Infinite intersection of compact, path connected, nested sets is path connected? I showed that given $A_1\supseteq A_2, ...$ compact, connected sets, $\bigcap_{i=1}^n A_n$ is  connected but is the statement true if we replace connected with path connected? Is there an counterexample of compact, path connected sets whose arbitrary intersection is not path connected? 
 A: Take $A$ to be the (closure of the) topologist's sine curve (so that $A$ includes the segment $\{0\} \times [-1, 1]$) and let $A_n = A \cup ([0, 1/n] \times [-1,1])$. The $A_n$ are a descending family of path-connected compact sets whose intersection is $A$, which is not path-connected.
Thanks to Travis for pointing out that the topologist's sine curve according to the first definition given in the Wikipedia definition is not closed.
A: No. Recall that the closed topologists' sine curve is the set
$$C := \left\{\left(x, \sin \tfrac{1}{x}\right) : x \in (0, 1)\right\} \cup (\{0\} \times [-1, 1])$$
Now, let $R_n$, $n = 1, 2, 3, \ldots$, denote the filled rectangle $\left[0, \tfrac{1}{n}\right] \times [-1, 1]$, and set $$A_n := C \cup R_n ,\qquad n = 1, 2, 3, \ldots .$$ Since the $R_n$ are nested, so are $A_n$. By construction, $A_n$ is the union of the compact, path-connected sets $R_n$ and $\left\{\left(x, \sin \tfrac{1}{x}\right) : \tfrac{1}{n} \leq x \leq 1\right\}$ that share a common point, so each $A_n$ is compact and path-connected. The infinite intersection $\bigcap_{n = 1}^{\infty} R_n$ of the $R_n$ is just $\{0\} \times [-1, 1] \subset C$, so the intersection $\bigcap_{n = 1}^{\infty} A_n$ of the $A_n$ is just $C$, which is not path-connected.
