony, you can consider $V_n=A_n-C_n$ (Annulus - Circle (part)), and you can check that $A_{n+1} \subset A_n$ and $C_{n+1} \supset C_n$. Jul 14 '19 at 22:14
• If I understand correctly, $$V_{n}=\{z\in\mathbb{C}: 0<arg(z)<2\pi-1/n, 1-1/n<|z|<1+1/n, |z|\not=1\}.$$ Take the point with arg=$1/(n+0.5)$: it is in $V_{n+1}$ but not in $V_{n}$(?) Jul 15 '19 at 1:41
• No. If you want to understand it with complex numbers, we have $$V_n=\{z\in \mathbb C: 1-\frac1n < |z| < 1+\frac1n \land (|z| \neq 1 \lor \arg(z) > 2\pi-\frac1n)\}$$ Jul 15 '19 at 8:22
• I've added a picture, maybe that help you understand what $V_n$ looks like. Jul 15 '19 at 8:35

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.