Let $\{A_r:\ r\in(0,1)\}$ be a collection of connected subsets in $\Bbb{C}$ such that $A_{r_1}\subseteq A_{r_2}$ if $r_1<r_2$. Prove that, $\bigcap\limits_{r\in(0,1)} A_r$ is connected.
In general, intersection of connected sets is not connected. But here the collection is increasing with $r\in (0,1)$. In this set up, if the intersection was finite or moreover if the indexing set is bounded below i.e. the bound attained in the indexing set, the intersection is $A_{r_0}$ where $r_0$ is the least member in the indexing set and we are done. But here the indexing set is open interval $(0,1)$ and the infimum of the set which is $0$ doesn't belong to that set.
Let $A:=\bigcap\limits_{r\in(0,1)} A_r$ and $f:A\to\{0,1\}$ be a continuous function which is onto. So there are $x,y\in A$ such that $f(x)=0$ and $f(y)=1$. Now I have to use the contentedness of $A_r$'s to reach a contradiction.
Can anyone help complete the proof? Thank you for your help and valuable comments in advance.
The statement is false. For $r \in (0,1)$, consider the sets
$$ A_{r} = \mathbb{C} - \{ z \in \mathbb{C} : 0 < \text{ Re }(z) < 1, -1/r < \text{ Im }(z) < 1/r\}. $$
Then, for every $r \in (0,1)$, $A_{r}$ is connected. Further, for all $r_1, r_2 \in (0,1)$ and $r_1 < r_2$, we have $A_{r_1} \subseteq A_{r_2}$. However, $$\bigcap\limits_{r\in(0,1)} A_r= \mathbb{C} - \{ z \in \mathbb{C} : 0 < \text{ Re }(z) < 1\},$$ which is not connected.