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

Let $(x^{(n)})$ be Cauchy for $d$, and fix a coordinate $i$. Every nonnegative summand is bounded by the whole sum:Given $\epsilon >0$, choose $N$ such that the right-hand side is less than $2^{-i}\epsilon /(1+\epsilon )$ for all $m,n\ge N$. Then strict monotonicity of $\varphi$ implies $d_i(x_i^{(m)},x_i^{(n)})<\epsilon$. Hence every coordinate sequence is Cauchy in its original metric $d_i$, with the index bound allowed to depend on the fixed coordinate.