proof that products of connected spaces are connected
Using the Axiom of Choice, one can straightforwardly show that each is surjective; they are continuous by definition, and the continuous image of a connected space is connected, so if is connected, then all are.
Let be connected topological spaces, and let be the product, with projection maps .
First note that each is an open map: If is open, then it is the union of open sets of the form where is a finite subset of and is an open set in . But 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 and , and suppose and are nonempty. Then there is an and an element and an element that differ only in the place. To see this, observe that for all but finitely many places , both and must be , so there are elements and that differ in finitely many places. But then since and are supposed to cover , if , changing in the place lands us in either or . If it lands us in , we have elements that differ in only one place. Otherwise, we can make a such that and which otherwise agrees with . Then by induction we can obtain elements and that differ in only one place. Call that place .
Note that if we do not assume the Axiom of Choice, the product may be empty, and hence connected, whether or not the are connected; by taking the discrete topology on some 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 are connected, then is.”
|Title||proof that products of connected spaces are connected|
|Date of creation||2013-03-22 14:10:05|
|Last modified on||2013-03-22 14:10:05|
|Last modified by||yark (2760)|