Past exam of the metric spaces and complex analysis course of the University of Oxford / 2023 / 1 / a

A metric space $(X,d)$ is a bounded set if there is a finite constant $M\ge 0$ such thatFor nonempty $X$, this is equivalent to containment in an open ball of finite radius: the triangle inequality converts a radius bound into a diameter bound, and conversely a diameter bound gives a ball of any slightly larger radius about a chosen point.

It is a complete metric space if every Cauchy sequence in $X$ has a limit of a sequence belonging to $X$. Explicitly, the hypothesis on $(x_n)$ isand completeness requires an $x\in X$ with $d(x_n,x)\to 0$. ThusBoundedness controls distances; completeness ensures that these limiting points remain in the space.

A subset $F\subseteq X$ is a closed set in $X$ when its complement is an open set in the metric topology. Equivalently,The open ball is taken inside the ambient metric space $X$; closedness always depends on that ambient space.

First suppose $F$ is a closed set. If $x\notin F$, the definition in part (ii) gives an open ball $B(x,r)$ disjoint from $F$. Thus every $y\in F$ has $d(x,y)\ge r$, and the distance to a set satisfies $\mathrm{dist}(x,F)\ge r>0$. Its contrapositive is the required implication.

Conversely, suppose that zero distance to a set forces membership. For $x\notin F$ we have $\mathrm{dist}(x,F)>0$. Choose $0<r<\mathrm{dist}(x,F)$. Then $B(x,r)\cap F=\varnothing$, proving $X\setminus F$ is an open set. HenceFor $F=\varnothing$, use the extended-real convention $\inf \varnothing =+\infty$: both assertions hold, with the implication vacuous. Part (b) only uses nonempty sets, so all its distances to a set are ordinary finite real numbers.