The above conditions imply that if is hemicompact with admissible sequence then because every point of is compact and lies in one of the .
A hemicompact space is clearly -compact. The converse is false in general. This follows from the fact that a first countable hemicompact space is locally compact (see below). Consider the set of rational numbers with the induced euclidean topology. is -compact but not hemicompact. Since satisfies the first axiom of countability it can’t be hemicompact as this would imply local compactness.
Proposition. Let be a first countable hemicompact space. Then is locally compact.
Proposition. Let be a locally compact and -compact space. Then is hemicompact.
By local compactness we choose a cover of open sets with compact closure (take a compact neighborhood of every point). By -compactness there is a sequence of compacts such that . To each there is a finite subfamily of which covers . Denote the union of this finite family by for each . Set . Then is a sequence of compacts. Let be compact then there is a finite subfamily of covering . Therefore for some . ∎
|Date of creation||2013-03-22 19:08:18|
|Last modified on||2013-03-22 19:08:18|
|Last modified by||karstenb (16623)|