Metric on $\mathbb{Q}$ for which the completion of $\mathbb{Q}$ is isomorphic to $\mathbb{Q}_2 \times \mathbb{Q}_3$. Is there a metric on $\mathbb{Q}$ for which the completion of $\mathbb{Q}$ is isomorphic to $\mathbb{Q}_2 \times \mathbb{Q}_3$?
 A: Let$$d(x, y) = \max\left( |x-y|_2,|x-y|_3\right).$$Obviously $d$ is a metric. We claim that completing $\mathbb{Q}$ with respect to $d$ is $\mathbb{Q}_2 \times \mathbb{Q}_3$ with metric$$\hat{d}((x, x'), (y, y')) = \max\left(|x-y|_2,|x' - y'|_3\right),$$where $x$, $y \in \mathbb{Q}_2$, $x'$, $y' \in \mathbb{Q}_3$. Clearly, $(\mathbb{Q}_2 \times \mathbb{Q}_3, \hat{d})$ is complete, and the diagonal map$$\mathbb{Q} \hookrightarrow \mathbb{Q}_2 \times \mathbb{Q}_3$$is an isometry. Let $(x, y) \in \mathbb{Z}_2 \times \mathbb{Z}_3$; then there exist sequences $(x_n)$, $(y_n)$ in $\mathbb{Z}$ with$$v_2(x_n - x),\text{ }v_3(y_n - y) \ge n.$$By the Chinese Remainder Theorem, there exists $z_n \in \mathbb{Z}$ with$$z \equiv \begin{cases} x_n \text{ (mod }2^n\text{)} \\ y_n \text{ (mod }3^n\text{)}\end{cases}$$$$\implies (z_n) \to x \text{ }\,2\text{-adically, }(z_n) \to y \text{ }\,3\text{-adically}$$$$\implies (z_n, z_n) \to (x, y) \text{ in }(\mathbb{Q}_2 \times \mathbb{Q}_3, \hat{d}).$$If $(x, y) \in \mathbb{Q}_2 \times \mathbb{Q}_3$, there exists $m$ such that $(6^mx, 6^my) \in \mathbb{Z}_2 \times \mathbb{Z}_3$, so there exists $z_n \in \mathbb{Z}$ with$$(z_n, z_n)_n \to (6^mx, 6^my)$$$$\implies \left({1\over{6^m}}z_n, {1\over{6^m}}z_n\right)_n \to (x, y).$$
