A resting shallow water layer has depth $h_0$ and wave speed $c_0=\sqrt{g{h}_0}$. A vertical wall receding into negative $x$ at speed $0<V<2c_0$ creates a simple wave with invariant $u-2c=-2c_0$. Inside the expansion fan $x/t=u+c$, proving the displayed profiles. The wall-adjacent state is $u=-V$, $c=c_0-V/2$; the fan occupies $c_0-3V/2<x/t<c_0$, with the resting state ahead. The wall lies at $x/t=-V$, strictly to the left of the fan tail because $c_0-V/2>0$. Thus the assumption on wall speed is exactly the condition for positive depth and continued contact between fluid and wall. The depth follows from $h=c^2/g$.

Suppose an affine deformation on each side of a plane with unit normal $N$ is continuous across it. Tangential derivatives agree, so $(A-B)t=0$ for every $t\perp N$. Decomposition of an arbitrary vector into its tangent component and $(x\cdot N)N$ then gives $(A-B)x=((A-B)N)(N\cdot x)$, proving the rank-one form. Conversely that form makes the two affine maps agree on the plane when their constants agree there.

Orientation preservation on both sides also requires $1-N\cdot A^{-1}a>0$: the matrix determinant lemma gives $\det B=\det A(1-N\cdot A^{-1}a)$. This inequality is additional geometric information supplied by the determinant signs, not by continuity alone.

The unrotated variants differ by $2\delta e_3\otimes e_1$, giving normal $e_1$. The rotation about $e_2$ with $\cos \theta =(1-{\delta }^2)/(1+{\delta }^2)$ and $\sin \theta =2\delta /(1+{\delta }^2)$ gives $U_{+}-\overline{R}{U}_{-}=[2\delta /(1+{\delta }^2)](e_1+\delta e_3)\otimes e_3$, giving the second normal $e_3$. If these are the only two compatible rotations, the only gradients in $SO(3)U_{+}$ compatible with $\overline{R}{U}_{-}$ are $U_{+}$ and $\overline{R}{U}_{+}$. Left multiplication by a rotation preserves rank, so the second classification follows directly from the first.

Let $A=R(k-1,\ell )$ and $B=R(k,\ell -1)$. From one vertex of any red-blue colored $K_{A+B}$, either at least $A$ incident edges are red or at least $B$ are blue. In the first case, apply the definition of the Ramsey number to its red neighbors: a red $K_{k-1}$ extends with the chosen vertex, and a blue $K_{\ell }$ is already enough. The blue case is analogous. This proves the recursion, with boundary values $R(k,2)=k$ and $R(2,\ell )=\ell$.

A material element is in local radiative equilibrium when its absorbed and emitted radiative powers balance, so radiation causes no net local heating. In a non-scattering grey atmosphere in local thermodynamic equilibrium, this reduces to equality of the angular mean specific intensity $J$ and the Planck function $B$. Integrating the radiative transfer equation over solid angle then shows that the radiative flux is constant through a plane-parallel atmosphere. The radiation can remain anisotropic; equilibrium constrains its angular mean, not every ray separately.

Let $T:X\to Y$ be a bounded linear operator with $X$ a Banach space. Suppose every $\Vert y\Vert <1$ has an approximation $Tx$ with $\Vert x\Vert <1$ and error less than $\delta \in (0,1)$. Repeatedly approximate the scaled residual to construct $x_n$ with $\Vert x_n\Vert <{\delta }^n$ and residual less than ${\delta }^{n+1}$. Completeness sums the series, and continuity gives $T\sum x_n=y$. Since $\Vert \sum x_n\Vert <(1-\delta {)}^{-1}$, scaling proves $B_Y(0,1-\delta )\subseteq T(B_X(0,1))$. Hence $T$ is a quotient operator, even without assuming completeness of $Y$.

One element that satisfies a relation simultaneously for every quantified input. A uniform witness also proves the corresponding pointwise statement, because it may be chosen for each particular input.

A witness whose existence is asserted separately for each input. The witness need not be the same element for different inputs, so this condition is weaker than existence of a uniform witness.

A projective triangle is an ordered triple of noncollinear projective points in a projective plane, together with their three joining projective lines. In the real projective plane this means that three chosen representative vectors are linearly independent. Pairing the vertices of two such triangles determines corresponding sides and the three corresponding-vertex joins.

For two projective triangles in a projective plane over a field, perspectivity from a point implies perspectivity from a line. The usual nondegenerate form assumes that the corresponding-side intersections are well defined and distinct. Its dual is the converse, by point-line duality. The result holds in the real projective plane; it is not an axiom valid in every abstract incidence plane.

One linear algebra proof explains the collinearity. Choose a representative $o$ of the center. Rescale the paired representatives to $p^{\prime }=p+\alpha o$, $q^{\prime }=q+\beta o$, and $r^{\prime }=r+\gamma o$, with nonzero coefficients in a nondegenerate configuration whose vertices differ from the center. The first corresponding-side intersection is represented by $x=\beta p-\alpha q=\beta p^{\prime }-\alpha q^{\prime }$. Similarly choose $y=\gamma q-\beta r$ and $z=\alpha r-\gamma p$. Then $\gamma x+\alpha y+\beta z=0$, so the three representatives are linearly dependent. Their projective points are consequently collinear.

In the real projective plane, represent a projective point by a nonzero vector and a projective line by its nonzero coefficient covector, identified with a vector in chosen coordinates. The join of distinct projective points represented by $p,q$ has coefficient vector $p\times q$, using the cross product. It is nonzero because the representatives are linearly independent, and its dot product with each representative vanishes. Dually, two distinct projective lines with coefficients $\ell ,m$ meet in the projective point represented by $\ell \times m$. Scaling an input scales the output and leaves the projective object unchanged. This makes joins, intersections, concurrency, and collinearity exact calculations with homogeneous coordinates.

For two triples of distinct points on two projective lines, the intersections $r_{ij}=p_i{q}_j\cap p_j{q}_i$ are collinear when the configuration is defined. Normalize a projective frame so $p_1=[0,1,0]$, $p_2=[1,1,0]$, $q_1=[0,0,1]$, $q_2=[1,0,1]$, and write $p_3=[1,\alpha ,0]$, $q_3=[1,0,\beta ]$. Then representatives of the three intersections are $(1,1,1)$, $(1,\alpha ,\beta )$, and $(\alpha \beta -1,\alpha (\beta -1),\beta (\alpha -1))$. The last is $\alpha \beta$ times the first minus the second, proving collinearity by linear dependence. This gives a coordinate proof over any field when the normalizations are available.

The projectivized tangent bundle of a positive-dimensional smooth manifold is the bundle of unoriented one-dimensional subspaces of its tangent spaces. A local tangent trivialization gives $P(TM){|}_U\cong U\times {\mathbb{RP}}^{n-1}$. If a base-chart transition is $F$, the line coordinates transform by $[v]\mapsto [D{F}_{x}v]$. In any projective chart, this is a ratio of smooth linear expressions with nonzero denominator. It therefore defines a smooth manifold of dimension $2n-1$.

For the two-sphere, a point consists of $x\in S^2$ and a line $L\subset x^{\perp }$. Choosing a unit vector $v\in L$ gives an oriented orthonormal basis $(x,v,x\times v)$. A unique rotation takes one chosen frame to another, so $S{O}_3$ acts transitively. Its stabilizer consists of the identity and the half-turn about $x$, which reverses $v$ while preserving its unoriented line. Thus $P(T{S}^2)\cong S{O}_3/C_2$.

Take $d$ independent copies of the simple random walk on the integer line, each stepping left or right with equal probability at every integer time. Their product Markov chain moves all coordinates at once; each of its $2^d$ diagonal increments has probability $2^{-d}$. At odd times a return to the origin is impossible. At time $2n$, independence gives the displayed return probability, whose asymptotic equivalence follows from Stirling's approximation. The recurrence criterion by return probabilities and the p-series therefore imply recurrence for $d=1,2$ and a transient Markov chain for $d\ge 3$.

For $d>1$ the chain from the origin visits only the sublattice whose coordinates all have the same parity. This is one communicating class; restricting to it gives an irreducible Markov chain. The period of a state in a Markov chain is two here. The conclusion applies to that class, without incorrectly claiming Markov chain irreducibility on all of ${\mathbb{Z}}^d$. This walk moves diagonally; the usual axis-step simple random walk has different transition probabilities, even though the same dimensional recurrence threshold holds.

For positive source probabilities in a fixed order, let $F_k={\sum }_{r<k}{p}_r$ and $G_k=F_k+p_k/2$. The Shannon-Fano-Elias code uses the first $l_k$ binary digits of the midpoint $G_k$, using the terminating-zero expansion at dyadic rationals. Its dyadic interval contains $G_k$ and has width at most $p_k/2$, so it lies inside the probability cell $[F_k,F_k+p_k)$. These cells are disjoint, whereas a codeword that is a prefix of another has a dyadic interval containing the other interval. The code is therefore a prefix code. Since $l_k<-{\log }_2{p}_k+2$, averaging proves $\mathbb{E}L<H(X)+2$. This is a direct cumulative construction and does not require sorting the probabilities or constructing a Huffman tree.