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

Put $\varphi (t)=t/(1+t)$. Every summand is nonnegative and at most $2^{-i}$, so the series defining $d$ converges and $d\le 1$. Symmetry follows from each $d_i$. If $d(x,y)=0$, every nonnegative summand is zero, hence every $d_i(x_i,y_i)=0$ and $x_i=y_i$. The converse is immediate.

For the triangle inequality, first observe that for $a,b\ge 0$,Monotonicity of $\varphi$ and the triangle inequality for $d_i$ therefore giveMultiply by $2^{-i}$ and sum the convergent nonnegative series. This proves the required triangle inequality for $d$, so $d$ is a metric. The construction combines a bounded metric transform with a countable product metric.

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.

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.