# 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 $\pi_{\beta}(u)\neq\pi_{\beta}(v)$, 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.”

Title proof that products of connected spaces are connected ProofThatProductsOfConnectedSpacesAreConnected 2013-03-22 14:10:05 2013-03-22 14:10:05 yark (2760) yark (2760) 13 yark (2760) Proof msc 54D05