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For local dichotomic outcomes define $A(a,\lambda ),B(b,\lambda )=\pm 1$ and a setting-independent normalized distribution $\rho (\lambda )$. At each $\lambda$,because exactly one bracket vanishes and the other is $\pm 2$. Integrating and using the triangle inequality yieldsFor stochastic local response functions the same bound follows by replacing outcomes with conditional means in $[-1,1]$, or by including local randomness in $\lambda$. The important locality condition is factorization of joint conditional outcome probabilities, together with independence of the setting distribution and $\lambda$.

The spin singlet instead predicts $E(a,b)=-a\cdot b$. Two orthogonal directions at one wing and directions halfway between them at the other give $|S|=2\sqrt{2}$. Therefore reproducing the quantum predictions requires giving up at least one of the stated premises. In the perfect-anticorrelation version, $B(a,\lambda )=-A(a,\lambda )$ gives Bell's three-setting form $|E(a,b)-E(a,c)|\le 1+E(b,c)$ by the same algebra.

This does not establish that all deterministic or hidden-variable theories are impossible: nonlocal guidance is a counterexample to that reading. Nor does it equate Bell locality with no-signalling; averaging quantum outcomes can preserve setting-independent local marginals despite nonfactorizing correlations. The theorem is significant because it turns a local-completeness dispute into incompatible quantitative correlation bounds, not because it refutes every possible meaning of physical reality.^[ref]

Set $K={\sum }_i\Vert A^i{\Vert }_{\mathrm{op}}^2$, where the norms are operator norms, and $D_m=y^{(m+1)}-y^{(m)}$. Independence of the Brownian coordinates and the vector Itô isometry giveInduction, as in part (b), yields the Factorial Picard estimate for a linear stochastic differential equationFor $t\le T$, the triangle inequality in $L^2$ therefore bounds any tail of the iterates by the corresponding tail ofThis numerical series converges by the ratio test: successive terms have ratio $\sqrt{KT/(m+2)}\to 0$. Thus the iterates are Cauchy in $L^2$, uniformly in the deterministic time parameter on each bounded interval. Completeness of Lp spaces gives a limit $y_t$. Once its continuous adapted version is chosen in part (d), the Itô isometry permits passage to the limit in the integral equation: the squared norm of each integral difference is bounded by $K{\int }_0^T\mathbb{E}|y_s^{(m)}-y_s{|}^{2}ds\to 0$. Consequently the Picard limit solves the stochastic differential equation.

Put $\rho =|z|$. The triangle inequality gives $1-\rho \le |z-e^{i\theta }|\le 1+\rho$, so the Poisson kernel for the unit disk obeysIf $\rho \le r$, replace the lower bound by $(1-r)/(1+r)$ and the upper bound by $(1+r)/(1-r)$. Integrating against the positive measure $h(e^{i\theta })d\theta /(2\pi )$ givesHere $u(0)$ is exactly the uniform boundary average from part (a). Positivity permits multiplication of the pointwise kernel bounds; continuity on the compact circle ensures all integrals are finite. This is the Harnack inequality on the unit disk.

The specified Bell-state phases giveThereforeThe sum contains complex squares, not squared moduli. Define $q={\sum }_j{b}_j^2$ and $\mathcal{C}=|q|$. Normalization gives ${\sum }_j|b_j{|}^2=1$, so the triangle inequality ensures $0\le \mathcal{C}\le 1$.

The characteristic polynomial is ${\lambda }^2-\lambda +{\mathcal{C}}^2/4$. HenceHere logarithms are natural and $0\ln 0$ means its limiting value zero. Using logarithms to base two gives entropy in bits instead. Product states have $\mathcal{C}=0$, spectrum $(1,0)$ and entropy zero; maximally entangled states have $\mathcal{C}=1$, spectrum $(1/2,1/2)$ and entropy $\ln 2$.

This phased Bell basis is a magic basis, and $\mathcal{C}$ is the pure-state concurrence. The paper's ordering is a permutation of the usual magic-basis ordering, which leaves $|{\sum }_j{b}_j^2|$ unchanged.^[ref]

Let $q_S:S^2\to S^2/{\sim }_S$ and $q_D:D^2\to D^2/{\sim }_D$ be the quotient maps. The continuous mapis a homeomorphism onto the upper hemisphere; its inverse function is the restriction of the projection map $(x,y,z)\mapsto (x,y)$. Every antipodal pair on $S^2$ meets that hemisphere. Moreover, two points of this hemisphere are antipodal only if both lie on the equator. ThusBy the universal property of the quotient topology, $q_S\circ h$ induces a continuous bijection

To justify continuity of the inverse function, the source is compact, being a continuous image of a compact space $D^2$; the Heine-Borel theorem gives compactness of $D^2$. The target is Hausdorff, as follows. For distinct equivalence classes $[u]\ne [v]$, putIn $S^2$, take the union $U$ of the two relative open balls of radius $\delta /3$ about $u,-u$, and similarly take $V$ about $v,-v$. Every distance between their respective centres is at least $\delta$, so the triangle inequality gives $U\cap V=\varnothing$. These open sets contain entire antipodal pairs, hence $q_S^{-1}(q_S(U))=U$ and similarly for $V$. The quotient topology therefore makes $q_S(U)$ and $q_S(V)$ disjoint open neighbourhoods of the two classes.

The compact-to-Hausdorff continuous bijection theorem now gives the required homeomorphism $H$.

The intended hypothesis is that $A$ is not connected. The original PDF has the same grammatical ambiguity as the transcription. If only $X$ were assumed not connected, the assertion would fail for the discrete space $X=\{0,1\}$ with its usual metric and the closed subset $A=\{0\}$: two disjoint sets cannot both meet that singleton.

Under the intended reading, choose a separation of a topological space $A=B\bigsqcup C$. Both $B$ and $C$ are nonempty and closed in $A$, because each is the complement of the other relatively open part. Since $A$ is closed in $X$, both are closed in $X$.

For a nonempty subset $F$ of the metric space $(X,d)$, the distance to a set satisfiesIndeed, the triangle inequality gives $d(x,F)\le d(x,z)+d(z,F)$, and swapping $x,z$ proves the bound. Thus the two distance to a set functions below are continuous. SetThese are disjoint open sets. For $b\in B$, we have $d(b,B)=0$ and $d(b,C)>0$, since $C$ is closed and $b\notin C$. Hence $B\subseteq U$. Similarly $C\subseteq V$. Therefore $A\subseteq U\cup V$ and both $U$ and $V$ meet $A$, as required. The construction is a concrete instance of the normality of every metric space.

Use natural logarithms; the same argument works for any fixed base greater than $1$. The function $\log \log x$ is increasing for $x>1$, since its derivative is $1/(x\log x)>0$. With natural logs the sole negative summand is at $i=2$. Put $C_0=|\log \log 2|$.

For all sufficiently large $n$ with $\log \log n\ge 1$, monotonicity and the triangle inequality giveThese are fixed constants and an eventual bound, provingFor another fixed logarithm base there are still only finitely many negative initial terms, which are absorbed in $C_0$.

Beyond the same common threshold, the triangle inequality givesThus $f_1-f_2=O(g_1+g_2)$. The absolute value matters because the difference need not itself be positive, even though the four original functions are.

Absorb the finitely many exceptional initial terms in the Big O notation assumption into a constant $C>0$, so that $|a_n|\le Cn$ for every $n\ge 1$. This is possible because each initial term is finite and each index is positive. We prove the stronger bound $|{}^k{b}_n|\le Cn$ for every $k,n\ge 1$.

The case $k=1$ is the chosen bound. If it holds at $k-1$, the triangle inequality givesMathematical induction completes the proof. Thus ${}^k{b}_n=O(n)$ for every fixed $k$; in fact the same constant works for all the iterated Cesàro means. This is the linear-growth bound for iterated Cesàro means.

Apply the preceding Taylor's theorem at $t=h$ and $t=-h$, with $h>0$. The even-order terms cancel in the central finite difference, givingBoth Lagrange remainders have magnitude at most $h^3/6$, so the triangle inequality gives the sharp central-secant derivative error

For five points, partition the unit square into four squares of side $1/2$. Assign boundary points to just one neighbouring piece. By the pigeonhole principle, two of the points lie in one piece, whose diameter is its diagonal, $\sqrt{(1/2{)}^2+(1/2{)}^2}=\sqrt{2}/2$.

For eight points, use seven smaller-diameter pieces rather than eight rectangles. Start with the convex pentagonTake its four reflected copies at the corners of the square. The remaining pieces are the left triangle with vertices $(0,1/4),(1/4,1/2),(0,3/4)$, its right-hand reflection, and the central diamond with vertices $(1/4,1/2),(1/2,1/4),(3/4,1/2),(1/2,3/4)$. These seven polygons cover the square with disjoint interiors. To verify coverage directly, for $0\le y\le 1/4$ the two bottom pentagons split the horizontal section at $x=1/2$. For $1/4\le y\le 1/2$, the five successive sections have endpoints $0,y-1/4,3/4-y,y+1/4,5/4-y,1$ and belong respectively to the left triangle, bottom-left pentagon, diamond, bottom-right pentagon and right triangle. Reflection gives the top half.

The diameter of a finite convex hull is the greatest distance between its vertices: if two points are convex combinations of vertices, their difference is a convex combination of vertex differences, and the triangle inequality bounds its norm by the largest such distance. Checking the five vertices of $P$ gives diameter $\sqrt{(1/2{)}^2+(1/4{)}^2}=\sqrt{5}/4$; reflections preserve that value. Each remaining triangle and the diamond has diameter $1/2$, which is smaller. After assigning shared boundaries to one piece, eight points in these seven pieces force a pair in a common piece. Hence some pair is at distance at most $\sqrt{5}/4$.

Both arguments are instances of a pigeonhole distance bound from a partition.

Let $h=(b-a)/n$ and let $M_4={\max }_{x\in [a,b]}|f^{(4)}(x)|$. A composite Simpson's rule panel covers two strips, so it has width $2h$ and absolute error at mostThere are $n/2$ panels. By the triangle inequality, the absolute error of their sum is at most the sum of their absolute errors. HenceThis assumes even $n$ and, for the usual Simpson remainder theorem, $f\in C^4[a,b]$. In particular its fourth derivative is bounded on the interval. It is the error bound for composite Simpson's rule, expressed in strip count rather than panel count.^[ref]

For $h\in L^1(0,1)$ with integral zero, $Ph$ is the unique absolutely continuous function with $(Ph{)}^{\prime }=-h$, equal endpoint values, and integral zero. The first formula gives the derivative; zero integral of $h$ gives equal endpoint values, while the added constant makes the mean zero. Uniqueness follows because the difference of two such primitives has zero weak derivative and therefore is constant, then its mean is zero. Its periodic extension has distributional derivative $-\overline{h}$, with no endpoint Dirac delta distribution.

Writing $Rh(x)=x{\int }_0^{1}h-{\int }_0^{x}h$, the same operator on zero-mean inputs is $Ph=Rh-{\int }_0^{1}Rh$. For arbitrary $h\in L^1(0,1)$,The first inequality follows from the integral kernel $x-1_{\{t<x\}}$, whose absolute value has integral $(t^2+(1-t{)}^2)/2\le 1/2$ in $x$. The second follows by the triangle inequality. This bound permits arbitrarily many iterations without increasing the L1 space norm.

Target: Rapid distributional decay of periodic oscillations. Parent: Weak convergence of distributions. Intended file: distribution-theory.bigb.

The strictly positive case states that a real square matrix $W$ with $W_{ij}>0$ has a positive simple eigenvalue $\lambda$ and positive left and right eigenvectors, with all other eigenvalues strictly smaller in modulus. The Brouwer fixed-point theorem applied to $v\mapsto Wv/(1^{T}Wv)$ on the nonnegative unit simplex gives $v>0$ and $Wv=\lambda v$. For another eigenvector $z$, maximize $|z_i|/v_i$; the triangle inequality gives $|\mu |\le \lambda$. Equality forces every modulus ratio and complex argument to agree because every entry is positive, so $z$ is proportional to $v$ and $\mu =\lambda$. Apply the same argument to $W^T$ for a positive left eigenvector; its positive pairing with $v$ rules out a generalized eigenvector at $\lambda$, establishing algebraic simplicity. For merely nonnegative matrices a nonnegative leading eigenvector exists. An irreducible nonnegative matrix has a positive leading eigenvector and simple Perron eigenvalue; a primitive nonnegative matrix has the strict modulus gap.

Positive real number lengths $a_1,\dots ,a_k$, with $k\ge 3$, are the sides of a nondegenerate polygon exactly when $2{\max }_i{a}_i<{\sum }_i{a}_i$. Necessity follows from the triangle inequality: the longest side must be shorter than the path formed by the remaining sides.

For a real or complex polynomial $p(t)={\sum }_{k\ge 0}{a}_k{t}^k$, this finite sum defines a norm: definiteness follows from uniqueness of polynomial coefficients, absolute homogeneity of a norm follows termwise, and the triangle inequality follows by summing $|a_k+b_k|\le |a_k|+|b_k|$. On $[0,1]$, its supremum norm satisfies $\Vert p{\Vert }_{\infty }\le \Vert p{\Vert }_{\mathrm{coef},1}$, since $|t^k|\le 1$. The reverse comparison has no constant independent of the degree of a polynomial: for $p_n(t)=(1-t{)}^n$, the two norms are $1$ and $2^n$ respectively, by the binomial theorem.

For a partition of a set of complete sample observations into groups of known sizes $n_g$ and total $N$, its sample mean is the weighted mean of the group sample means. Indeed, dividing ${\sum }_g{\sum }_{i\in g}{y}_i={\sum }_g{n}_g{\overline{y}}_g$ by $N$ proves the identity. If the printed group means $m_g$ each have absolute rounding error at most ${\delta }_g$, their compatible pooled mean lies inThis follows because all the weights are nonnegative. Equivalently, the pooled absolute rounding error is at most ${\sum }_g(n_g/N){\delta }_g$, by the triangle inequality. A printed overall mean must have a rounding interval intersecting this interval. Disjoint intervals prove incompatibility, rather than merely a difference of displayed decimals. This is a necessary consistency check for means of the same complete sample; it cannot silently equate a full-sample mean with a different complete-case mean or establish a unique repair to inconsistent output.

The vertical field of a disk has kernel $Gz/(|\mathbf{R}-{\mathbf{R}}^{\prime }{|}^2+z^2{)}^{3/2}>0$. The triangle inequality therefore bounds a signed-surface density field by the field of its absolute surface density. If that surface density is everywhere smaller than a nonnegative comparison surface density, their vertical fields satisfy the displayed strict comparison, provided the integrals exist and the strict bound holds on positive measure. This is useful when comparing gravity-equivalent magnetic surface density with actual disk mass.

For unit-modulus prime values, pretentious distance is the Euclidean metric on the finite vectors $(f(p)/\sqrt{2p}{)}_{p\le X}$. On whole arithmetic functions it is only a pseudometric: it sees neither values on prime powers nor primes above the cutoff. Even for every cutoff, the functions one and ${\mu }^2$ have distance zero but differ on nonsquarefree integers. The norm representation proves the triangle inequality and identifies exactly the quotient on which separation holds.

For metric spaces $(X_1,d_1)$ and $(X_2,d_2)$, the sum metric on $X_1\times X_2$ is the displayed formula. Nonnegativity, symmetry and the triangle inequality follow by adding the respective inequalities in the two factors, and zero distance forces equality in both coordinates. It induces the finite product topology, as does the maximum product metric, since their distances satisfy $d_{\max }\le d_{\mathrm{sum}}\le 2d_{\max }$. A sequence converges for this metric exactly when both coordinate sequences converge. Two successive extractions of a convergent subsequence therefore show that the product of two sequentially compact spaces is sequentially compact.