Let \(X_n\) be a metric space with metric \(d_n\), for \(n \in \mathbb{Z}_+\).
- Show that
\[ \rho(x, y) = \max\ts{d_1(x_1, y_1), \dots, d_n(x_n, y_n)} \]
is a metric for the product space \(X_1 \times \cdots \times X_n\).
- Let \(\bar{d}_i = \min\ts{d_i, 1}\). Show that
\[ D(x, y) = \sup\ts{\bar{d}_i(x_i, y_i)/i} \]
is a metric for the product space \(\prod X_i\).
Solution
- Let \[ \rho(x,y)=\max_{1\le i\le n}d_i(x_i,y_i). \] Nonnegativity, symmetry, and definiteness are immediate. For each \(i\), \[ d_i(x_i,z_i)\le d_i(x_i,y_i)+d_i(y_i,z_i)\le \rho(x,y)+\rho(y,z), \] so taking the maximum gives the triangle inequality.
A \(\rho\)-ball of radius \(r\) is exactly \[ \prod_{i=1}^n B_{d_i}(x_i,r), \] so the metric topology equals the finite product topology.
- Put \(\bar d_i=\min(d_i,1)\) and \[ D(x,y)=\sup_i\frac{\bar d_i(x_i,y_i)}{i}. \] Each \(\bar d_i\) is a metric, and the same coordinatewise triangle inequality followed by supremum proves the triangle inequality for \(D\). Definiteness follows because \(D(x,y)=0\) forces every coordinate distance to vanish.
We compare topologies. Given \(\varepsilon>0\), choose \(N\) so large that \(1/i<\varepsilon\) for \(i>N\). The set of \(y\) with \[ d_i(x_i,y_i)<i\varepsilon\qquad(1\le i\le N) \] is a basic product neighborhood of \(x\). For such \(y\), \(\bar d_i(x_i,y_i)/i<\varepsilon\) for \(i\le N\), and \(\bar d_i(x_i,y_i)/i\le1/i\le1/(N+1)<\varepsilon\) for \(i>N\), so \(D(x,y)<\varepsilon\). Hence this neighborhood lies in the \(D\)-ball of radius \(\varepsilon\) about \(x\).
Conversely, let \[ U=\prod_iU_i \] be a basic product neighborhood of \(x\), with \(U_i=X_i\) except for the finite set \(F\). For each \(i\in F\), choose \(r_i\in(0,1]\) such that \[ B_{d_i}(x_i,r_i)\subseteq U_i. \] Choose \[ 0<\varepsilon<\min_{i\in F}\frac{r_i}{i}. \] If \(D(x,y)<\varepsilon\), then for \(i\in F\), \[ d_i(x_i,y_i)=\bar d_i(x_i,y_i)<i\varepsilon<r_i, \] so \(y\in U\). Thus the \(D\)-topology is exactly the product topology.