compactness is preserved under a continuous map
The inclusion map shows that the requirement for to be surjective cannot be omitted. If is compact and is continuous we can always conclude, however, that is compact, since is continuous (http://planetmath.org/IfFcolonXtoYIsContinuousThenFcolonXtoFXIsContinuous).
Thus is an open cover for . Since is compact, there exists a finite subset such that is a finite open cover for . Since is a surjection, we have for any (see this page (http://planetmath.org/InverseImage)). Thus
Thus is an open cover for , and is compact.
Shorter proof. Suppose is a collection of closed subsets of with the finite intersection property. Then is a collection of closed subsets of with the finite intersection property, because if is finite then
which is nonempty as is a surjection. As is compact, we have
and so . Therefore is compact.
|Title||compactness is preserved under a continuous map|
|Date of creation||2013-03-22 13:55:50|
|Last modified on||2013-03-22 13:55:50|
|Last modified by||yark (2760)|