How to prove $X=\prod X_{\alpha}$ is locally connected iff $X_{\alpha}$ is locally connected and only finite of them can be disconnected? I have difficulties with the finite part.
BTW,I showed the $X_{\alpha}$ part like this:
Assume $X_1$ is not locally connected, so $\exists x_1$ and its neighborhood $U_1$, there is no connected neighborhood $V_1$ such that $x\in V_1 \subset U_1$.
$X$ is locally connected, so take $U= U_1 \times \ldots$, there exists a connected open set $O= O_1 \times \ldots$ such that $x \in O \in U$. So $O_1 \subset U_1$. So $O_1$ is not connected.
Then take the seperation of $O_1$ as $B\cup C$,then $B\times O_2\times...$ and $C\times O_2\times...$ is the seperation of O,meanwhile O is connected.
Is this proof right?
 A: I don't see your proof as correct, and it anyway suffers from some problems with ambiguous notation. What does $U_1 \times \ldots$ mean? Does it mean $U_1 \times X_2 \times X_3 \times X_4 \times \ldots$? The notational problems are exacerbated if the indexing set is not the positive integers.
Here are a few observations which might help you prove that infinitely many disconnected $X_\alpha$ leads to a product which is not locally connected.

*

*the image of a locally connected space by an open map is locally connected,

*a separation of a disconnected space gives an open map onto $\{0,1\}$,

*the coordinate projections of a product space are open surjections,

*a product of open maps is open.

Maybe you can combine these observations to construct an open surjection from $\prod X_\alpha$ to the Cantor space $\{0,1\}^\mathbb{N}$ in the case where a countable infinity of the factors are disconnected?
A: If $X = \prod_{i \in I} X_i$ is locally connected, then all $X_i$, as they are images of $X$ under the open and continuous projection map $p_i: X \to X_i$ are also locally connected.
Also let $x \in X$ be  arbitrary, there is a basic open $\prod_i O_i$ (so all $O_i$ are open and $F:= \{i \in I \mid O_i \neq X_i\}$ is finite so that $C \subseteq \prod_i O_i$ for some connected neighbourhood $C$. It follows that $\pi_i[C] = X_i$ for all $i \notin F$ so all but finitely many $X_i$ are connected, being the continuous image of a connected set.
