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For a finite graph $G$, its two-colour Ramsey number $R(G)$ is the least $n$ such that every red-blue colouring of $K_n$ contains a monochromatic copy of $G$. It exists because a monochromatic clique on $|V(G)|$ vertices contains a copy of $G$.

The diagonal Ramsey number $R(t)$ is the least $n$ such that every red-blue edge-colouring of $K_n$ contains a monochromatic $K_t$.

Fix positive integers $r,k,m$ with $m\ge r$. There is an integer $n\ge m$, depending only on $r,k,m$, such that every finite coloringhas a subset $H\subseteq [n]$ of size $m$ for which the restriction of a function $c$ to $\left(\genfrac{}{}{0ex}{}{H}{r}\right)$ is a constant function. Here $[n]=\{1,\dots ,n\}$, and $\left(\genfrac{}{}{0ex}{}{S}{r}\right)$ denotes the collection of all $r$-element subsets of $S$, rather than its binomial coefficient cardinality. The notation $[n]⟶(m{)}_k^r$ abbreviates this property. Thus $H$ is a homogeneous set for a colouring; when $r>1$, the monochromatic set consists of its $r$-element subsets, not its individual elements. This is the finite assertion in Ramsey's original Theorem B, p. 267.

For $r=1$, this is the pigeonhole principle, with sharp threshold $n=k(m-1)+1$. For instance, every red-blue coloring of five objects has three of one color; four objects can split two and two. For $r=2$, $k=2$ and $m=3$, the sharp threshold is six, as shown by the Ramsey number for triangles. Coloring pairs makes this a problem about edges of a complete graph. For larger $r$, the analogous viewpoint uses a hypergraph.

The finite Ramsey theorem lets an existence argument discard most of a large finite set while retaining a prescribed amount of uniform structure. Its threshold applies to every coloring; one does not need an assumption that colors were assigned randomly. A diagonal compactness argument deduces finite existence from Ramsey's theorem, but does not by itself identify the smallest threshold. For pairs, a Ramsey number makes that optimization explicit.

For an outward ray through a nonscattering atmosphere,

In local thermodynamic equilibrium, matter populations are thermal at the local temperature and the source function is the Planck function, $S_{\nu }=B_{\nu }(T)$.

Electromagnetic duality replaces the electric dipole $p$ by $m/c$ for a magnetic dipole moment $m$. Therefore

This is the magnetic-dipole term of the multipole radiation expansion.

For a localized source whose size is small compared with its radiation wavelength, the radiation fields of its electric dipole moment $p$ areTheir total instantaneous power is

An isotropic source of average power $P$ has intensityat distance $r$. For monochromatic vacuum radiation, its electric-field amplitude is $E_0=\sqrt{2⟨S⟩/({ϵ}_{0}c)}$.

If the pressure at $r=b(t)$ is $p_{\infty }$, the Unsteady Bernoulli equation gives

Weyl curvature produces relative acceleration between neighbouring freely falling trajectories and therefore tidal deformation. A freely falling frame removes gravity at one event, as required by the equivalence principle, but cannot remove curvature over an extended region.

With the $1/n$ Rademacher convention, an empirical risk minimizer has expected excess risk at most twice the Rademacher complexity of its loss class.

For $h_M(x)=x^{T}Mx$, $M⪰0$, $\mathrm{tr}M\le s$, and $\Vert x\Vert \le C$, the complexity is at most $C^{2}s/\sqrt{n}$.

Independent signs cancel cross terms in $\mathbb{E}\Vert {\sum }_i{\sigma }_i{A}_i{\Vert }_F^2$, leaving ${\sum }_i\Vert A_i{\Vert }_F^2$.

Composition with an $L$-Lipschitz scalar map increases Rademacher complexity by at most the factor $L$ after centering at zero.

A linear functional has the same supremum over a class and its convex hull. Thus taking convex combinations does not increase empirical Rademacher complexity.

For $|D_i|\le M$ and a binary class $H$ of VC dimension at most $p$, the vectors $(D_{i}h(X_i){)}_{i=1}^n$ have norm at most $M\sqrt{n}$ and at most $(n+1{)}^p$ distinct values. The Massart finite-class lemma gives

For a finite $A\subseteq {\mathbb{R}}^n$ with ${\max }_{a\in A}\Vert a{\Vert }_2\le R$, its empirical Rademacher complexity is at most $R\sqrt{2\log |A|}/n$.

Identify the vertical sides of $[0,1{]}^2$, then collapse each horizontal side separately to a point. The resulting quotient is homeomorphic to $S^2$ throughEquivalently, the quotient is the suspension of the circle.

Identify $x,y\in \mathbb{R}$ when $x-y\in \mathbb{Q}$. Every nonempty saturated open subset of $\mathbb{R}$ meets every other one because rational translates of any open interval meet any prescribed nonempty open interval. The quotient $\mathbb{R}/\mathbb{Q}$ has multiple points but no two nonempty disjoint open subsets, so it is not Hausdorff.

If $q:X\to X/R$ is the quotient map and a continuous map $f:X\to Y$ is constant on each equivalence class, there is a unique continuous map $F:X/R\to Y$ satisfying $Fq=f$. It is given by $F([x])=f(x)$; continuity follows becausefor every open $U\subseteq Y$.