$\forall \beta \in I$, $A\_{\beta }$ 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！ 



Union of connected sets
I assume that you intend $I$ is linearly ordered by $<$. The hypothesis is impossible when $I$ has a least element $\beta$ — in particular, it is impossible when $I$ is nonempty and well-ordered — since if
$\beta$ is the least element, then
$\bigcup\_{\alpha<\beta}A\_\alpha$ is empty itself. But if we insist on that requirement only when $\beta$ is not the least element of $I$, 
then your conclusion holds if and only if $I$ is well-ordered (or nonempty and well-ordered, depending on whether you consider $\emptyset$ to be connected). 


**Theorem.** The following are equivalent for any linearly ordered
set $\langle I,<\rangle$.


1. $\langle I,<\rangle$ is a nonempty well-order.
2. Whenever $X$ is a topological space and $A\_\alpha\subset X$ is
a connected subspace for every $\alpha\in I$ and whenever
$\beta\in I$ is not least, then
$\bigcup\_{\alpha<\beta}A\_\alpha\cap A\_\beta\neq\emptyset$, then
$\bigcup\_{\alpha\in I}A\_\alpha$ is connected.


**Proof.** ($1\to 2$) Assume $I$ is nonempty and well-ordered by $<$. We prove
that $\bigcup\_{\alpha\in I}A\_\alpha$ is connected by transfinite
induction on the order type of $I$. Suppose that $U\sqcup V$ is a
nontrivial open separation of $\bigcup\_{\alpha\in I}A\_\alpha$. Let
$\beta$ be least such that $\bigcup\_{\alpha\leq\beta}A\_\alpha$ has
points from both $U$ and $V$. But this union is equal to
$\bigcup\_{\alpha<\beta}A\_\alpha\cup A\_\beta$, which by induction
is the union of two overlapping connected subspaces, and hence is
connected. So it cannot have points from both sides of the
separation, a contradiction. So there is no nontrivial open separation of $\bigcup\_{\alpha\in I}A\_\alpha$, and so it is connected.


($2\to 1$) We prove the contrapositive. For the main case, suppose that $I$ is not a well-order. Thus, it has a
strictly descending sequence $\alpha\_0>\alpha\_1>\alpha\_2>\cdots$ and so on.
If $\alpha$ is below all the $\alpha\_n$, then let $A\_\alpha=\{0\}$
have just the point $0$. Otherwise, $\alpha\_n\leq
\alpha$ for some smallest $n$, and in this case, we let $A\_\alpha$
be $\{0\}$, if $n$ is even, and otherwise $\{1\}$. So each
$A\_\alpha$ is connected, and furthermore, for every $\beta\in I$
(except the least element, if there is one), we have
$\bigcup\_{\alpha<\beta}A\_\alpha\cap A\_\beta\neq\emptyset$, since
below the descending sequence, both of these sets are just
$\{0\}$, and above any point in the descending sequence, the union
set is $\{0,1\}$, which meets $A\_\beta$. Meanwhile, the union
$\bigcup\_{\alpha\in I}A\_\alpha$ is $\{0,1\}$, which is not
connected, by design. Lastly, if $I$ is empty, then the hypothesis of 2 is vacuous, but the union set is empty, which is technically disconnected. **QED**


