Past exam of the metric spaces and complex analysis course of the University of Oxford / 2017 / 1 / b / iii

By part (ii) and completeness of each factor, choose $x_i\in X_i$ with $x_i^{(n)}\to x_i$. Let $x=(x_i)$. Given $\epsilon >0$, choose $N$ such that ${\sum }_{i>N}{2}^{-i}=2^{-N}<\epsilon /2$. Coordinate convergence gives a single $n_0$ for which $d_i(x_i^{(n)},x_i)<\epsilon /2$ for every $i\le N$ and $n\ge n_0$. Since $\varphi (t)\le t$,Thus $x^{(n)}\to x$ in the product metric, proving the product is complete. The finite head uses coordinate convergence; the summable tail is controlled uniformly, regardless of how large the remaining original coordinate distances are. This is completeness of a countable product metric.