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A compact real tree is a compact metric space $(T,d)$ such that any $x,y\in T$ are joined by a unique arc, and that arc is isometric to $[0,d(x,y)]$. The multiplicity of a point in a real tree $a\in T$ is the number of connected components of $T\setminus \{a\}$.

For nonempty compact subsets $A,B$ of a metric space $(Z,d)$, the Hausdorff distance isFor compact metric spaces $X,Y$, the Gromov-Hausdorff distance iswhere $\phi$ and $\psi$ range over isometric embeddings into a common metric space $Z$.

The collection of compact real trees is not compact in the Gromov-Hausdorff topology. Indeed, the intervals $T_n=[0,n]$ are compact real trees andTheir diameters are unbounded, so $(T_n)$ has no convergent subsequence in the Gromov-Hausdorff topology.

Let $A,B$ be disjoint closed subsets of the metric space $(X,d)$; the cases where one is empty are immediate. The distance functionsare continuous, because each is 1-Lipschitz. Their sum is strictly positive: if both distances vanished, closedness would put $x$ in $A\cap B$. Thereforeis continuous, equals zero on $A$, and equals one on $B$. The setsare disjoint open neighbourhoods of $A$ and $B$. Thus every metric space is a normal topological space; this is the normality of every metric space.

Weak convergence of probability measures ${\mu }_n\to \mu$ on a metric space means thatfor every bounded continuous real function $f$. Weak convergence of random variables means weak convergence of their probability distributions; equivalently,for every bounded continuous $f$.

A computational problem in the SCI hierarchy is a quadrupleHere $\Omega$ is the primary set of inputs, $\Lambda$ is the set of permitted evaluation functions, $(\mathcal{M},d)$ is the output metric space, and $\Xi :\Omega \to \mathcal{M}$ is the problem function. The Solvability complexity index is defined from the minimum height of a tower of algorithms that computes $\Xi$ from finite subsets of $\Lambda$.

For the classical computational spectral problem, takeandSince the spectrum of a bounded operator is a nonempty compact subset of the complex numbers, one may take $\mathcal{M}$ to be the nonempty compact subsets of $\mathbb{C}$ with the Hausdorff distance. The Attouch--Wets topology gives an equivalent convenient formulation on bounded spectral sets and also extends naturally to unbounded closed sets. This choice of $\mathcal{M}$ makes convergence mean convergence of the whole spectrum as a set, including both the absence of persistent spectral pollution and the approximation of every genuine spectral point.

The closure $\overline{A}$ is the smallest closed subset of $M$ containing $A$, equivalently the intersection of all such closed subsets. In a metric space it is characterized byIf some ball misses $A$, its centre lies outside a closed superset of $A$; conversely the complement of any closed superset is an open neighbourhood disjoint from $A$. This verifies the equivalence of the two descriptions.

The Heine-Borel theorem says that a subset $K$ of finite-dimensional Euclidean space is compact exactly when it is both closed and bounded:Here compactness refers to the subspace topology induced by the usual Euclidean metric, and closedness and boundedness are measured in the ambient Euclidean space. This characterization does not hold for arbitrary metric spaces.

The points $(n,0)$ belong to the indicated subset for every positive integer $n$. They form an unbounded sequence, so this metric space is not bounded. Therefore it is not compact. Shrinking the vertical width does not bound its horizontal extent.

A metric space is path-connected if every pair $x,y$ can be joined by a path: a continuous map $\gamma :[0,1]\to X$ with $\gamma (0)=x$ and $\gamma (1)=y$.The path must remain in $X$, including at its endpoints.

A metric space is connected when it has no separation of a topological space: it cannot be written as the union of two disjoint nonempty open subsets. Equivalently, the only subsets that are both open and closed are the empty set and the whole space.The open sets here are those of the metric topology.

The sequence converges to $x\in X$ ifEquivalently $d(x_n,x)\to 0$. The limit must belong to the specified metric space, not just to a larger space containing it.

A sequence $(x_n)$ in a metric space $(X,d)$ is a Cauchy sequence ifThe index $N$ may depend on $\epsilon$, but must work for every pair of sufficiently late terms. No candidate limit is part of this definition.

Solutions to all six questions of the University of Oxford Part A metric space and complex analysis paper, sat on 14 June 2021. Source: released examination PDF, compilation pages 54–58; paper pages 1–5. The examination counted the best four answers; all six are covered here. The printed request in 6(c)(i) is impossible for the stated planar domain: its solution proves this and gives the precise map obtained when the missing point at infinity is accounted for.

A metric space $(X,d)$ is a complete metric space if every Cauchy sequence in $X$ converges to a point of $X$. Explicitly, the hypothesis isand the conclusion is that some $x\in X$ satisfies $d(x_n,x)\to 0$.

It is a sequentially compact space if every sequence in $X$ has a convergent subsequence whose limit of a sequence belongs to $X$.

It is a connected space if there are no disjoint nonempty open subsets $U,V$ of $X$ with $X=U\cup V$. Here openness refers to the metric topology of $X$.

It is a path-connected space if, for every $x,y\in X$, there is a continuous map $\gamma :[0,1]\to X$ with $\gamma (0)=x$ and $\gamma (1)=y$. Such a map is a path. Every path-connected space is a connected space: a separation would pull back along a path joining its two sides to a separation of the interval $[0,1]$.

Solutions to all six questions of the University of Oxford Part A metric space and complex analysis paper, sat on 12 June 2023. Source: released examination PDF, compilation pages 65–68; paper pages 1–4. The solutions retain the paper's labels. In 2(b)(i), the printed zero case $m\ne 0$ is inconsistent with its one case; the solution explicitly uses the intended $m\ne n$. The extra closing parenthesis in the printed definition of the supremum norm metric has no mathematical effect.

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.

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 metric space $X$ is a compact space if every open cover $X={\bigcup }_{\alpha \in I}{U}_{\alpha }$, with all $U_{\alpha }$ open in $X$, has a finite subcover: finitely many indices ${\alpha }_1,\dots ,{\alpha }_r$ cover $X$.

It is a sequentially compact space if every sequence $(x_n)$ in $X$ has a convergent subsequence $(x_{n_k})$ whose limit of a sequence belongs to $X$. In formulas,A finite subcover is the defining test for compactness; a convergent subsequence is the defining test for sequential compactness.

A metric space is a set $X$ with a function $d:X\times X\to [0,\infty )$ satisfying $d(x,y)=0⟺x=y$, symmetry and the triangle inequality. A subset $U$ is open if every $x\in U$ has some open open ball $B(x,r)$ contained in $U$.

For finitely many open sets $U_1,\dots ,U_k$ and $x$ in their intersection, choose radii $r_j>0$ with $B(x,r_j)\subseteq U_j$. The positive minimum radius gives a ball contained in the intersection. Thus every finite intersection is open; the empty intersection is $X$, also open.

For distinct points put $\delta =d(x_1,x_2)>0$. The balls $B(x_1,\delta /3)$ and $B(x_2,\delta /3)$ are disjoint: a common point would make $d(x_1,x_2)<2\delta /3$ by the triangle inequality. This proves the Hausdorff condition.

A singleton $\{x\}$ is closed because each $y\ne x$ has a ball of radius $d(x,y)/2$ avoiding it, making its complement open. Finite unions of closed sets are closed, as their complements are finite intersections of open sets. Hence every finite subset is closed, including the empty subset.

In a metric space, a Cauchy sequence has arbitrarily small distances between all sufficiently late terms:This condition compares terms to each other and does not assume that a limit exists in the space.