Metric topology

Metrizable implies Hausdorff

Let be a metric space. Then is Hausdorff under its metric topology. #m/thm/topology

Proof

Let with (if no such exist then the conclusion is trivially satisfied). Let . Then and are disjoint open neighbourhoods of and respectively. Since if then which is a contradiction.

#state/tidy | #lang/en | #SemBr