Show that every metric space is Hausdorff in its metric topology.
Solution
1. Let \(x\ne y\) in a metric space \((X,d)\), and set\[ r=\frac{d(x,y)}{2}>0. \]
2. The metric balls \(B(x,r)\) and \(B(y,r)\) are open neighborhoods of \(x\) and \(y\).
3. They are disjoint.
If \(z\in B(x,r)\cap B(y,r)\), then the triangle inequality gives \[ d(x,y)\le d(x,z)+d(z,y)<r+r=d(x,y), \] a contradiction.
4. Thus every two distinct points admit disjoint open neighborhoods, so \(X\) is Hausdorff.