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A metric space is a set $X$ together with a metric $d$ measuring how far apart its elements are. The elements can be positions, functions, or other objects: what matters is that the metric obeys the same basic rules. This lets one discuss convergence in a metric space, continuous maps, and Cauchy sequences without first assigning Cartesian coordinates. The useful abstraction is that many arguments require only comparisons of distances, rather than the particular objects being compared.

Even the same set can support several informative metrics. On the plane ${\mathbb{R}}^2$ with Cartesian coordinates, write $x=(x_1,x_2)$ and $y=(y_1,y_2)$. Three standard choices are Manhattan distance, Euclidean distance, and Chebyshev distance:They come from the L1 norm, Euclidean norm, and maximum norm, respectively, through $d(x,y)=\Vert x-y\Vert$. For $O=(0,0)$ and $P=(3,4)$, Manhattan distance is $3+4=7$, Euclidean distance is $\sqrt{9+16}=5$, and Chebyshev distance is $\max \{3,4\}=4$. All three answers are correct for their respective metrics.

The choice encodes the question being asked. Manhattan distance models travel along a rectangular street grid. Euclidean distance measures unrestricted travel along a line segment. Chebyshev distance measures the time needed when two Cartesian coordinates can change simultaneously, each at speed at most one. Thus choosing a metric is part of modelling, rather than a consequence of knowing the endpoints. These examples have the same metric topology, but different metric path lengths and different collections of metric geodesics.

A historical reference for the general study of metric spaces is Maurice Fréchet's 1906 paper Sur quelques points du calcul fonctionnel, published in volume 22 of the Rendiconti del Circolo Matematico di Palermo. ^[ref]

For any estimator $T$ in a metric parameter space, $d(T,f{)}^2+d(T,g{)}^2\ge d(f,g{)}^2/2$ by the triangle inequality. Average the two risk functions and replace their probability density functions by their minimum. This proves $\max (R_f,R_g)\ge d(f,g{)}^2\int \min (p_f,p_g)/4$. The overlap equals one minus total variation distance. The same proof applies to squared error for a real-valued statistical functional of the parameter, without requiring that functional to be injective.

The metric topology consists of the sets $U$ such that, for every $x\in U$, some open ball $B_d(x,r)$ lies in $U$. Thus open sets describe which sufficiently small changes remain inside a region; they do not specify the numerical size of those changes.

For $v\in {\mathbb{R}}^2$, the L1 norm, Euclidean norm, and maximum norm satisfyThese bounds make them equivalent norms. For example, $B_{\infty }(a,r/2)\subseteq B_1(a,r)\subseteq B_{\infty }(a,r)$: every open ball contains a sufficiently small open ball for either other metric. Hence all three define the same open sets, convergent sequences, and continuous maps. More generally, any two norms on a real or complex finite-dimensional vector space are equivalent norms; see H. A. Priestley, Functional Analysis I, §§0.5–0.7 and Theorem 4.3.

The metric topology therefore does not determine numerical distances. Multiplying a metric by $100$ preserves its open sets while multiplying every distance by $100$. Nor does the general equivalence of metrics require constant-factor bounds: that stronger comparison holds for the three norms used here.

If two parameters are separated by at least $2r$ in a metric, any estimator gives a test by deciding whether its distance from the first parameter is below $r$. For $0<\eta <1$, the event $|Z-1|\le \eta$ and Markov inequality give the displayed minimax risk bound, where $Z$ is the product likelihood ratio. The unrestricted claim for every positive $\eta$ is false when both factors on the right are negative.

A topological vector space is metrizable when its topology is induced by a metric. At the origin it then has a countable neighborhood basis, so continuity of a linear map can be tested on sequences.

For measurable functions $f,g$ with values in the real numbers or complex numbers on the same measure space $(X,\mathcal{A},\mu )$, and $1\le p\le \infty$, membership in the same Lp space givesThus adding two functions cannot make their combined Lp norm exceed the sum of their separate Lp norms. The measure need not be finite or a probability measure. The hypotheses require finite Lp norms; $p=\infty$ uses the essential supremum of the absolute value, not an ordinary pointwise supremum. For complex numbers, the absolute value is the complex modulus. As usual, Lp spaces identify functions equal almost everywhere. ^[ref]

The Minkowski inequality makes the Lp norm a workable notion of size: it supplies the triangle inequality needed to turn $d_p(f,g)=\Vert f-g{\Vert }_p$ into a metric on Lp space. Applied to $(f-h)+(h-g)$, it says that travelling through an intermediate function $h$ cannot shorten the direct distance measured by this metric. Applied to two error functions, it gives a bound on the total error even when the errors reinforce one another; Minkowski inequality for error aggregation gives explicit examples.

The geometry is already visible in the Euclidean plane. Under counting measure on two coordinates, the closed unit balls for the L1 norm, Euclidean norm, and L-infinity norm are respectively $|x|+|y|\le 1$, $x^2+y^2\le 1$, and $\max (|x|,|y|)\le 1$. Every one is a convex set: if $u,v$ have Lp norm at most one, the Minkowski inequality and absolute homogeneity of a norm give $\Vert tu+(1-t)v{\Vert }_p\le t+(1-t)=1$ for $0\le t\le 1$. The Euclidean norm has no flat boundary segments, whereas the other two norms do. This is the distinction captured by a strictly convex normed space and developed in equality in Minkowski inequality.

A short Hölder's inequality proof outline handles $1<p<\infty$. First $|f+g{|}^p\le 2^{p-1}(|f{|}^p+|g{|}^p)$, so $f+g$ has finite Lp norm before any division. With conjugate exponents $p$ and $q=p/(p-1)$, set $h=|f+g{|}^{p-1}$; then $\Vert h{\Vert }_q=\Vert f+g{\Vert }_p^{p-1}$. The scalar triangle inequality followed by Hölder's inequality givesDivide if $\Vert f+g{\Vert }_p>0$; otherwise the desired bound is immediate. For $p=1$, integrate $|f+g|\le |f|+|g|$. For $p=\infty$, the almost everywhere bounds $|f|\le \Vert f{\Vert }_{\infty }$ and $|g|\le \Vert g{\Vert }_{\infty }$ give the result by the essential supremum. The same scalar triangle inequality works for real numbers and complex numbers.

Let $a$ be an integer not divisible by $p$. Its modular inverse modulo $p^n$ has a unique least nonnegative representative $r_n$. Reducing the inverse equation modulo $p^n$ shows that the representatives are compatible. Since $a$ is a p-adic unit,so the sequence converges to the inverse $a^{-1}$ in ${\mathbb{Z}}_p$. This computes a fraction through finite integer operations.

For $a=3$ and $p=2$, the first eight representatives are $1,3,3,11,11,43,43,171$. Each satisfies $3r_n\equiv 1(\mathrm{mod}{2}^n)$. The resulting expansion isThe geometric series has ratio $4$, whose 2-adic absolute value is $1/4$, and its sum is $1+2/(1-4)=1/3$. In the ordinary real number absolute value, the representatives grow. Different metrics impose different meanings of a good approximation. The periodic digits here are a specific instance of the rational expansions studied in Conrad's authored note.^[ref]

A normed vector space is a vector space equipped with a norm, whose induced metric is $d(x,y)=\Vert x-y\Vert$. The norm must be positive on nonzero vectors, scale by the absolute value of a scalar, and satisfy the triangle inequality. These conditions make distance compatible with addition and scaling.

On $C([0,1])$, both the supremum norm $\Vert f{\Vert }_{\infty }={\max }_{0\le x\le 1}|f(x)|$ and the L2 norm $\Vert f{\Vert }_2=({\int }_0^1|f(x){|}^{2}dx{)}^{1/2}$ are genuine norms. A nonzero continuous function stays nonzero on an interval of positive length, so its squared integral cannot vanish. For $f(x)=x$, they give $1$ and $1/\sqrt{3}$, respectively. Always $\Vert f{\Vert }_2\le \Vert f{\Vert }_{\infty }$, but the shrinking peaks below show that no uniform reverse bound exists. Unlike the finite-dimensional equivalence of norms, infinite-dimensional vector spaces can support inequivalent norm topologies.

Fix a prime number $p$. Write a nonzero rational number as $x=p^{k}a/b$, where $k\in \mathbb{Z}$ and the nonzero integers $a,b$ are not divisible by $p$. The exponent $k$ is uniquely determined. Its P-adic valuation is this signed exponent $v_p(x)=k$: equivalently, for any nonzero integer numerator $A$ and denominator $B$,Factors of $p$ in the numerator decrease the p-adic absolute value; factors in the denominator increase it. For $p=2$ the computations are:

This is an absolute value on a field measuring divisibility, rather than the magnitude on a real number number line. It satisfiesThe proof comes directly from P-adic valuations: exponents add in a product of numbers, and extracting a common power of $p$ from a sum gives $v_p(x+y)\ge \min \{v_p(x),v_p(y)\}$. If the two P-adic valuations differ, the remaining term with smaller exponent cannot cancel modulo $p$, so equality holds. Thus the p-adic absolute value is a Non-Archimedean absolute value. ^[ref]

The metric $d_p(x,y)=|x-y{|}_p$ makes two rational numbers close when their difference contains a high power of $p$. This packages successive modular congruences into a numerical error estimate, allowing convergence in a metric space and convergent series to encode exact arithmetic information.

For a prime number $p$, a p-adic number is an element of the field ${\mathbb{Q}}_p$ obtained by completing the rational numbers for the metric $d_p(x,y)=|x-y{|}_p$. Here $|\cdot {|}_p$ is the p-adic absolute value. The Cauchy-sequence construction of a metric completion supplies limits for every Cauchy sequence, and addition, multiplication, and the p-adic absolute value extend to this complete metric space. The rational numbers form a dense subset. ^[ref]

For a concrete example with $p=2$, consider the partial sums of a geometric series:The first values $1,3,7,15,31$ have distances $1/2,1/4,1/8,1/16,1/32$ from $-1$ in the 2-adic absolute value. ConsequentlyThis is a statement about convergence in a metric space with the 2-adic absolute value. In the ordinary real number absolute value, the same partial sums grow without bound. Choosing a metric changes which approximation errors shrink.

More generally every nonzero P-adic number has a unique digit expansionOnly finitely many negative powers occur, while positive powers can continue indefinitely. Written with the highest powers on the left, the digits extend infinitely to the left; for example the P-adic number $-1$ at $p=2$ is written $\dots 111_2$. The p-adic integers are exactly the elements with $|x{|}_p\le 1$, whose digit expansions use only nonnegative powers. ^[ref]

The construction can be understood through modular arithmetic. Successively choose the low-order digits to agree modulo $p$, $p^2$, and higher powers. After fixing digits below $p^n$, the remaining error has p-adic absolute value at most $p^{-n}$; the partial sums form a Cauchy sequence, so completeness supplies a limit. If two expansions first differ at position $j$, their difference has P-adic valuation $j$, proving uniqueness. This turns compatible finite congruence classes into one exact number. In particular, Hensel's lemma upgrades suitable approximate roots of a polynomial modulo powers of $p$ into roots in the p-adic integers.

Complete the square in the real-valued complex plane equation:Dividing by $\alpha$ gives center and radiusIf the radicand is zero, the locus is the single point $c$, a degenerate circle. If it is negative, the locus is empty. The absolute value of $\alpha$ is essential even when $\alpha <0$.

For the requested sketch, the five centers and radii are respectively $(-1/2,\sqrt{3}/2)$, $(-1,\sqrt{2})$, $(-2,\sqrt{6})$, $(2,\sqrt{2})$ and $(1,0)$, all centers on the real axis.

The original PDF's map is $f(z)=(\beta /\bar{\beta })\bar{z}$; the converted TeX drops the conjugate on its denominator. Write $\beta =|\beta |e^{i\theta }$, so $f(z)=e^{2i\theta }\bar{z}$. This is the reflection symmetry of a complex circle equation. It fixes $c=\beta /\alpha$ and satisfiesThus it maps every point of the circle into the same circle. Conversely, for any point $w$ on that circle, $z=f(w)$ also lies on it and $f(z)=w$. This proves $f(C)=C$, including the surjectivity required by the question.

For any three nonempty circles, fix $u\in C_3$. The triangle inequality gives, for every $z\in C_1,w\in C_2$,Taking the maximum over $z,w$ proves the requested inequality.

For centers $c_1,c_2$ and radii $r_1,r_2$, the maximum separation of two circles is bounded above by $|c_1-c_2|+r_1+r_2$. If the centers differ, choose $z=c_1+r_{1}v$ and $w=c_2-r_{2}v$, where $v=(c_1-c_2)/|c_1-c_2|$, to attain the bound. If the centers coincide, any unit $v$ gives opposite points attaining $r_1+r_2$. ThereforeThe function $m$ is not a metric on positive-radius circles, since $m(C,C)=2r$, even though it satisfies this triangle-type inequality.

Assume $a<b$. On continuous functions on this compact interval, both proposed distances are finite and symmetric, and vanish on identical functions. For the supremum norm, zero distance plainly implies pointwise equality. For the L2 norm, if a continuous difference is nonzero at any point, it stays bounded away from zero on a subinterval of positive length. Its squared integral is then positive, so zero distance also implies equality.

For the first metric, the pointwise triangle inequality followed by the supremum gives $d_1(x,z)\le d_1(x,y)+d_1(y,z)$. For the second, put $u=x-y$, $v=y-z$. The Cauchy-Schwarz inequality givesTaking square roots proves the required triangle inequality. Thus both are metrics.

The continuous-function space is complete in the uniform metric. If $(x_n)$ is a Cauchy sequence in $d_1$, at each $t$ its scalar values have a limit $x(t)$, by completeness of the scalar field. Given $\epsilon >0$, choose $N$ with $\Vert x_n-x_m{\Vert }_{\infty }<\epsilon$ for $m,n\ge N$. Letting $m\to \infty$ gives $|x_n(t)-x(t)|\le \epsilon$ uniformly in $t$. Hence there is uniform convergence. To see that the limit is continuous at $t_0$, approximate it uniformly by one continuous $x_n$ and useThe outer terms can be made small uniformly, and the middle term is small near $t_0$. This proves the needed uniform limit theorem and completeness.

For the second metric, let $H$ be zero on $(-1,0)$ and one on $(0,1)$; its value at zero is immaterial to the integral. The supplied ramp functions are continuous and satisfyAlthough $H$ is not in the continuous-function space, the integral triangle inequality yields $d_2(x_m,x_n)\le (3m{)}^{-1/2}+(3n{)}^{-1/2}\to 0$. Thus the sequence is Cauchy. If it converged in $d_2$ to a continuous $x$, the same inequality would give $\Vert x-H{\Vert }_2=0$. Continuity would then force $x=0$ on the negative half-interval and $x=1$ on the positive half-interval, contradicting continuity at zero. Therefore the continuous-function space is not complete in $d_2$; its limit in the larger L2 space is the discontinuous step.