proof that products of connected spaces are connected

Let {Xα for αA} be topological spacesMathworldPlanetmath, and let X=Xα be the productPlanetmathPlanetmathPlanetmath, with projection maps πα.

Using the Axiom of ChoiceMathworldPlanetmath, one can straightforwardly show that each πα is surjectivePlanetmathPlanetmath; they are continuousMathworldPlanetmathPlanetmath by definition, and the continuous image of a connected space is connected, so if X is connected, then all Xα are.

Let {Xα for αA} be connected topological spaces, and let X=Xα be the product, with projection maps πα.

First note that each πα is an open map: If U is open, then it is the union of open sets of the form βFπβ-1Uβ where F is a finite subset of A and Uβ is an open set in Xβ. But πα(Uβ) is always open, and the image of a union is the union of the images.

Suppose the product is the disjoint unionMathworldPlanetmath of open sets U and V, and suppose U and V are nonempty. Then there is an αA and an element uU and an element vV that differ only in the α place. To see this, observe that for all but finitely many places γ, both πβ(U) and πβ(V) must be Xγ, 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 πβ(u)πβ(v), changing u in the β 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 uU such that πβ(u)=πβ(v) and which otherwise agrees with u. Then by inductionMathworldPlanetmath we can obtain elements uU and vV that differ in only one place. Call that place α.

We then have a map ρ:XαX such that παρ is the identity map on Xα, and (ρπ)(u)=u. Observe that since πα is open, ρ is continuous. But ρ-1(U) and ρ-1(V) are disjoint nonempty open sets that cover Xα, 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 Xα are connected; by taking the discrete topology on some Xα we get a counterexampleMathworldPlanetmath to one direction of the theoremMathworldPlanetmath: 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 argumentMathworldPlanetmath below works unchanged. So without the Axiom of Choice, this theorem becomes “If all Xα 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