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If a sequence in a metric space has ${\sum }_{j=1}^{\infty }d(x_{j+1},x_j)<\infty$, the triangle inequality gives $d(x_m,x_n)\le {\sum }_{j=n}^{m-1}d(x_{j+1},x_j)$ for $m>n$. The tail of this convergent series tends to zero and bounds every later pair, proving the Cauchy sequence condition. In particular, an increment bound $d(x_{j+1},x_j)\le C{r}^j$ with $C>0$ and $0<r<1$ gives $d(x_m,x_n)\le C{r}^n/(1-r)$ by the geometric series. Unlike checking a small latest step, this bounds all remaining movement.

This sufficient condition is not necessary. The real sequence $x_n=(-1{)}^n/n$ converges to zero and is therefore a Cauchy sequence, although $|x_{n+1}-x_n|=1/(n+1)+1/n$ has a divergent sum. Cancellation allows convergence despite infinite accumulated absolute values of the successive changes.

Let $X$ be a complete metric space and let $F_n$ be nonempty closed bounded subsets with $F_{n+1}\subseteq F_n$ and Hausdorff distance $d_H(F_n,F_{n+1})\le a_n$, where $a_n\ge 0$ and ${\sum }_n{a}_n<\infty$. Then $G={\bigcap }_n{F}_n$ is nonempty and closed, and $d_H(F_n,G)\le {\sum }_{k=n}^{\infty }{a}_k\to 0$.

To prove this, start with any $x_n\in F_n$ and choose $x_{k+1}\in F_{k+1}$ with $d(x_k,x_{k+1})<a_k+\eta 2^{n-k-1}$. The triangle inequality makes the resulting sequence a Cauchy sequence. Its limit exists by completeness and belongs to every $F_j$ by nesting and closedness. Its distance from the starting point is at most ${\sum }_{k=n}^{\infty }{a}_k+\eta$. Let $\eta \downarrow 0$ and take the supremum over starting points. The reverse directed distance is zero since $G\subseteq F_n$. Unlike the usual shrinking-diameter form of Cantor's intersection theorem, the sets need not have diameters tending to zero.

Inside both convergence discs, the triangle inequality gives absolute convergence of the sum. Suppose $R<S$ and the sum had radius $T>R$. At any modulus between $R$ and $\min (T,S)$, subtracting the absolutely convergent second series from the absolutely convergent sum would give absolute convergence of the first series outside its radius of convergence, a contradiction. Interchanging the two series covers $S<R$.

For $p_u\in \partial J(u)$ and $p_v\in \partial J(v)$, add the two Bregman distances to get $D_J^{p_v}(u,v)+D_J^{p_u}(v,u)=⟨p_u-p_v,u-v⟩\ge 0$. The choice of subgradients matters. This is generally not a metric: it need not separate points or satisfy a triangle inequality.

A topological vector space is a vector space $E$ over $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$, equipped with a topology such that vector addition and scalar multiplication are jointly continuous:The domains carry the product topology, and the scalar field carries its usual topology. Joint continuity permits both inputs to vary at once. Here the definition does not impose the Hausdorff condition; we state it separately when needed. ^[ref].

Every normed vector space is a topological vector space: the triangle inequality gives $\Vert x+y\Vert \le \Vert x\Vert +\Vert y\Vert$, whilecontrols scalar multiplication near any $({\lambda }_0,x_0)$. More generally, a seminorm topology supplies many measurements of smallness, rather than one norm. Translation by any vector is a homeomorphism, so a neighbourhood basis at zero determines all neighbourhoods by translation.

The compatibility matters: on a nonzero real vector space, the discrete topology makes vector addition continuous, but not scalar multiplication. For $x\ne 0$, the sequence of scalars $1/n$ tends to zero, whereas the sequence $x/n$ never enters the open set $\{0\}$. Thus choosing a topology arbitrarily need not respect the linear algebra.

This framework lets functional analysis choose the appropriate notion of limit of a sequence. Norm topology measures one global error, weak topology tests linear functionals, and the Fréchet space of smooth functions controls every order of derivative. In each case the operations forming linear combinations are continuous, making approximation compatible with the underlying equations.

Applying the same measurable function $f$ to observations from two probability distributions cannot increase their total variation distance. Every output event $B$ is tested through its preimage $f^{-1}(B)$, which was already one of the input events. ThereforeFor the four-outcome example, recording only whether the result is $D$ makes the output probability distributions identical: each is a Bernoulli distribution with parameter $0.3$. The original total variation distance was $0.3$, but this coarse measurement leaves zero discrepancy. Discarding information can make two sources less distinguishable.

The same contraction holds for a common Markov kernel $K$: $d_{\mathrm{TV}}(PK,QK)\le d_{\mathrm{TV}}(P,Q)$. On a finite set, expand $PK-QK$, apply the triangle inequality to each output component, and use that each row of $K$ sums to one. A common randomized postprocessing step cannot increase distinguishability either.

The total variation of a function $f:[a,b]\to \mathbb{R}$ measures its accumulated rises and falls, rather than just its change between the endpoints. For each partition of an interval $\Pi =(a=t_0<\cdots <t_n=b)$, add the absolute values of its sampled increments. Then take the supremum over all such partitions of an interval:The function has bounded variation when $V_a^b(f)<\infty$. This definition uses its value at every point of the interval, including both endpoints.^[ref]

For a nondecreasing function, all increments are nonnegative and the sum telescopes to $f(b)-f(a)$. More generally, the triangle inequality gives $|f(b)-f(a)|\le V_a^b(f)$. Equality can fail dramatically when the function changes direction: an object can travel a long distance and finish near its starting position. Here we measure movement of the function's value; arc length of its function graph also includes horizontal movement.

The dual of the trace-class operators, and also of the normed space of finite-rank operators with the trace norm, is isometrically the space of bounded operators. The pairing is $F_B(T)=\mathrm{Tr}(BT)$. For $R_{x,y}z=⟨z,y⟩x$, one has $\Vert R_{x,y}{\Vert }_1=\Vert x\Vert \Vert y\Vert$ and $F_B(R_{x,y})=⟨Bx,y⟩$, using an inner product linear in its first argument. A bounded functional on these rank-one operators gives a bounded sesquilinear form and hence a unique $B$ by the Riesz representation theorem. The singular value decomposition proves $\Vert F_B\Vert =\Vert B\Vert$ and the triangle inequality for the trace norm.

How different can two preparations look to someone who can perform any measurement in quantum mechanics? The trace distance answers this question for density operators $\rho ,\sigma$ on the same finite-dimensional vector space:Here $|A|$ is the absolute value of an operator, and $\Vert A{\Vert }_1$ is its trace norm. Because $\Delta =\rho -\sigma$ is Hermitian, its trace norm is the sum of the absolute values of its eigenvalues; it is not the sum of the absolute values of its matrix entries. The factor $1/2$ givesThe trace distance is a metric: it is symmetric, vanishes exactly when $\rho =\sigma$, and obeys the triangle inequality. Identical preparations have trace distance zero; preparations with orthogonal occupied eigenspaces have trace distance one and can be identified perfectly in a single measurement in quantum mechanics.

Operationally, the trace distance is the largest total variation distance between the outcome probability distributions obtainable by a POVM. Equivalently, the variational characterization of trace distance says it is the largest possible probability gap for one event in a binary measurement in quantum mechanics. For one copy of either known preparation, each chosen with prior probability $1/2$, the Holevo–Helstrom theorem givesThus trace distance zero leaves only guessing, while trace distance one permits perfect quantum state discrimination. The optimum ranges over measurements; different measurement bases can expose different amounts of distinguishability. ^[ref]

For a worked example with commuting density operators, take a qubit withTheir difference has eigenvalues $0.5,-0.5$, so $D=(|0.5|+|-0.5|)/2=0.5$. Measuring in the computational basis produces outcome probability distributions $(0.8,0.2)$ and $(0.3,0.7)$, whose total variation distance is also $0.5$. Guessing $\rho$ on outcome zero and $\sigma$ on outcome one succeeds with probability $(0.8+0.7)/2=0.75$. A projective measurement in the Hadamard basis instead produces $(0.5,0.5)$ for both preparations and gives no discrimination advantage.

The trace-distance contraction under quantum channels ensures that a common quantum channel, including the partial trace, cannot increase the trace distance. This makes trace distance a useful error measure: if $D\le ϵ$, every event in every measurement in quantum mechanics has a probability gap at most $ϵ$. Losing access to a subsystem can hide distinguishability, but cannot create more of it.

For $1\le p<\infty$ and $f\in L^p({\mathbb{R}}^d)$, translations converge to $f$ in the Lp space norm. First prove this for continuous compactly supported functions, using uniform continuity and a common bounded support. Their density in Lp space, together with the norm-preserving translation map and the triangle inequality, proves the assertion for arbitrary $f$. It implies convergence of averages over shrinking intervals, a basic approximate identity argument. The corresponding assertion fails for general $L^{\infty }$ functions, such as an indicator function with a jump.

For speed $v=k{r}^2$ in the punctured plane, the travel-time element is $ds/(k{r}^2)$. Under the inversion in a circle with radial coordinate $u=1/r$, this becomes $k^{-1}\sqrt{d{u}^2+u^{2}d{\theta }^2}$, the arc length in Euclidean space divided by $k$. Thus a shortest travel-time route corresponds to a straight line segment in the inverted plane, whenever that segment avoids the origin and lies in the allowed region. Between $(a,0)$ and $(0,a)$ the inverted segment has equation $u(\cos \theta +\sin \theta )=1/a$. The route is $r=a(\cos \theta +\sin \theta )$, and its minimum travel time is $\sqrt{2}/(ka)$. The triangle inequality for arc length proves global minimality.

A triangle is the convex hull of three noncollinear points: the three vertices must not lie on one line. Its boundary consists of the three joining line segments; depending on context, "triangle" can mean this boundary or the filled region it encloses. Here a triangle's area refers to the filled region. The noncollinearity assumption excludes a collapsed shape with zero area, for which ordinary interior angles are not defined.

For example, the positions $(0,0)$, $(4,0)$ and $(0,3)$ form a right triangle with a right angle at the origin. Taking the side of length $4$ as its base gives perpendicular height $3$, so its area is $4\cdot 3/2=6$. The height is measured at a right angle to the chosen base; it is not generally the length of another side. In this example it happens to be a side because the triangle is a right triangle.

By contrast, side lengths $3,4,6$ define a valid triangle: each side is shorter than the sum of the other two, as required by the strict triangle inequality. But $3^2+4^2=25\ne 36=6^2$, so this is not a right triangle and the Pythagorean theorem cannot calculate its longest side from the other two. The law of cosines instead gives the opposite interior angle $C$ throughThis example separates the definition of a triangle from the additional assumption of a right angle.

Every triangle in the Euclidean plane has interior angles summing to $\pi$ radians, a result proved in Euclid's Elements I.32.^[ref] It lets two measured angles determine the third, and underlies the angle-angle criterion for similar triangles. The qualification "Euclidean plane" matters: the earlier spherical triangle demonstrates a different sum.

A metric satisfies $d(x,z)\le d(x,y)+d(y,z)$: visiting an intermediate point cannot beat the direct endpoint distance. A norm satisfies the corresponding inequality $\Vert u+v\Vert \le \Vert u\Vert +\Vert v\Vert$.

Take $O=(0,0)$, $Q=(3,0)$, and $P=(3,4)$. For Euclidean distance, the triangle inequality reads $5\le 3+4=7$. The direct line segment has metric path length $5$, while the route through $Q$ has metric path length $7$. For Manhattan distance, the same comparison becomes $7=3+4$: both routes are shortest. The line segment and the bent route both become metric geodesics when progress is measured by metric path length. For Chebyshev distance, the direct line segment has metric path length $4$, while the same bent route still costs $7$.

Repeated application of the triangle inequality bounds the endpoint distance by the sum along any finite chain of intermediate points. This explains why metric path length is always at least the distance between endpoints. Equality reveals a shortest route, but its shape and uniqueness depend on the chosen metric.

The ultrametric inequality, also called the strong triangle inequality, bounds a direct distance by the larger of the two distances through an intermediate point:A metric space satisfying this rule is an ultrametric space. For a Non-Archimedean absolute value, the corresponding inequality is $|a+b|\le \max \{|a|,|b|\}$, and the associated metric is $d(x,y)=|x-y|$. The ordinary triangle inequality follows because the maximum of two nonnegative real numbers is at most their sum.

For example, the 2-adic absolute value gives the three points $0,1,8$ distancesEach distance is bounded by the maximum of the other two. The two largest distances are equal: every triangle in an ultrametric space has this property. Otherwise its unique longest side would exceed the maximum of the other two, contradicting the ultrametric inequality. The drawn lengths of a triangle on paper need not represent these distances.

The ultrametric inequality gives a useful exact rule when $|a|\ne |b|$:To see why, suppose $|a|>|b|$. The upper bound is $|a+b|\le |a|$; applying the ultrametric inequality to $a=(a+b)-b$ forces the reverse bound. Equal sizes can instead cancel: $|1{|}_2=|1{|}_2=1$, but $|1+1{|}_2=1/2$. This makes a uniquely largest term decisive in a sum, a principle used in the Newton polygon and in arguments about roots of a polynomial.

A sequence of functions $f_n:E\to \mathbb{R}$ has a uniform limit if and only if, for every $\epsilon >0$, there is $N$ such that $|f_n(x)-f_m(x)|<\epsilon$ for every $m,n\ge N$ and every $x\in E$. Necessity follows from the triangle inequality. For sufficiency, completeness of the real numbers first gives a pointwise limit $f$. Letting $m\to \infty$ in the bound with tolerance $\epsilon /2$ gives $|f_n(x)-f(x)|\le \epsilon /2$ uniformly in $x$. The same proof works for a complete metric space as codomain; completeness is essential.

Suppose continuous functions $f_n$ converge by uniform convergence to $f$ on a domain of a function $X$, and $x_n\to x$ with all these inputs in $X$. The uniform limit theorem makes $f$ a continuous function, and the triangle inequality givesThe first term handles the changing approximation and the second handles the changing input. For $f_n(x)=x^n$ on $[0,1]$, the inputs $x_n=2^{-1/n}$ instead give $x_n\to 1$, $f_n(x_n)=1/2$, and $f(1)=1$. Pointwise convergence cannot supply the first uniform bound, and this pointwise limit also lacks continuity at one.

A uniform limit of continuous functions is a continuous function. No compactness assumption on the domain of a function is required.^[ref]

For a proof, fix an input $x$ and tolerance $\epsilon >0$. Uniform convergence supplies one $f_N$ with $\Vert f_N-f{\Vert }_{\infty }<\epsilon /3$. Its continuity at $x$ supplies $\delta >0$ with $|f_N(y)-f_N(x)|<\epsilon /3$ for allowed inputs satisfying $|y-x|<\delta$. The triangle inequality givesOne approximating function controls both endpoints of this estimate. Under pointwise convergence, selecting an index for $x$ supplies no common control at nearby $y$. This explains why the polynomial functions $x^n$ can have a discontinuous pointwise limit on $[0,1]$.

A uniformly continuous function has one input tolerance valid throughout its domain of a function. For $f:D\to \mathbb{R}$, with $D\subseteq \mathbb{R}$, uniform continuity meansChoose the output tolerance $\epsilon$ first, then a positive $\delta$ valid for every pair of function inputs. It may depend on $\epsilon$, the function and its domain of a function, but not on the pair’s location. The absolute values measure input and output distance.^[ref]

Compare the quantifier order with the epsilon-delta definition of continuity:Both lines end with $|x-a|<\delta \Rightarrow |f(x)-f(a)|<\epsilon$. In the first, $\delta$ can depend on the point $a$; in the second, it is chosen before $a$. Thus uniform continuity implies continuity at every point. Each requested $\epsilon$ gets its own shared tolerance.

For an exact example, take the polynomial function $f(x)=x^2$ on the closed interval $D=[0,2]$. For every pair of function inputs in that interval, polynomial factorization givesConsequently $\delta =\epsilon /4$ works everywhere on this domain of a function. With $\epsilon =0.4$, the common tolerance is $\delta =0.1$. Near the right endpoint, for example, $x=1.92$ and $y=2$ have input distance $0.08$ and output distance $4-1.92^2=0.3136<0.4$. The bound is sufficient; it does not claim the largest admissible input window at each point.

On the whole real line, the same polynomial function is a continuous function but fails uniform continuity. Take the sequences $x_n=n$ and $y_n=n+1/n$, for positive integers $n$. ThenFor any proposed $\delta >0$, choose an integer $n>1/\delta$. The inputs are closer than $\delta$, yet their outputs violate the fixed tolerance $\epsilon =1$. The sequences need not themselves have a finite limit: their mutual distance is what tends to zero. This is the obstruction captured by the sequential criterion for uniform continuity. Restricting the domain of a function to $[0,2]$ removes these escaping pairs.

A uniformly continuous function need not be a bounded function. The identity function $f(x)=x$ on the real line has $|f(x)-f(y)|=|x-y|$, so $\delta =\epsilon$ always works, although its image of a function is the unbounded real line. Uniform continuity controls changes between nearby inputs, not the overall size of outputs. Likewise, an unbounded domain of a function does not by itself prevent uniform continuity.

The square root $g(x)=\sqrt{x}$ supplies a different common tolerance on the interval $[0,\infty )$. Assuming $x\ge y\ge 0$, put $u=\sqrt{x}$ and $v=\sqrt{y}$. Since $0\le u-v\le u+v$,Interchanging the inputs when necessary provesThus $\delta ={\epsilon }^2$ guarantees uniform continuity on the entire domain of a function, including zero. For $\epsilon =0.4$, any input distance below $0.16$ suffices. The pairs $(0,0.09)$ and $(4,4.09)$ both have input distance $0.09$; their output distances are respectively $0.3$ and $0.09/(\sqrt{4.09}+2)<0.0225$. The common guarantee accommodates quite different local sensitivities.

This Hölder continuity bound allows a curved tolerance rule. It is weaker than a Lipschitz bound: at zero the output change divided by the input change isNo finite Lipschitz constant works near zero, even though uniform continuity does. A Lipschitz function does have uniform continuity, choosing $\delta =\epsilon /L$ when its Lipschitz bound $L$ is positive; when $L=0$ it is a constant function.^[ref]

For general metric spaces, replace the absolute values by the respective distance functions. The Heine-Cantor theorem gives an existence guarantee: every continuous function on a compact metric space is uniformly continuous. In particular, continuity on a finite closed interval provides a common tolerance, even when no simple formula for it is known.^[ref]

For a short proof outline, fix $\epsilon >0$. Continuity gives each point $a$ an open ball of radius $r_a>0$ within which outputs differ from $f(a)$ by less than $\epsilon /2$. The open balls of half these radii still cover the domain of a function. Compactness supplies a finite subcover. Take $\delta$ to be the smallest half-radius in this finite set. Two inputs closer than $\delta$ lie together in one of the original open balls, so the triangle inequality bounds their output distance by $\epsilon$. The beautiful step is that the compact set turns local choices into finitely many choices, whose minimum stays positive.

A shared tolerance also makes uniformly continuous functions preserve Cauchy sequences: after choosing $\delta$ for $\epsilon$, all sufficiently late input pairs are closer than $\delta$, so all corresponding output pairs are closer than $\epsilon$. Unlike a convergent sequence argument, this does not require an input limit inside the domain of a function. It is why uniform continuity is useful for extending a function toward missing endpoints and for the decay argument below.

If $X_n$ converges in probability to both $X$ and $Y$, then $X=Y$ almost surely. Indeed, the triangle inequality and a union bound show that $\mathbb{P}(|X-Y|>\epsilon )=0$ for every $\epsilon >0$.

A wall of marginal stability is a locus in the parameter or vacuum space where the complex central charges in supersymmetry of possible decay products align in phase. The triangle inequality then becomes an equality, $|Z_1+Z_2|=|Z_1|+|Z_2|$, permitting a BPS state to reach the threshold for decay into other BPS states. Away from such alignment, the strict triangle inequality can prohibit the decay. Supersymmetry protects the mass-charge relation of an existing short state; it does not guarantee that the state exists or remains stable everywhere in parameter space.