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In ZF, $\omega$ is the least nonzero limit ordinal. Ordinality is expressed by a bounded formula in set theory asserting transitivity and strict total membership order, with well-foundedness supplied by the Axiom of foundation. Put $L(x)=\mathrm{Ord}(x)\wedge x\ne \varnothing \wedge \forall y\in x\exists z\in x(y\in z)$. Then $x=\omega$ is equivalent over ZF to $L(x)\wedge \forall y\in x\neg L(y)$, again bounded. Therefore nested transitive models of ZF have the same $\omega$.

For normalized Legendre polynomials $p_n=\sqrt{(2n+1)/2}{P}_n$, parity makes the sign function orthogonal to every even index. Integrating the Legendre differential equation gives ${\int }_0^1{P}_n=P_n^{\prime }(0)/(n(n+1))$. The differentiated generating function $z(1+z^2{)}^{-3/2}$ gives the derivatives at zero and hence the displayed odd coefficients. The Parseval identity and squared norm two then evaluate ${\sum }_{m\ge 0}((2m)!{)}^2(4m+3)/[2^{4m}(m!(m+1)!{)}^2]=4$.

For orthonormal Legendre polynomials on $[-1,1]$, the polynomial $q_n$ represents evaluation at zero on polynomials of degree at most $n$: $⟨r,q_n⟩=r(0)$. This follows by expanding $r$ in the orthonormal basis, or from the Riesz representation theorem. With $M=\lfloor n/2\rfloor$ and $c_m=\left(\genfrac{}{}{0ex}{}{2m}{m}\right)/4^m$, orthonormality givesThe last identity telescopes using $c_m/c_{m-1}=(2m-1)/(2m)$. The norms are unbounded: $c_m\ge 1/(2m+1)$ makes the sum dominate a divergent harmonic series. Thus the evaluation identities hold eventually for every fixed polynomial but cannot extend to convergence for every $L^2$ vector, by the Uniform boundedness principle. Point evaluation is also not defined on $L^2$ equivalence classes.

For $D=\{(x,y):x^2+y^2<1\}$, the product order gives least upper bound $(1,1)$, because each coordinate has supremum one. The lexicographic order instead has no least upper bound. Every point $(1,v)$ is an upper bound, but $(1,v-1)$ is a strictly smaller upper bound; a point with first coordinate below one cannot bound the disc. The same geometric set therefore has different order-completeness behavior under the two partial orders. Neither conclusion is determined by the usual drawing alone.

The energy per unit mass required to change a saturated liquid to vapour at a stated pressure and coexistence temperature. In a steam-condensation calorimetry experiment, sensible cooling of the condensate must be subtracted from the measured receiver heat gain.

With constant lateral supply $M$, a discharge law of the displayed form gives equilibrium area $[(m+1)(Q_{\mathrm{in}}+Ms)/c{]}^{1/(m+1)}$ and characteristic speed $c{A}^m$. Under a Chezy law, a wetted-perimeter relation $P_b\propto A^q$ gives $m=(1-q)/2$. Small area perturbations are transported downstream with amplitude inversely proportional to this characteristic speed; finite compressive disturbances can still form shocks because the flux is convex.

Let $H(t)={\int }_0^{t}E(r)dr$, increasing until $t^{\ast }$ and decreasing afterwards. For dry initial data and zero upstream depth, a characteristic emitted at time $\tau$ has $h=H(t)-H(\tau )$ and position $s=\frac{5}{3}{\int }_{\tau }^t[H(r)-H(\tau ){]}^{2/3}dr$, for as long as its depth stays nonnegative. The height follows by integrating the source along a characteristic, and the position follows from the derivative of the Manning flux.

When $E<0$ and $H(t)>0$, define the earlier emission time ${\tau }_d(t)\le t^{\ast }$ by $H({\tau }_d)=H(t)$. The front where this water is just exhausted hasIt moves downstream because $s_d^{\prime }=-(10/9)E(t){\int }_{{\tau }_d}^t[H(r)-H(t){]}^{-1/3}dr>0$. Behind it depth is zero; ahead, unexhausted characteristic labels determine the wet profile. This construction presumes removal limited by water availability: continuing a negative source unchanged at zero depth would violate nonnegativity. If $H$ reaches zero, the remaining initially generated water has also vanished.

For a real symmetric positive-definite matrix $A$, the energy norm is $\Vert x{\Vert }_A=(x^{T}Ax{)}^{1/2}=\Vert A^{1/2}x{\Vert }_2$. In the defining supremum of the induced matrix norm, substitute $y=A^{1/2}x$. The matrix square root is an invertible matrix, so this ranges over every nonzero vector and proves the displayed identity. This matrix similarity is useful because contraction in the energy norm becomes an ordinary Euclidean operator norm estimate after a change of coordinates.

For each row, the weighted Cauchy-Schwarz inequality gives $|(Bx{)}_i{|}^2\le ({\sum }_j|b_{ij}|)({\sum }_j|b_{ij}||x_j{|}^2)$. Summing over rows bounds the result by the maximum absolute row sum times the maximum absolute column sum times $\Vert x{\Vert }_2^2$. Taking square roots and the supremum proves the displayed induced matrix norm inequality. It gives a Euclidean estimate directly from entries without computing singular values.

A first-order clause containing at most one positive first-order literal. A clause with exactly one positive literal is definite; with no positive literal it is a constraint or goal clause.

For ${\omega }^3=1$ primitive and $k=0,1,2$, these nine points are the inflection points of every nonsingular finite member of the Hesse pencil. On the cubic, substituting the equation into its Hessian curve gives $-4({\lambda }^3+1)xyz=0$. Nonsingularity forces one coordinate to vanish, and the remaining two cubes sum to zero, producing exactly the nine displayed points. They are the fixed base configuration of the pencil, independently of its smooth parameter.

Ore's theorem, named after Øystein Ore, says that a finite simple graph with $n\ge 3$ vertices is Hamiltonian if every nonadjacent pair has degree sum at least $n$.

For a proof, enlarge a hypothetical counterexample to an edge-maximal non-Hamiltonian graph. Adding a missing edge $uv$ creates a Hamiltonian cycle, hence the original graph has a Hamiltonian path $u=v_1,\dots ,v_n=v$. If both $u{v}_{i+1}$ and $v{v}_i$ are edges, the path closes to a cycle by reversing its final segment. Thus the index sets of these two kinds of edges are disjoint in $\{1,\dots ,n-1\}$, giving $d(u)+d(v)\le n-1$, a contradiction. The degree condition is sufficient and need not follow from a lower bound of $n/2$ on every individual degree.

Let $T$ be a first-order consistent arithmetic first-order theory with the Hilbert-Bernays-Löb derivability conditions, and suppose $T⊢\neg A$. Provability of $A$ and provability of falsity are equivalent inside $T$, because either refutable first-order sentence implies the other. Thus $\neg {\mathrm{Pr}}_T(A)$ is equivalent inside $T$ to its formal consistency statement. The second incompleteness theorem gives $T⊬\neg {\mathrm{Pr}}_T(A)$.

For integers $m,n\ge 0$, the standard Rodrigues' formula $L_n^{(m)}=e^x{x}^{-m}{D}^n(x^{n+m}{e}^{-x})/n!$ has the endpoint value and squared norm displayed above. The lowest-power term after $n$ differentiations is $(n+m)!x^m/m!$, giving the endpoint value after division by $n!x^m$. For the norm, integrate by parts $n$ times against the same polynomial; its leading coefficient is $(-1{)}^n/n!$, so the final derivative is $(-1{)}^n$. All boundary terms vanish and the remaining Gamma function integral is $(n+m)!/n!$. Normalizing the value at zero to one divides the polynomial by $\left(\genfrac{}{}{0ex}{}{n+m}{n}\right)$ and its squared norm by the square of that factor. Only when $m=0$ or $n=0$ do these normalizations already coincide.

For a vector state on a C-star algebra, $N=\{a:a\eta =0\}$ and $⟨[a],[b]⟩=⟨a\eta ,b\eta ⟩$. Thus $U_0[a]=a\eta$ is a well-defined linear isometry from the quotient pre-Hilbert space onto $A\eta$. It extends to an isometry from the completed GNS space onto $K=\overline{A\eta }$: its range is complete and therefore closed, and contains the dense subspace $A\eta$. The space $K$ is invariant under both $A$ and its adjoints. On quotient vectors, $U\pi (a)[b]=ab\eta =aU[b]$, proving the displayed unitary equivalence after continuity. The cyclic vector maps to $\eta$.

If $A=\mathcal{B}(H)$, then $A\eta =H$, since the rank-one map $v\mapsto ⟨v,\eta ⟩\zeta$ sends $\eta$ to any chosen $\zeta$. In this case, for the rank-one orthogonal projection $p_{\eta }$, $a\eta =0$ is equivalent to $a{p}_{\eta }=0$, by evaluating at $\eta$ and using $(a{p}_{\eta })v=⟨v,\eta ⟩a\eta$.

For the state on a C-star algebra $a\mapsto \mathrm{Tr}(\rho a)$ on $M_d(\mathbb{C})$, the GNS construction has the displayed Hilbert-space dimension. Here $\rho$ is a density matrix. Its null space for the GNS seminorm consists of $b$ with $b{\rho }^{1/2}=0$, because $\mathrm{Tr}(\rho b^{\ast }b)=\Vert b{\rho }^{1/2}{\Vert }_{\mathrm{HS}}^2$. Thus the GNS space is the space of matrices $b{\rho }^{1/2}$ with the Hilbert-Schmidt inner product. If $\rho$ has rank $r$, diagonalization identifies these matrices with $r$ arbitrary columns, giving dimension $dr$. Left multiplication is the representation and ${\rho }^{1/2}$ is its cyclic vector. In particular a pure rank-one matrix state gives the defining $d$-dimensional representation, while a full-rank state gives a $d^2$-dimensional space.

On the C-star algebra of convergent complex sequences, a functional in the displayed coordinate representation is a state on a C-star algebra exactly when all its weights are nonnegative and sum to one. Coordinate projections test positivity of $w_j$ at finite indices. Positive sequences with the first $N$ terms zero and the rest one test $w_{\infty }$ after passage to the limit. The known dual isometry makes the norm the sum of the absolute weights, proving necessity and sufficiency.

If only finitely many weights are positive, the GNS construction identifies the quotient by $\{\alpha :F({\alpha }^{\ast }\alpha )=0\}$ with the values at precisely those supporting indices. Its inner product is the weighted coordinate inner product. Passing to the standard Hilbert norm gives diagonal evaluation operators and cyclic vector whose entries are the square roots of the weights. Arbitrary finite coordinate values and the limiting value can be assigned independently by a convergent sequence, which proves cyclicity. For weights $1/2$ at infinity and $1/4$ at each of the first two indices, the space is ${\mathbb{C}}^3$ and the cyclic vector is $(1/\sqrt{2},1/2,1/2)$.

Let $R$ be the valuation ring of a complete non-Archimedean valued field whose residue field has characteristic $p$, and let $\mathfrak{m}=\{x:|x|<1\}$. For a one-dimensional formal group law over $R$, repeated addition gives $[p](X)=pX+{\sum }_{j\ge 2}{c}_j{X}^j$ with $c_j\in R$. Hence $|[p](x)|\le \max (|p||x|,|x{|}^2)$. The series converges for $x\in \mathfrak{m}$ because its coefficients are bounded and $|x|<1$.

For $x_0=x$ and $x_{n+1}=[p](x_n)$, choose $\rho =\max (|p|,|x|)<1$. The values remain in the ball $|x_n|\le |x|$, so $|x_n|\le {\rho }^n|x|\to 0$. Associativity identifies $x_n=[p^n](x)$. The argument also works in equal characteristic $p$, where the linear coefficient vanishes, and does not require a discrete valuation. It proves contraction of prime-power multiplication near the formal identity without assuming a characteristic-zero formal logarithm.

Let a one-dimensional formal group law be defined over the valuation ring of a complete discretely valued non-Archimedean valued field of characteristic zero and residue characteristic $p$. If $z$ in the maximal ideal has exact formal order $p^n$, its absolute value satisfies the displayed lower bound. Use the coefficient splitting $[p](T)=pf(T)+g(T^p)$, where $f(T)=T+O(T^2)$ and $g(0)=0$. On the maximal ideal, $|f(z)|=|z|$ and $|g(z^p)|\le |z{|}^p$. If $|z|<|p{|}^{1/(p-1)}$, the first term strictly dominates, so $|[p]z|=|p||z|$ and repeated multiplication never reaches zero. Every nonzero $p$-power torsion point is therefore at least that large.

On this remaining radius range, $|[p]z|\le |z{|}^p$. For $n=1$, cancellation in $[p]z=0$ gives $|z{|}^{p-1}\ge |p|$. For larger $n$, apply induction to the point $[p]z$ of order $p^{n-1}$ and take a $p$th root. This yields the stated exponent. Over ${\mathbb{Q}}_p$ for odd $p$, the radius already exceeds $p^{-1}$, excluding torsion from the formal group on $p{\mathbb{Z}}_p$.

The lowest even quantum harmonic oscillator quantum state has density $\rho (y)=e^{-y^2/{\tau }^2}/(\sqrt{\pi }\tau )$, and the lowest odd quantum state has density $2y^2\rho (y)/{\tau }^2$. Since ${\rho }^{\prime }=-2y\rho /{\tau }^2$, integration by parts gives ${\mathbb{E}}_{\mathrm{F}}[f]={\mathbb{E}}_{\mathrm{B}}[f+y{f}^{\prime }]$ whenever the boundary terms vanish. For $f(y)=y^n$, this proves the displayed identity. Odd moments vanish in both quantum states by parity; even moments quantify the larger spatial spread in the odd quantum state. In particular the relative-coordinate variance is three times as large, and the odd density vanishes at particle coincidence.