proof that products of connected spaces are connected

Let $\{X_{\alpha }\text{ for }\alpha \in A\}$ be topological spaces, and let $X=\prod X_{\alpha }$ be the product, with projection maps ${\pi }_{\alpha }$.

Using the Axiom of Choice, one can straightforwardly show that each ${\pi }_{\alpha }$ is surjective; they are continuous by definition, and the continuous image of a connected space is connected, so if $X$ is connected, then all $X_{\alpha }$ are.

Let $\{X_{\alpha }\text{ for }\alpha \in A\}$ be connected topological spaces, and let $X=\prod X_{\alpha }$ be the product, with projection maps ${\pi }_{\alpha }$.

First note that each ${\pi }_{\alpha }$ is an open map: If $U$ is open, then it is the union of open sets of the form ${\bigcap }_{\beta \in F}{\pi }_{\beta }^{-1}{U}_{\beta }$ where $F$ is a finite subset of $A$ and $U_{\beta }$ is an open set in $X_{\beta }$. But ${\pi }_{\alpha }(U_{\beta })$ is always open, and the image of a union is the union of the images.

Suppose the product is the disjoint union of open sets $U$ and $V$, and suppose $U$ and $V$ are nonempty. Then there is an $\alpha \in A$ and an element $u\in U$ and an element $v\in V$ that differ only in the $\alpha$ place. To see this, observe that for all but finitely many places $\gamma$, both ${\pi }_{\beta }(U)$ and ${\pi }_{\beta }(V)$ must be $X_{\gamma }$, so there are elements $u$ and $v$ that differ in finitely many places. But then since $U$ and $V$ are supposed to cover $X$, if ${\pi }_{\beta }(u)\ne {\pi }_{\beta }(v)$, changing $u$ in the $\beta$ place lands us in either $U$ or $V$. If it lands us in $V$, we have elements that differ in only one place. Otherwise, we can make a $u^{\prime }\in U$ such that ${\pi }_{\beta }(u^{\prime })={\pi }_{\beta }(v)$ and which otherwise agrees with $u$. Then by induction we can obtain elements $u\in U$ and $v\in V$ that differ in only one place. Call that place $\alpha$.

We then have a map $\rho :X_{\alpha }\to X$ such that ${\pi }_{\alpha }\circ \rho$ is the identity map on $X_{\alpha }$, and $(\rho \circ \pi )(u)=u$. Observe that since ${\pi }_{\alpha }$ is open, $\rho$ is continuous. But ${\rho }^{-1}(U)$ and ${\rho }^{-1}(V)$ are disjoint nonempty open sets that cover $X_{\alpha }$, which is impossible.

Note that if we do not assume the Axiom of Choice, the product may be empty, and hence connected, whether or not the $X_{\alpha }$ are connected; by taking the discrete topology on some $X_{\alpha }$ we get a counterexample to one direction of the theorem: we have a connected (empty!) space that is the product of non-connected spaces.For the other direction, if the product is empty, it is connected; if it is not empty, then the argument below works unchanged.So without the Axiom of Choice, this theorem becomes “If all $X_{\alpha }$ are connected, then $X$ is.”