Is infinite(include countable and uncountable) product of connected spaces still connected? Let $\{X_{\alpha}\}_{\alpha \in J}$ be a collection of connected spaces, $X=\prod_{\alpha \in J}X_{\alpha}$.
I know $X$ usually  isn't connected with box-topology, with an example of $R^{\omega}$.
So I wonder will any product with box-topology be not connected? If this is true,how to prove it?
 A: If all the $X_\alpha$-s are irreducible (and axiom of choice holds for their open sets), then the box topology on the product is irreducible as well (hence connected). For let $\Omega_1,\Omega_2$ be two non-empty open subsets. By definition, there are $\left\{U^1_\alpha\right\}_{\alpha\in J}$ and $\left\{U^2_\alpha\right\}_{\alpha\in J}$ open subsets of the $X_\alpha$-s such that $\emptyset\ne\prod_{\alpha\in J}U_\alpha^i\subseteq\Omega_i$ for $i=1,2$. Since each $X_\alpha$ is irreducible and $U^1_\alpha$ and $U_\alpha^2$ are non-empty and open in $X_\alpha$, for all $\alpha$ we have $U^1_\alpha\cap U_\alpha^2\ne\emptyset$. By the axiom of choice, $$\emptyset\ne\prod_{\alpha\in J}(U^1_\alpha\cap U^2_\alpha)\subseteq\left(\prod_{\alpha\in J}U^1_\alpha\right)\cap\left(\prod_{\alpha\in J}U^2_\alpha\right)\subseteq\Omega_1\cap\Omega_2$$
A: In the Handbook of Set-theoretic Topology, chapter Box Products, it is shown that any infinite product of non-discrete Tychonoff spaces is non-connected. So this accounts for "most" box products. There are special cases (e.g. a box product of infinite cofinite spaces), where the box product is connected due to spcicial properties of the factors, but these are quite rare.
