a space is Hausdorff if and only if is closed
First, some preliminaries: Recall that the diagonal map is defined as . Also recall that in a topology generated by a basis (like the product topology), a set is open if and only if, for every point , there’s a basis element with . Basis elements for have the form where are open sets in .
Now, suppose that is Hausdorff. We’d like to show its image under is closed. We can do that by showing that its complement is open. consists of points with equal coordinates, so consists of points with and distinct.
For any , the Hausdorff condition gives us disjoint open with . Then is a basis element containing . and have no points in common, so contains nothing in the image of the diagonal map: is contained in . So is open, making closed.
Now let’s suppose is closed. Then is open. Given any , there’s a basis element with . lying in implies that and are disjoint.
If we have in , then is in . The basis element containing gives us open, disjoint with . is Hausdorff, just like we wanted. ∎
|Title||a space is Hausdorff if and only if is closed|
|Date of creation||2013-03-22 14:20:47|
|Last modified on||2013-03-22 14:20:47|
|Last modified by||mathcam (2727)|