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 $n\geq N$?

  • $\begingroup$ By decreasing sequence of sets what exactly do you mean? $\endgroup$
    – irh
    Jul 14 '19 at 1:49
  • $\begingroup$ $V_{n+1}\subset V_{n}$ for all $n$. $\endgroup$
    – M. Rahmat
    Jul 14 '19 at 2:21
  • $\begingroup$ Can you show us an example of such a sequence, please? $\endgroup$
    – Daniel W.
    Jul 14 '19 at 5:36
  • 3
    $\begingroup$ Yes. Set $V_{n}=(0,1/n)$. $\endgroup$
    – M. Rahmat
    Jul 14 '19 at 5:40
  • $\begingroup$ Yes. What did you try to solve this? $\endgroup$ 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 $\forall n \ge 1$

$$V_n=\{(r,\phi) \in \mathbb R_{\ge 0}\times [0,2\pi): 1-\frac1n < r < 1 + \frac1n\} - \{1\}\times[0,2\pi-\frac1n] .$$

enter image description here $V_n$ 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 $\partial{V_n} \supseteq F\; \forall n \ge N$.

  • $\begingroup$ 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 $V_{n}$? $\endgroup$
    – M. Rahmat
    Jul 14 '19 at 19:10
  • $\begingroup$ 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 $C_{n+1} \supset C_n$. $\endgroup$
    – Ingix
    Jul 14 '19 at 22:14
  • $\begingroup$ 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 $V_{n}$(?) $\endgroup$
    – M. Rahmat
    Jul 15 '19 at 1:41
  • $\begingroup$ No. If you want to understand it with complex numbers, we have $$V_n=\{z\in \mathbb C: 1-\frac1n < |z| < 1+\frac1n \land (|z| \neq 1 \lor \arg(z) > 2\pi-\frac1n)\}$$ $\endgroup$
    – Ingix
    Jul 15 '19 at 8:22
  • $\begingroup$ I've added a picture, maybe that help you understand what $V_n$ looks like. $\endgroup$
    – Ingix
    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.


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