Past exam of the metric spaces and complex analysis course of the University of Oxford / 2014 / 2

Nonnegativity and symmetry follow from the corresponding metric axioms in the two factors. The maximum is zero exactly when both coordinate distances are zero, hence exactly when the ordered pairs are equal.

For the triangle inequality, put $p=(m_1,n_1)$, $q=(m_2,n_2)$ and $r=(m_3,n_3)$. ThenTherefore the maximum formula defines a metric, the maximum product metric.

Each coordinate distance is at most the maximum, so convergence of the pairs implies convergence of both coordinates. Conversely, for any $ϵ>0$, coordinate convergence gives indices $N_M,N_N$ after which the respective distances are below $ϵ$. Beyond their maximum, both are below $ϵ$, and so is the maximum product metric.The same distance comparison will also handle Cauchy sequences.

Let $(x_n,y_n)$ be a Cauchy sequence in the product. Since each coordinate distance is bounded by $\delta$, both $(x_n)$ and $(y_n)$ are Cauchy in their respective factors. By completeness they have limits $x\in M$, $y\in N$. Part (ii) then gives convergence of the pairs to $(x,y)$.

Thus the product of two complete metric spaces is complete under the maximum product metric. This checks every Cauchy sequence, rather than assuming beforehand that it has a coordinate limit.

For coordinate differences $a,b$, the elementary inequalitiesgiveThe identity map between the two versions of the square is bijective, and both it and its inverse are Lipschitz continuous maps by these inequalities. It is therefore a homeomorphism. In fact the same comparison works on all of ${\mathbb{R}}^2$; restriction to the square causes no difficulty.

Take $M=\mathbb{R}$ and the constant contraction mapping $f(x)=0$. For $p=(0,0)$, $q=(0,1)$,No Lipschitz constant less than one can work, so $F$ need not be a contraction. More generally, differing only in the second input coordinate already prevents a strict contraction whenever $M$ has two points.

If $f$ has contraction constant $0\le c<1$, two iterations giveand consequentlyThus $F^2$ is a contraction mapping. The swap lift of a contraction shows why an iterate can contract even when one step does not.