Suppose that $\{A_i:i\in \mathbb{N}\}$ are connected subspaces in $\mathbb{R}^2$ and $A_{i+1} \subset A_i$. Is $\cap_{i\in \mathbb{N}} A_i$ necessarily connected?
I have seen several posts on compact and connected subspaces with the above hypothesis that give a positive answer. This makes me think that this is untrue but I am yet to come up with a counterexample.
Any thoughts?
Let $A_n=\{(x,y)\in\mathbb{R}^2\mid 0\leq x\lt 1$, $0\leq y\leq x^n\}\cup\{(1,1)\}$.
This is the region between the $x$-axis and the graph of $y=x^n$ on $[0,1)$, plus the $(1,1)$ point. It is not hard to verify that $A_n$ is connected, and that $A_{n+1}\subseteq A_n$ for all $n$.
What is $A=\bigcap\limits_{n=1}^{\infty}A_n$? It certainly contains $[0,1)\times\{0\}$; it also contains $(1,1)$. I claim that is all it contains. To verify that, let $(x,y)\in A_1$ with $0\lt x\lt 1$, $y\gt 0$. Since $0\lt x\lt 1$, $\lim_{n\to\infty}x^n = 0$, so there exists $N\gt 0$ such that $x^N\lt y$. Thus, $(x,y)\notin A_N$, so $(x,y)\notin A$.
But $\Bigl( [0,1)\times\{0\}\Bigr)\cup\{(1,1)\}$ is not connected.