I am having some trouble trying to prove the following statement:$$$$
Let $(X,d)$ be a metric space and $\mathcal A$ a family of connected sets in $X$ such that for every pair of subsets $A,B \in \mathcal A$ there exist $A_0$,…,$A_n \in \mathcal A$ that satisfy $A_0=A$, $A_n=B$ and $A_i \cap A_{i+1} \neq \emptyset$ for every i=0,…,n-1. Prove that $\bigcup_{A \in \mathcal A}A$ is connected.
$$$$I've tried to prove it by the absurd: Suppose the union is disconnected, then there exist $U$ and $V$ nonempty disjoint open sets such that $\bigcup_{A \in \mathcal A}A=U \cup V$. Then, there is $A \in \bigcup_{A \in \mathcal A}A$ : $A \subset V$ and $A \cap V=\emptyset$. The same argument applies for $B \in \bigcup_{A \in \mathcal A}A$ with $B \subset U$ ($A$ and $B$ both nonempty). By hypothesis, $A=A_0$ and $B=A_n$. In this part I got stuck. I know I have to use the fact that the intersection of $A_i$ and $A_{i+1}$ is nonempty and that all the sets in $\mathcal A$ are connected, but I don't know where to use that.
[Math] Family of connected sets, proving union is connected
analysisconnectedness
Best Answer
Recall that a (topological or metric) space $X$ is connected if and only if the only continuous functions $f:X\to\{0,1\}$ are the constant functions, where $\{0,1\}$ is endowed with the discrete metric.
Now consider a continuous function $f:\cup\mathcal A\to\{0,1\}$, and prove that this function is constant by using the fact that $f$ restricted to each member of $\mathcal A$ is constant (by connectedness of such members) plus your hypothesis.