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Let $(X,Y)$ be a regular pair of vertex sets with parameter $ϵ$, equal class sizes $n$, density greater than $ϵ$, and every vertex having at least $ϵn$ neighbors across the pair. Then it has a perfect matching. To prove Hall's condition, take $S\subseteq X$. If $0<|S|\le ϵn$, one vertex already supplies at least $|S|$ neighbors. If $ϵn<|S|\le (1-ϵ)n$ and $|N(S)|<|S|$, the two large sets $S$ and $Y\setminus N(S)$ have density zero, contradicting regularity and the positive density surplus. Finally, if $|S|>(1-ϵ)n$, every vertex of $Y$ has a neighbor in $S$, so $N(S)=Y$. The empty subset satisfies Hall's condition directly.

Let $0<ϵ<1$ and suppose an $N$-vertex graph has induced-edge discrepancies $|e(G[W])-p(\genfrac{}{}{0ex}{}{|W|}{2})|<{ϵ}^5{N}^2/9$ whenever $|W|\ge {ϵ}^{2}N$. Any two disjoint vertex sets of size at least $ϵN$ then form an epsilon-regular pair. Indeed, inclusion-exclusion of the induced edge counts giveswhenever both sets have size at least ${ϵ}^{2}N$. The original pair has density within ${ϵ}^3/3$ of $p$, while each pair of allowed relative subsets has density within $ϵ/3$ of $p$. Their densities differ by less than $(ϵ+{ϵ}^3)/3<ϵ$.

Thus induced subgraph counts control bipartite regularity, even when the two original sets have different sizes. Equal partitions with a small exceptional remainder follow for each fixed $k<1/ϵ$ once $N$ is sufficiently large to absorb integer rounding. That size condition cannot be discarded if all $k$ nonexceptional classes must be nonempty.

For a regular outer measure and measurable $M$, the forward implication is countable additivity. For the reverse, choose a measurable hull of $A$ and intersect it with $M$ to obtain $A\subseteq B\subseteq M$ and $\mu (B)={\mu }^{\ast }(A)$. Measurability of $B$ splits $M\setminus A$ into $M\setminus B$ and $B\setminus A$, soThe assumed equality and finiteness force ${\mu }^{\ast }(B\setminus A)=0$. Every outer-measure-null set is measurable by Caratheodory measurability criterion, so $A=B\setminus (B\setminus A)$ is measurable. Finiteness prevents invalid subtraction of infinite masses.

For a regular outer measure, its Caratheodory measurability criterion defines a maximal algebra of sets on which that outer measure is finitely additive. Indeed, suppose an additive algebra contains all measurable sets and also $A$. Given $E$ of finite outer measure, take a measurable hull $H\supseteq E$ of equal mass. Finite additivity splits $H$ by $A$, and monotonicity gives ${\mu }^{\ast }(E)\ge {\mu }^{\ast }(E\cap A)+{\mu }^{\ast }(E\setminus A)$. Outer subadditivity gives equality. For infinite mass that equality already follows from outer subadditivity. Thus $A$ was measurable. No sigma-finite measure assumption is necessary.

An irreducible Markov chain is null recurrent when every state is a null recurrent state: it returns to that state almost surely, but its mean recurrence time is infinite. Such a countable-state chain has no stationary distribution.

If $S_0=a$ and $S_n=S_{\lfloor n/2\rfloor }+m$ for positive integers $n$, each halving removes one binary digit until zero is reached. There are $1+\lfloor {\log }_{2}n\rfloor$ such steps. The formula follows by strong induction, using $\lfloor {\log }_2\lfloor n/2\rfloor \rfloor =\lfloor {\log }_{2}n\rfloor -1$ for $n\ge 2$ and checking $n=1$ separately. With fixed positive $m$, this is $\Theta (\log n)$ as $n\to \infty$; the zero case retains its separately defined value.

Prove a proposed bound by mathematical induction, substituting the inductive hypothesis for smaller arguments into the recurrence relation. Choose its constants so the resulting inequality implies the proposed bound at the current argument, and verify all base cases. A leading-order guess may need an additional lower-order term to make the induction close. The entropy potential for a divide-and-conquer recurrence is an example using a linear correction to an $n\log n$ bound.

On vertices $1,\dots ,6$, take the simplicial complex generated by $123,124,135,146,156,236,245,256,345,346$. Every edge occurs in two facets. The vertex links, listed cyclically, are $23564$, $13654$, $12645$, $12536$, $13426$, and $14325$, respectively. Thus each vertex link is a cycle graph, and the realization is a compact connected surface without boundary. Its Euler characteristic is $6-15+10=1$, so the classification theorem for surfaces identifies it with the real projective plane. All fifteen possible edges occur, and the listed triangles are distinct three-element sets, so this is an honest abstract simplicial complex, not merely a polygon presentation.

In a triangle whose two sides at an apex are glued with that apex fixed, cross-sections become circles and the apex becomes their cone point. This quotient is a disc. A second triangle with boundary word $aa{b}^{-1}$ becomes a Möbius band after the $a$ sides are identified. Gluing the remaining $b$ boundary to the first disc caps the band and gives the real projective plane. The resulting two-face cell structure has two vertices, three edges and Euler characteristic one.

For $u_t-u_{xx}+u=0$ with homogeneous mixed Dirichlet and Neumann boundary conditions, testing the Backward Euler method finite element equation with the new solution gives the displayed identity. Both the squared derivative norm and the squared function norm are required in the dissipation, because diffusion and reaction contribute separately. Replacing the full $H^1$ norm by its seminorm omits the reaction term. Positivity gives unconditional decay of the $L^2$ norm.

In a homogeneous isotropic elastic half-space $Y>0$, a Rayleigh wave with factor $e^{ik(X-ct)}$ has two depth factors $e^{-b_{L}Y}$ and $e^{-b_{T}Y}$. Substitution into the Navier-Cauchy equation factors the amplitude determinant into longitudinal and transverse factors, giving the displayed exponents. Strict exponential decay requires $0<c<\min (c_L,c_T)$. The longitudinal constituent has $(A,B)\propto (ik,-b_L)$; the transverse constituent has $ikA-b_{T}B=0$. The free-surface traction conditions select a combination of both constituents.

For a Rayleigh wave at a free surface of a homogeneous isotropic elastic half-space, define $s=c^2/c_T^2$ and $\eta =c_T^2/c_L^2$. Cancelling the two surface tractions between the constituents with Rayleigh wave decay exponents gives the displayed equation with $0<s<1$. Equivalently $(b_T^2+k^2{)}^2=4k^2{b}_L{b}_T$. Squaring gives a factor $s$ times $P(s)=s^3-8s^2+(24-16\eta )s-16(1-\eta )$. For stable Lamé parameters, $0<\eta <3/4$, so $P(0)<0$, $P(1)=1$, and $P^{\prime \prime }(s)=6s-16<0$ on $(0,1)$. Its concavity ensures exactly one crossing in this interval. Both sides of the unsquared equation are positive, so this root gives the physical speed. The zero root is a degenerate static cancellation, not a propagating Rayleigh wave.

Let $D$ be a rational vector subspace of the real numbers and let $t\notin D$. Every element of $D+\mathbb{Q}t$ has a unique expression $d+qt$. For any prescribed real number $a$, the rule $h(d+qt)=f(d)+qa$ extends a rational-linear map $f$ on $D$ and remains rational-linear. The uniqueness follows because a nonzero rational multiple of $t$ cannot lie in $D$.

Let $t$ be transcendental over a field $k$ and let $\tau$ be the field automorphism sending $t$ to $1/t$. The rational function $u=t+1/t$ is fixed, and $t$ satisfies $X^2-uX+1=0$. Thus the extension over $k(u)$ has degree at most two. It is not one because the nontrivial automorphism $\tau$ fixes $k(u)$. The fixed field of $\tau$ also has index two by the fixed-field theorem, so it equals $k(u)$.

No nonconstant polynomial in $k[t]$ is fixed by inversion. For a polynomial of positive degree, substitution of $1/t$ creates negative powers, whereas the original has positive powers. Uniqueness of the coefficients of a Laurent polynomial rules out equality. Consequently this fixed field cannot have a generator that is a polynomial in the original variable, even though it has the simple rational-function generator above. The statement is valid in characteristic two as well: $t$ and $1/t$ remain distinct transcendental rational functions.

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.