Topological analysis

Every continuous self-map of a closed disc has a fixed point.

If a continuous map $g:{\mathbb{R}}^n\to {\mathbb{R}}^n$ has $\Vert g(x)-x\Vert \le K$, then it is onto. For a target $y$, choose $R\ge K+\Vert y\Vert$, $R>0$, and map the unit ball to itself by $f(x)=x-(g(Rx)-y)/R$. A fixed point gives $g(Rx)=y$.

Let $V$ be a finite-dimensional real inner-product space and let $F:V\to V$ be continuous. If $⟨F(x),x⟩<0$ on the sphere $\Vert x\Vert =R$, then $F$ has a zero in the open ball. Otherwise the radial projection of $F$ gives a map of the ball that contradicts the Brouwer fixed-point theorem.

If a polynomial equation can be rearranged as $z=F(z)$ with a continuous $F$ mapping a closed disk into itself, Brouwer's theorem guarantees a root in that disk.

If each component of a continuous map on a box has opposite weak signs on the corresponding pair of faces, the map has a zero.

Projecting a sign-corrected displacement map back to a square turns Brouwer's fixed point into a zero of the displacement, proving that paths joining opposite side pairs intersect.

Triangulate a triangle and label every vertex by $1$, $2$, or $3$, forbidding label $i$ on the side opposite vertex $i$. Then the number of small triangles carrying all three labels is odd, and in particular at least one such triangle exists. Counting edges whose endpoint labels are $1$ and $2$ proves the parity statement: boundary incidences are odd, while a small triangle has odd incidence exactly when it has all three labels.

No closed disc retracts continuously onto its boundary. This is equivalent to Brouwer fixed-point theorem by the standard ray construction and the antipodal map.

A metric space is a set equipped with a nonnegative symmetric distance satisfying definiteness and the triangle inequality.

The supremum of pointwise distances defines a metric on maps when it is finite, for example if the target metric space is bounded. Convergence in this metric is uniform convergence. If the target is complete, Cauchy sequences have pointwise limits and the Cauchy estimate makes convergence uniform.

A Cauchy sequence in the uniform metric has a pointwise limit in the complete target, and the common Cauchy bound makes convergence uniform. The triangle inequality with one fixed continuous approximant shows that the limit is continuous. Thus the space of all continuous maps into a bounded complete metric space is complete, with no compactness assumption on the domain.

A subset is $r$-separated if distinct points have distance at least $r$. Such a subset is a packing by points; small balls around them have disjoint interiors at suitable radius. A large separated set supplies many distinguishable alternatives in statistical lower-bound arguments.

An inclusion-maximal separated set is one to which no further point of the ambient metric space can be added while keeping the same separation. It need not have the largest possible cardinality. In a finite space it can be obtained by repeatedly adding an admissible point.

If an inclusion-maximal separated set has separation at least $r$, every point of the ambient metric space is at distance strictly less than $r$ from a selected point. Otherwise that point could be added. Thus upper bounds on the sizes or measures of radius-$r$ balls give lower bounds on the size of the separated set.

In a metric space $(M,d)$, the closed ball centred at $x$ of radius $r\ge 0$ is $\{y\in M:d(x,y)\le r\}$. It is a closed set, since the distance to $x$ is a continuous function. Its boundary need not be the whole corresponding distance sphere in an arbitrary metric space.

An $\epsilon$-net of a metric space $(T,d)$ is a subset $U\subseteq T$ such that every $x\in T$ has $d(x,u)\le \epsilon$ for some $u\in U$. Its least possible finite size is the metric covering number with closed balls. A maximal set with pairwise distances greater than $\epsilon$ is an $\epsilon$-net.

A cover of the unit ball by balls of radius $\delta /2$ gives a radius-$\delta$ metric net on the unit sphere: discard balls missing the sphere and choose a sphere point in each remaining ball. Every sphere point lies within $\delta$ of the selected point in its ball. The argument keeps the number of balls and ensures that the selected points have unit length, regardless of the initial covering centers.

For a symmetric matrix $H$ and an $\eta$-metric net $U$ of the unit sphere, with $0<\eta <1/2$, approximate a maximizing unit vector $x$ by $u\in U$. Expanding $x^{\top }Hx-u^{\top }Hu=(x-u{)}^{\top }Hx+u^{\top }H(x-u)$ bounds its absolute value by $2\eta \Vert H{\Vert }_{\mathrm{op}}$. Rearrangement proves the bound. The volumetric bound for Euclidean metric nets limits the number of needed directions, allowing a union bound to control a random matrix.

A subset of the radius-$R$ Euclidean ball admits an $\epsilon$-metric net of size at most $(1+2R/\epsilon {)}^n$. Take a maximal separated set: its disjoint radius-$\epsilon /2$ Euclidean balls lie in the radius-$(R+\epsilon /2)$ ball. Comparing Lebesgue measures gives the estimate. In particular, the unit sphere admits a net of radius one with at most $3^n$ points.

For a metric space or pseudometric space $(T,d)$, this is the minimum number of balls of radius $\epsilon >0$, with centers in $T$, needed to cover $T$. The displayed grid bound uses closed balls. For open balls, replacing the radius by a fixed constant factor gives the same entropy estimates. For $T=[0,1]$ and $d(s,t)=c|s-t{|}^H$, a uniform grid gives $N(\epsilon ,T,d)\le 1+\lceil (c/\epsilon {)}^{1/H}\rceil$. These polynomial growth bounds make the Dudley entropy integral finite.

An $\epsilon$-net of a metric space is a set $G$ such that every point lies within distance $\epsilon$ of a point of $G$. A Cartesian grid of coordinate spacing $\eta$ is a finite $\sqrt{d}\eta$-net of a bounded cube in the Euclidean norm. A Lipschitz bound transfers control on the grid to control over the whole cube.

Suppose $|Z(u)-Z(v)|\le L\Vert u-v\Vert$ with a random finite $L$. On $L\le \ell$, a finite net with radius $\eta$ bounds the full supremum by the grid maximum plus $\ell \eta$. The Markov inequality controls the exceptional event using $\mathbb{E}L$, and a union bound controls the grid maximum. Choosing the net spacing balances these two costs.

For a continuous path $\gamma :[a,b]\to X$ in a metric space, take the supremum over finite partitions $a=t_0<\cdots <t_n=b$ of ${\sum }_{i=1}^{n}d(\gamma (t_i),\gamma (t_{i-1}))$. This definition allows $+\infty$ and needs no derivative. A metric geodesic parametrized on $[a,b]$ has length $b-a$. In a Riemannian manifold, it agrees with the usual arc length integral for a smooth curve.

A metric space in which every two points are joined by a metric geodesic segment. Equivalently, they admit a continuous path whose metric path length equals their distance. In this general setting no Riemannian metric or covariant derivative is required.

A unit-speed metric geodesic in a metric space is an isometric embedding $\gamma :I\to X$ of a real interval. Every restricted segment realizes the distance between its endpoints. A one-point interval permits a constant segment. In a unit-edge graph, shortest edge paths parametrized by metric path length are metric geodesics; no smooth structure is required. This is stronger than the differential definition of a geodesic in a Riemannian manifold, which need only minimize locally: on a unit circle, an arc of angle $3\pi /2$ is a Riemannian geodesic but its endpoint distance is only $\pi /2$.

A metric geodesic triangle consists of three points of a geodesic metric space and a chosen metric geodesic segment between each pair. Multiple choices are allowed, and coincident vertices give degenerate triangles. A Gromov-hyperbolic metric space has a uniform bound on the distance from each side to the other two sides for every such choice. This definition applies to Cayley graphs and trees without tangent vectors or curvature assumptions; it does not invoke the surface angle formula for a geodesic triangle.

The metric topology consists of the sets $U$ such that every $x\in U$ contains some open ball $B(x,r)$ lying in $U$.

A subset $A$ of a metric space is bounded if it lies in some finite-radius ball: there are $x$ and $R<\infty$ such that $d(x,a)\le R$ for every $a\in A$.

For a point $x$ and a nonempty closed set $A$ in a metric space, the distanceis a continuous function of $x$. It vanishes exactly on $A$.

A metric on a set $X$ is a function $d:X\times X\to [0,\infty )$ that is symmetric, satisfies the triangle inequality, and obeys $d(x,y)=0$ exactly when $x=y$.

A metric is compatible with a given topology when its open balls generate exactly that topology. Compatible metrics on a compact topological space are uniformly equivalent: the identity maps between the resulting compact metric spaces are uniformly continuous. Consequently they define the same proximality and the same asymptotic-pair relation for a fixed continuous map.

This metric identifies $\mathbb{R}$ isometrically with the open interval $(-\pi /2,\pi /2)$ in the usual real distance. It induces the usual topology but is incomplete: the integer sequence has transformed limit $\pi /2$ outside that interval. This illustrates that completeness depends on the metric, not solely on its topology.

The discrete metric sets $d(x,x)=0$ and $d(x,y)=1$ for distinct points. It induces the discrete topology, and its Cauchy sequences are exactly the eventually constant sequences.

A pseudometric satisfies the axioms of a metric except that distinct points may have distance zero.

An ultrametric is a metric satisfying the strong triangle inequality

A compact metric space is a metric space that is compact. Equivalently, it is complete and totally bounded.

A metric space is compact exactly when it is both complete and totally bounded. For the nontrivial direction, successively choose a point from a nested sequence of finite $2^{-n}$-nets; a diagonal construction gives a Cauchy subsequence of every sequence, and completeness gives its limit.

A metric space is totally bounded when, for every $\epsilon >0$, it is covered by finitely many open balls of radius $\epsilon$.

A sequence $(x_n)$ converges to $x$ in a metric space when $d(x_n,x)\to 0$, equivalently when every neighbourhood of $x$ contains all sufficiently late terms.

A map between metric spaces is continuous exactly when $x_n\to x$ always implies $f(x_n)\to f(x)$. If an inverse image of an open set were not open, points outside it could be chosen within distance $1/n$ of a point inside it, proving the converse by contradiction.

Every sequence in a compact metric space has a convergent subsequence. Conversely, sequential compactness and compactness are equivalent for metric spaces.

Every metric space is normal. For disjoint closed sets $A,B$, the functionis continuous, equals zero on $A$, and equals one on $B$; inverse images of disjoint neighbourhoods of zero and one separate the sets.

A metric space $X$ is compact exactly when every continuous function $X\to \mathbb{R}$ is bounded. For the converse, a noncompact metric space contains a closed discrete sequence; the unbounded function assigning its $n$th point the value $n$ extends to $X$ by the Tietze extension theorem.

A metric space is complete when every Cauchy sequence converges to a point of the space.

A metric satisfies $d(x,z)\le d(x,y)+d(y,z)$. A norm satisfies the corresponding inequality $\Vert u+v\Vert \le \Vert u\Vert +\Vert v\Vert$.

In a metric space, applying the triangle inequality in both directions gives $|d(x,z)-d(y,z)|\le d(x,y)$. In a normed vector space, its corresponding form is $|\Vert x\Vert -\Vert y\Vert |\le \Vert x-y\Vert$. In particular, distance to a fixed point is a Lipschitz continuous real function.

For four points $x_1,x_2,y_1,y_2$ in a metric space,Both steps follow from the triangle inequality. The first also holds with an absolute value around the difference, by the reverse triangle inequality. This transports a difference of distances measured from one point into a sum measured from another.

For an integrable function $f$ with values in a finite-dimensional normed vector space,For the Euclidean norm, pair the integral with a unit vector in its direction and apply the Cauchy-Schwarz inequality pointwise.

For nonzero complex numbers $x,y$, equality $|x+y|=|x|+|y|$ holds exactly when $x/y$ is a positive real number, equivalently when the two numbers have the same complex argument modulo $2\pi$.

For every integrable real or complex function,

The Euclidean distance between $x,y\in {\mathbb{R}}^d$ is

The diameter of a subset $S$ of a metric space is

Uniform continuity requires one input tolerance to work at every point of the domain for a chosen output tolerance.

Choose the uniform continuity input tolerance corresponding to output tolerance one. A totally bounded domain is covered by finitely many balls of smaller radius centred in the domain. Every function value differs by less than one from the value at one of their finitely many centres. Taking the largest centre value bounds the whole function. Compactness or closedness of the domain is unnecessary.

A product of two uniformly continuous functions need not have uniform continuity on an unbounded domain. The functions $f(x)=g(x)=x$ give the counterexample $x^2$, since inputs $n$ and $n+1/n$ approach each other while their squared values remain separated. For bounded factors the estimate $|f(x)g(x)-f(y)g(y)|\le \Vert f{\Vert }_{\infty }|g(x)-g(y)|+\Vert g{\Vert }_{\infty }|f(x)-f(y)|$ restores uniform continuity.

If a uniformly continuous function $u:{\mathbb{R}}^d\to \mathbb{R}$ belongs to Lp space for some $1\le p<\infty$, then $u(x)\to 0$ as $|x|\to \infty$. Otherwise choose points escaping to infinity with $|u(x_j)|\ge 2\theta >0$. Uniform continuity gives one radius $\rho >0$ on which $|u|\ge \theta$ around every $x_j$. Extract disjoint such balls. Their contributions to $\int |u{|}^p$ are each at least ${\theta }^p|B_{\rho }|$, a contradiction. The finite-$p$ condition is essential: the constant function one lies in $W^{1,\infty }$ and does not vanish at infinity.

Every continuous $f:\mathbb{R}\to \mathbb{R}$ can be approximated with pointwise convergence by uniformly continuous functions. Set $c_n(x)=\max (-n,\min (x,n))$ and $f_n=f\circ c_n$. The map $c_n$ is Lipschitz continuous with constant one and $f{|}_{[-n,n]}$ is uniformly continuous by the Heine-Cantor theorem, so $f_n$ is uniformly continuous on the whole line. On every fixed bounded interval, $f_n=f$ eventually.

The modulus of continuity of a function $f$ isIt quantifies uniform continuity by recording the largest oscillation of $f$ across pairs of points at distance at most $\delta$.

Set $\mu (0)=0$ and extend $\mu (r)=1$ for $r\ge 1$. This positive, nondecreasing modulus of continuity is weaker than a linear Lipschitz continuity bound but still has ${\int }_0^{1}dr/\mu (r)=\infty$. At small distances it is equivalent to $r\log (1/r)$. A signed $r\log r$ cannot be an upper modulus at distances less than one. The planar vorticity velocity kernel maps $L^1\cap L^{\infty }$ into velocity fields with this modulus.

Every continuous map from a compact metric space to a metric space is uniformly continuous.

A uniform limit of uniformly continuous maps between metric spaces is uniformly continuous. Approximate the limit at both endpoints by one fixed member of the sequence and use its uniform continuity between them.

A function between metric spaces is uniformly continuous exactly when $d(x_n,y_n)\to 0$ implies $d(f(x_n),f(y_n))\to 0$ for every two sequences $(x_n)$ and $(y_n)$. Failure of uniform continuity supplies pairs with input distance below $1/n$ but output distance bounded below.

Two metrics are equivalent when they induce the same topology. Global two-sided constant bounds imply equivalence, but equivalence alone need not supply such bounds.

On $\mathbb{R}$, the usual metric and $|\arctan x-\arctan y|$ induce the same topology, but their ratio along $(0,n)$ tends to zero. Thus no positive global lower comparison constant exists.

Topological equivalence does not determine uniform convergence. For the usual and arctangent metrics on $\mathbb{R}$, maps that change the value $n$ to $2n$ at a single domain point have arctangent sup-error tending to zero while their usual sup-error diverges.

For a metric space, $d(x,S)={\inf }_{s\in S}d(x,s)$ is $1$-Lipschitz and vanishes on the closure of $S$.

A point $x$ of a metric space is isolated when some open ball about $x$ contains no other point of the space, equivalently when the singleton $\{x\}$ is open.

A metric space is complete when every Cauchy sequence converges to a point of the space.

A metrizable space is topologically complete when some metric inducing its topology is complete. The given metric need not be complete: the open interval $(0,1)$ is an example. A Polish space additionally requires separability.

A subspace of a complete metric space admits a compatible complete metric if and only if it is a G-delta set. For the reverse direction, adjoining the coordinates $1/d(x,X\setminus U_n)$ to the ambient metric prevents Cauchy sequences from approaching any excluded closed boundary. For the forward direction, small-diameter relative open covers in a compatible complete metric produce the ambient open sets.

A Polish space is a topological space whose topology can be induced by a complete separable metric. The completeness requirement concerns some compatible metric; it need not hold for every metric inducing the topology. Euclidean spaces are examples.

A metric space is complete if and only if every decreasing sequence of nonempty closed sets whose diameters tend to zero has nonempty intersection. In that case the intersection contains exactly one point.

Every complete metric space is a Baire space: every countable intersection of open dense subsets is dense. Equivalently, no nonempty open subset is a countable union of nowhere-dense subsets.

To prove the dense-intersection form, let $V$ be nonempty and open and let $G_1,G_2,\dots$ be open dense sets. Successively choose closed ballsstarting with a ball contained in $V\cap G_1$. The centres form a Cauchy sequence. Its limit belongs to every closed ball, and hence to $V\cap {\bigcap }_n{G}_n$.

A set containing a countable intersection of open dense subsets. The intersection itself is also called a residual set. In a Baire space it is dense. This topological notion of genericity does not by itself assert full measure.

A dense additive subgroup of a complete metrizable topological group cannot be proper if it is comeagre. Every translate is comeagre and meets the subgroup, so every translating element is a difference of two subgroup elements. Applied to a topologically complete normed space inside its completion, this proves that its original norm is complete.

No infinite-dimensional Banach space has a countable Hamel basis. If $(e_n)$ were such a basis, the space would be the countable union of the finite-dimensional subspaces $\mathrm{span}(e_1,\dots ,e_n)$. Each is closed and has empty interior, contradicting the Baire category theorem.

A nowhere differentiable continuous function is continuous at every point and has no finite derivative anywhere. A Baire-category argument shows that such functions form a dense subset of $C([0,1])$ in the uniform norm.

A Baire space is a topological space in which every countable intersection of open dense sets is dense, equivalently one in which no nonempty open set is a countable union of nowhere-dense sets. Every compact Hausdorff space and every complete metric space is a Baire space.

A Baire function belongs to the smallest hierarchy of real-valued functions containing the continuous functions and closed under pointwise sequential limits.

A Baire class one function is a pointwise limit of continuous functions. On a complete metric domain, its set of discontinuities is meagre; in particular, it has a point of continuity on every nonempty perfect closed subset.

The indicator of the rational numbers in the real numbers is discontinuous everywhere because both the rationals and irrationals are dense. The discontinuity theorem for Baire class one functions therefore shows that it is not Baire class one.

Let $S$ be a nonempty closed, convex, symmetric subset of a Banach space and suppose ${\bigcup }_{n\ge 1}nS=X$. Baire's theorem puts a ball $B(x,r)$ inside some $nS$. Symmetry also puts $B(-x,r)$ there, and convexity puts the midpoint of $x+z$ and $-x+z$, namely $z$, in $nS$ whenever $\Vert z\Vert <r$. Hence $B(0,r/n)\subset S$.

Convexity is essential. In $\mathbb{R}$, letThis set is closed, symmetric, and has gaps arbitrarily near zero. Given $x\ne 0$, choose $k$ so small-scale that the interval
$[|x|/(2\cdot 4^{-k}),|x|/4^{-k}]$ has length at least one; it contains an integer $n$, and then $x/n\in S$. Thus ${\bigcup }_{n}nS=\mathbb{R}$, although $S$ is not a neighbourhood of zero.

Every nonempty countable complete metric space has an isolated point. Otherwise it is the countable union of its singleton subsets, each of which is nowhere dense, contradicting the Baire category theorem.

If such a space has infinitely many points, it has infinitely many isolated points. Indeed, after deleting any finite collection of isolated points, the remaining set is closed, complete, countable, and nonempty. It therefore has a point isolated in the remaining space; because the deleted set is finite, that point is also isolated in the original space.

A subset $A$ of a topological space is nowhere dense when the interior of its closure is empty. Equivalently, every nonempty open set contains a nonempty open subset disjoint from $A$.

A meagre, or first-category, set is a countable union of nowhere dense sets.

An uncountable subset of the real numbers with countable intersection with every meagre set. This is the category counterpart of a Sierpiński set.

The complement of a comeagre set is a meagre set, a countable union of nowhere dense sets. A dense G-delta set is comeagre. The Baire category theorem ensures that two comeagre subsets of a nonempty complete metric space cannot be disjoint.

A Lusin set is an uncountable subset $A$ of the real numbers such that $A\cap M$ is countable for every meagre set $M$. Every meagre subset has a meagre $F_{\sigma }$ cover, so testing those covers suffices. The Lusin set construction under CH gives an example. A Lusin set has no uncountable meagre subset of itself.

Under the Continuum hypothesis, enumerate the meagre $F_{\sigma }$ covers by ${\omega }_1$. At stage $\alpha$ choose a new real outside all earlier covers and the previous choices. The excluded union is meagre, so the Baire category theorem makes the choice possible. Each cover then contains only countably many selected points. Arbitrary meagre sets need not themselves have only continuum many possible choices; the coded covers are what is enumerated.

In $C([0,1])$ with the uniform norm, the continuous functions that are monotone on some interval of positive length form a meagre set. Indeed, it is enough to use intervals with rational endpoints. For each such interval, the nondecreasing and nonincreasing functions form closed sets with empty interior: a sufficiently small local triangular perturbation breaks the relevant inequality. The Baire category theorem therefore shows that a dense set of continuous functions is monotone on no nontrivial interval.

Let $f\in C^{\infty }(\mathbb{R})$ and suppose that for every $x$ there is an $n=n(x)\ge 0$ such that $f^{(n)}(x)=0$. Then $f$ is a polynomial.

If an entire function $f$ is not a polynomial, then there is a point $z_0\in \mathbb{C}$ for which $f^{(n)}(z_0)\ne 0$ for every nonnegative integer $n$. Consequently every coefficient in the Taylor series of $f$ about $z_0$ is nonzero.

For each $n$, the function $f^{(n)}$ is not identically zero, since otherwise $f$ would be a polynomial. Its zero set is closed and has empty interior by the identity theorem, hence is a nowhere dense set. The Baire category theorem says that the complete metric space $\mathbb{C}$ cannot be the union of these countably many zero sets.

Set $S_n=\{x:f^{(n)}(x)=0\}$ and $E_n=\mathrm{int}{S}_n$. The sets $S_n$ are closed, $E_n\subset E_{n+1}$, and Baire's theorem on each interval shows that $\Omega ={\bigcup }_n{E}_n$ is dense. On every connected component of $\Omega$, compactness and the increasing cover by the $E_n$ show that one derivative vanishes locally with a uniform finite order, so $f$ agrees there with one polynomial.

If $F=\mathbb{R}\setminus \Omega$ were nonempty, it would be closed and have no isolated points: polynomial pieces on both sides of an isolated point would have matching derivatives of every order and hence join into one polynomial piece. Applying Baire's theorem to the cover $F={\bigcup }_n(F\cap S_n)$ gives an interval $U$ and an index $k$ for which $F\cap U$ is nonempty and contained in $S_k$. Because $F$ has no isolated points, difference quotients show that every derivative of order at least $k$ vanishes on $F\cap U$. Each polynomial component of $\Omega$ meeting $U$ has an endpoint in $F\cap U$, so its degree is less than $k$. Thus $f^{(k)}=0$ throughout $U$, contradicting $F\cap U\ne \varnothing$. Hence $\Omega =\mathbb{R}$, and its sole connected component carries one polynomial.

The space $C^{\infty }([0,1])$ becomes a complete metric space underConvergence in this metric is exactly uniform convergence of every derivative separately.

If $(f_n)$ is Cauchy in the smooth-function metric, each derivative sequence $(f_n^{(r)})$ converges uniformly to a continuous function $g_r$. Passing to the limit inshows that $g_r^{\prime }=g_{r+1}$. Hence $g_0$ is smooth and $f_n\to g_0$ in the metric.

Outside a meagre set in $C^{\infty }([0,1])$, every rational $q\in (0,1)$ and every positive integer $M$ admit $m\ge M$ such thatFor fixed $q,M$, the union over $m\ge M$ of these strict-inequality sets is open. It is dense because a perturbation $\delta \cos (K(x-q)-m\pi /2)$ can make the $m$th derivative arbitrarily large while keeping any prescribed finite collection of lower derivatives arbitrarily small. The Baire category theorem applied over the countable pairs $(q,M)$ gives the claim.

The generic derivative bound implies that at every rational $q\in (0,1)$,The Cauchy-Hadamard theorem therefore gives radius zero to the Taylor series at every such $q$. If a Taylor series represented $f$ on a neighborhood of any point, that neighborhood would contain a rational point at which $f$ was real analytic, a contradiction.

A closed subspace of a complete metric space is complete. A Cauchy sequence in the subspace converges in the ambient space, and closedness keeps its limit in the subspace.

For nonempty bounded subsets of a metric space,This is a metric on the nonempty closed bounded subsets. Without closedness it is only a pseudometric: a set and its closure have distance zero.

The map $x\mapsto \{x\}$ is an isometry into the hyperspace of nonempty closed bounded subsets. Its image is closed because a Hausdorff limit of singletons has diameter zero and is therefore a singleton.