Union of connected sets $\forall \beta \in I$, $A_{\beta }$ is connected, and $\left ( \bigcup_{\alpha < \beta }A_{\alpha } \right )\cap A_{\beta }\neq \varnothing$ .  Is $\bigcup_{\alpha \in I}A_{\alpha } $connected?
For the index set $I$ , when it is countable ,the answer is obvious. I want to know the general conclusion.
Thanks！ 

Yes. This assumes - what is not explicit in your problem statement - that 


*

*$I$ is wellordered by $<$ and that 

*the intersection condition is not postulated for $\beta=\min I$ (as it would postulate $\emptyset\cap A_{\min I}\ne\emptyset$ anyway)


Let $\bigcup_{\alpha\in I}A_\alpha$ be covered by disjoint open sets $U,V$. 
Assume $U\cap\bigcup_{\alpha\in I}A_\alpha\ne\emptyset$ and $V\cap\bigcup_{\alpha\in I}A_\alpha\ne\emptyset$.
Let $\beta\in I$ be minimal with $A_\beta\cap U\ne\emptyset$ and $\gamma\in I$ minimal with $A_\gamma\cap V\ne\emptyset$. Wlog. $\beta\le\gamma$. Since $A_\gamma$ is connected, we cannot have $\beta=\gamma$. Hence $\gamma>\min I$.
Since $A_\gamma$ is connected, $U\cap A_\gamma=\emptyset$ and hence $A\gamma\cap\bigcup_{\alpha<\gamma}A_\alpha\ne\emptyset$ implies  $V\cap\bigcup_{\alpha<\gamma}A_\alpha\ne\emptyset$, contradicting minimality of $\gamma$.

What if $<$ is merely a total order on $I$?
Even for the countable case $I=\mathbb Z$ with standard order we obtain a counterexample as follows:
For $n\in \mathbb Z$ let $A_n=[n-1,n+1]\times\{(-1)^n\}\subset\mathbb R^2$. Then the intersection condition holds because $A_n\cap A_{n-2}\ne\emptyset$; but $\bigcup_{\alpha\in I}A_\alpha=\mathbb R\times\{-1,1\}$ is not connected.
