# Example of a sequence of closed and connected sets whose intersection is disconnected [duplicate]

How can I find an example of subsets which satisfies;

for every $i \in \mathbb{N}$
$c_1 \supset c_2 \supset c_3 \supset \cdots \supset c_n$ which are closed and connected subsets of $\mathbb{R^k}$ then $\bigcap_{i=1}^\infty c_i$ is not connected.

until now I have

• $c_n=\{[n,\infty) , n \in \mathbb{N}\}$ $\Rightarrow$ $\bigcap_{i=1}^\infty c_i=\varnothing$ which is I couldn't decide whether if it is connected or not connected.
• $d_n=\{ ( [-\frac{1}{n},\frac{1}{n}] \times \{0\}) \cup (\{0\}\times ([-1,1]\setminus \{0\})), n\in \mathbb{N}\}$ $\Rightarrow$ $\bigcap_{i=1}^\infty d_i= \{0\} \times [-1,1]$ which is connected.
(here I want to seperate the vertical one from the point $(0,0)$ for not path connected implies not connected. If i rearrange the $[-1,1] \times \{0\}$ element -for not path connecting result- as $(-1,1) \times \{0\}$ the set will be not closed but intersection is not connected.)

• Empty sets are connected (vacuously), so this causes problems with $c_n$. With $d_n$, where is the dependence on $n$? May 11, 2017 at 11:07

$$e_n=(\mathbb{R}\times\{0\})\cup (\mathbb{R}\times\{1\})\cup\{(x,y):x\geq n,0\leq y\leq 1\}.$$
Another example. Let $c_n =\mathbb{R}^2 \setminus \{(x,y)| x \in (-1,1),y \in (-n,n)\}$. Complement is open so $c_n$ is closed. They are clearly nested and $c = \cap_i^\infty c_n = \mathbb{R}^2\setminus((-1,1)\times Y)$ is disconnected.