metric spaces are Hausdorff
Say we’ve got distinct . Since is a metric, . Then the open balls and are open sets in the metric topology which contain and respectively. If we could show and are disjoint, we’d have shown that is Hausdorff.
We’d like to show that an arbitrary point can’t be in both and . Suppose there is a in both, and we’ll derive a contradiction. Since is in these open balls, and . But then , contradicting the triangle inequality.
So and are disjoint, and is Hausdorff.
|Title||metric spaces are Hausdorff|
|Date of creation||2013-03-22 14:21:29|
|Last modified on||2013-03-22 14:21:29|
|Last modified by||waj (4416)|