$(A_i)_{i\in E}$ family of connected sets such that $\bigcap\limits_{i\in E} A_i \neq \emptyset$ then $\bigcup\limits_{i \in E} A_i$ is connected

Question

My idea so far is, since $\bigcap\limits_{i\in E} A_i \neq \emptyset$, then exists $p \in\bigcap\limits_{i\in E} A_i$

if $\bigcup\limits_{i \in E} A_i$ is not connected, then exists $A$ and $B$ open sets such that $A \bigcup B = \bigcup\limits_{i \in E} A_i$ with $A \bigcap B = \emptyset$. Then $p \in A$ or $p \in B$. I want to prove that if $p \in A$ the $B = \emptyset$

Thanks

Answer 1

Lemma: If $X$ is any topological space, $Z\subseteq X$ is both open and closed, $V\subseteq X$ is a connected set then either $V\subseteq Z$ or $V\subseteq X\setminus Z$.

I'll leave the proof of the lemma to you because it is almost trivial. Now let's use it for your exercise. Let $Z\subseteq\cup_{i\in E} A_i$ be a set which is both open and closed. For each $i\in E$ the set $A_i\subseteq \cup_{i\in E}A_i:=Y$ is connected and hence by the lemma $A_i\subseteq Z$ or $A_i\subseteq Y\setminus Z$. But now suppose that there are $i,j\in E$ such that $A_i\subseteq Z$ and $A_j\subseteq Y\setminus Z$. That means there is no element which is both in $A_i$ and $A_j$ which is a contradiction. So we conclude that either $A_i\subseteq Z$ for all $i\in E$ or $A_i\subseteq Y\setminus Z$ for all $i\in E$. If $A_i\subseteq Z$ for all $i\in E$ then the union is also a subset of $Z$ and hence $\cup_{i\in E}A_i=Z$. And if $A_i\subseteq Y\setminus Z$ for all $i\in E$ then $Z$ is empty.

Answer 2

Hint:

Use that a topological space $(X,\mathcal{T})$ is connected iff every continuous function

$$f: (X, \mathcal{T}) \to (\{0,1\}, 2^{\{0,1\}})$$ is constant.