# 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.”

Title | proof that products of connected spaces are connected |
---|---|

Canonical name | ProofThatProductsOfConnectedSpacesAreConnected |

Date of creation | 2013-03-22 14:10:05 |

Last modified on | 2013-03-22 14:10:05 |

Owner | yark (2760) |

Last modified by | yark (2760) |

Numerical id | 13 |

Author | yark (2760) |

Entry type | Proof |

Classification | msc 54D05 |