Let $(S_n)_{n=1}^\infty$ be a sequence of decreasing sets ($S_{n+1}\subset{}S_{n}$) that are compact and connect. I am trying to show that the limit set $$S=\bigcap_{i=1}^{\infty}S_n$$ is also connected.
I know that $S$is non-empty since each $S_n$ is compact and they're decreasing.
Suppose for a contradiction $S$ is disconnected. Then $S=A\cup{}B$ where $A,B$ are non-empty, disjoint, open subsets of $S$.
I don't know what to do from here, but my intuition tells me a contradiction should work.
The intersection of a nested sequence of nonempty compact sets is nonempty. (Proof : If it is empty then there is an open cover of $S$ by the increasing sequence $\big\{S − S_i\big\}_{i=1}^\infty$ . This must have a finite subcover, so $S_i = ∅$ for some $i$, which is a contradiction.)
Now uppose that $\bigcap_{i=1}^\infty S_i$ is not connected. Let $A$ and $B$ be two disjoint nonempty closed sets so that$\bigcap_{i=1}^\infty =S_i = A ∪ B$. Find disjoint open sets $U$ and $V$ so that $A ⊂ U$ and $B ⊂ V$ . Put $F_i= S_i − (U ∪ V )$. Then $\{F_i\}_{i=1}^\infty$ is a nested sequence of compact sets, whose intersection is empty. Thus $F_i = ∅$ for some $i$. That is, $S_i ⊂ U ∪ V$. However, $S_i$ intersects both $U$ and $V$ , since $S_i ∩A\neq ∅$ and $S_i ∩B\neq ∅$.Contradiction