Suppose that \(d,d'\) are metrics on the set \(X\) with \(d(x,y)\le d'(x,y)\) for every \(x,y\in X\). Show that the metric topology \((X,d)\) is coarser than the metric topology \((X,d')\).
Solution
Let \(\mathcal T_d\) and \(\mathcal T_{d'}\) be the two metric topologies.
1. \(B_{d'}(x,r)\subseteq B_d(x,r)\) for every \(x\in X\) and \(r>0\).
If \(d'(x,y)<r\), then \(d(x,y)\le d'(x,y)<r\).
2. Q.E.D.
Let \(U\in\mathcal T_d\) and \(x\in U\), and choose \(r>0\) with \(B_d(x,r)\subseteq U\). By step 1, \(B_{d'}(x,r)\subseteq U\), so \(U\in\mathcal T_{d'}\). Hence \(\mathcal T_d\subseteq\mathcal T_{d'}\).