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The complete metric space $\mathbb{R}$ supplies a counterexample: $A_n=[n,\infty )$ are descending nonempty closed sets, but no real number belongs to them all.

Now assume shrinking diameters. Choose $x_n\in A_n$. Given $\epsilon >0$, choose $N$ so that $\mathrm{diam}(A_N)<\epsilon$. Whenever $m,n\ge N$, both chosen points lie in $A_N$, so $d(x_m,x_n)<\epsilon$. Thus $(x_n)$ is a Cauchy sequence. By completeness it converges to some $x\in X$. For each fixed $N$, the tail lies in the closed set $A_N$, so its limit lies there too. Hence $x\in {\bigcap }_n{A}_n$. Any two points of this intersection have distance at most every $\mathrm{diam}(A_n)$, so their distance is zero. The shrinking intersection contains exactly one point. This is the Cantor intersection theorem.

For the dense-open conclusion, begin inside an arbitrary nonempty open set $O\subseteq X$; taking $O=X$ gives the requested existence. Since $U_1$ is a dense subset, choose $x_1\in O\cap U_1$ and $0<r_1\le 1/2$ small enough that the closed ball $A_1=\overline{B}(x_1,r_1)$ is contained in $O\cap U_1$. Here $\overline{B}$ denotes the closed ball, not an assumption about the closure of the open ball.

Inductively, the open ball $B(x_n,r_n)$ meets the dense open set $U_{n+1}$. Choose $x_{n+1}$ in that intersection and a positive radius $r_{n+1}\le 2^{-(n+1)}$ so small thatSuch a radius exists by openness and the triangle inequality. These nonempty closed sets descend and have diameters at most $2r_n\to 0$. The preceding result gives a point in every $A_n$, hence in $O$ and every $U_n$. Therefore ${\bigcap }_n{U}_n$ is nonempty; in fact it is dense in $X$. This proves the relevant Baire category theorem directly, without assuming closed balls are compact.

Suppose $K=\ker \phi$ for a group homomorphism $\phi :G\to H$. Its identity belongs to $K$, and for $k,l\in K$ we have $\phi (k{l}^{-1})=1$, so it is a subgroup. Moreover, for $g\in G$ and $k\in K$,Hence $gK{g}^{-1}\subseteq K$; applying this also to $g^{-1}$ gives equality, so $K$ is a normal subgroup.

Conversely, if $K$ is a normal subgroup, multiplication of cosets in the quotient group $G/K$ is well-defined. The map $g\mapsto gK$ is a group homomorphism and has kernel of a group homomorphism exactly $K$. This proves both directions without requiring $\phi$ to be injective or surjective.

For the matrix application, reduce all entries modulo $p$:Reduction respects matrix multiplication and the determinant, so this is a group homomorphism. Its kernel of a group homomorphism consists exactly of matrices congruent to the identity modulo $p$, which are the given matrices. Therefore $K=\ker \rho$ is a normal subgroup, the level-$p$ principal congruence subgroup. Surjectivity of reduction is not needed for this argument.

Associate to $Q$ its symmetric bilinear form by the polarization identity,We first construct a diagonal basis by induction on the dimension. A zero quadratic form is already diagonal. Otherwise choose $v$ with $Q(v)\ne 0$. Every $x$ has the decompositionwhose second term lies in $W=\{w:B(w,v)=0\}$. Also $\mathbb{R}v\cap W=\{0\}$, so $V=\mathbb{R}v\oplus W$ and $\dim W=n-1$. Apply the induction hypothesis to the restricted quadratic form on $W$. The cross terms with $v$ vanish, giving a diagonal basis on $V$. Rescale each vector with diagonal coefficient $a\ne 0$ by $1/\sqrt{|a|}$, and reorder positive, negative and zero coefficients. This yieldsNo nondegeneracy assumption was used.

For uniqueness, let $V_{+}$ be the positive coordinate subspace and $V_{\le 0}$ the span of all negative and zero coordinate vectors. Their dimensions are $p$ and $n-p$. Any subspace $W$ on which $Q$ has positive definiteness intersects $V_{\le 0}$ only in zero. The dimension formula for a sum of subspaces therefore gives $\dim W\le p$. Conversely $V_{+}$ itself has dimension $p$ and positive restriction. Thus $p$ is the largest possible dimension of a positive subspace, an intrinsic description independent of the basis. Applying the same argument to $-Q$ shows that $q$ is the largest dimension of a negative subspace. This proves Sylvester's law of inertia, including uniqueness of the zero count $n-p-q$.

The extremal subspaces themselves need not be unique. For $Q(x,y)=x^2-y^2$, the coordinate axes give one positive and one negative line. For any $t\ne 0$, putThen $Q(u)=1$, $Q(v)=-1$, $B(u,v)=0$, and the determinant of the change of basis is one. The distinct lines $\mathbb{R}u$ and $\mathbb{R}v$ give another positive/negative decomposition, with the same intrinsic counts $p=q=1$.

For a Markov chain started in its stationary distribution, reversibility means that for every $n$, the joint distribution of $(X_0,\dots ,X_n)$ equals that of $(X_n,\dots ,X_0)$. More generally, stationarity then makes every finite segment invariant under reversal of its time order.

If the Markov chain is reversible, take $n=1$. Comparing the events $(X_0,X_1)=(i,j)$ and $(j,i)$ gives the detailed balance equationsConversely, an irreducible Markov chain with an invariant probability distribution has ${\pi }_i>0$ for every state. Under detailed balance, a transition is positive exactly when its reverse is positive. For a path with positive probability, substitute $P_{ij}={\pi }_j{P}_{ji}/{\pi }_i$ in each factor to obtainThe factors of $\pi$ telescope. If a path has a zero transition, its reverse does too, so the equality still holds. These are the respective forward and reversed path probabilities. Summing over any collection of paths proves equality of the joint distributions, hence a reversible Markov chain. The first factor in the PDF's balance equation is ${\pi }_i$; the converted TeX incorrectly changes it to ${\pi }_j$.

For the first sampling design, treat the fixed groups of 60 and 90 people as two independent, representative samples from the respective populations. Within each sample, assume independent individuals with common eye-colour probabilities and mutually exclusive, exhaustive categories. This gives two independent multinomial distributions. The null hypothesis is that the two populations have the same eye-colour probabilities; its estimated common probabilities are $40/150$, $70/150$ and $40/150$.

The expected counts under the null hypothesis areThe Pearson chi-squared test of homogeneity usesThere are $2(3-1)=4$ unrestricted category parameters and $3-1=2$ under the null hypothesis, so the asymptotic chi-squared distribution has two degrees of freedom. All expected counts exceed five, making the usual large-sample approximation reasonable. Reject equality at the 5% level, but not at the 1% level: $5.99<T<9.21$. The approximate p-value is $\mathbb{P}({\chi }_2^2\ge T)=e^{-25/7}\simeq 0.0281$. Independence and representative sampling, rather than merely the count sizes, are essential to population inference; the result does not establish a causal explanation.

In the second design, the total sample size is fixed but both row and column classifications are observed for each sampled person. Assume an independent, representative sample with a common joint distribution of handedness and eye colour. The six cell counts then have one multinomial distribution. The null hypothesis is independence of the two classifications, so joint probabilities factor into row and column probabilities. Their maximum-likelihood estimates are the observed marginal proportions, again giving exactly $E$ above.

The unrestricted joint distribution has five free parameters; independence leaves $(2-1)+(3-1)=3$. The Pearson chi-squared test of independence again has two degrees of freedom, the same statistic, and the same significance conclusion. The numerical analysis is unchanged, but the sampling model changes from two fixed-row samples to one joint sample. If handedness groups had instead been deliberately sampled with fixed row sizes, the first, homogeneity formulation would apply to those groups too.

The general continuity equation is ${\rho }_t+\nabla \cdot (\rho u)=0$, or $D\rho /Dt+\rho \nabla \cdot u=0$. An incompressible flow has $D\rho /Dt=0$, so for positive density $\boxed{\nabla \cdot u=0}$.

For the displayed velocity field, on $x^2+y^2>0$,The sum is zero. A stream function with the required sign convention isIts derivatives give exactly $u=({\psi }_y,-{\psi }_x)$. The origin is excluded because the prescribed velocity is singular there.

For a real potential, the Time-dependent Schrödinger equation and its complex conjugate giveThe potential terms cancel. The expression on the right is a total derivative, givingFor the plane wave, ${\psi }_t=-i\omega \psi$ and ${\psi }_{xx}=-k^2\psi$. With zero potential, substitution givesIt is a free-particle momentum eigenstate with momentum $\hslash k$, energy $\hslash \omega$ and uniform probability flux. On the whole real line it is a generalized state, not a normalized square-integrable wavefunction; wave packets or box normalization give the corresponding physical interpretation.

In vacuum SI notation, Maxwell's equations areTaking the divergence of the last equation, using that the divergence of a curl vanishes and then the first equation, givesThis is local conservation of electric charge. For spatially uniform electrical conductivity, Ohm's law $J=\sigma E$ gives $\nabla \cdot J=\sigma \rho /{ϵ}_0$, soFor a homogeneous dielectric conductor with permittivity $ϵ$, replace ${ϵ}_0$ by $ϵ$. This charge relaxation is a statement about bulk charge; charge transported to a boundary is still accounted for by the conservation law.

In three dimensions, the general rank-two isotropic tensor is $a{\delta }_{ij}$ and the general rank-three tensor invariant under proper rotations is $b{ϵ}_{ijk}$. Here $\delta$ is the Kronecker delta and $ϵ$ the Levi-Civita symbol. The latter is orientation-sensitive: an ordinary third-rank tensor invariant under all orthogonal transformations is zero instead. This distinction does not change the stress conclusion.

For an isotropic material, the constitutive tensor must consequently have $A_{ijk}=b{ϵ}_{ijk}$ in the proper-rotation convention. Hence ${\sigma }_{ij}=b{ϵ}_{ijk}{B}_k=-{\sigma }_{ji}$. Mechanical stress in the model is also symmetric, so ${\sigma }_{ij}=-{\sigma }_{ij}$ and every component vanishes. Thus nonzero stress linear in the field requires anisotropic material coefficients. This is the isotropic linear magnetostriction vanishes restriction, not a claim that all higher-order magnetostriction is impossible in isotropic materials.

For a piecewise continuously differentiable curve $\gamma (t)=(x(t),y(t))$ in the upper half-plane model, the given Riemannian metric gives hyperbolic lengthThe metric matrix has determinant $y^{-4}$, so its area element is $y^{-2}dxdy$. Thus a measurable region $D$ has hyperbolic area ${\int }_D{y}^{-2}dxdy$. For the specified strip the integral isThe infinite Euclidean height is compatible with finite hyperbolic area because the area density decreases quadratically with height.

The normalizer is $N_G(H)=\{g\in G:gH{g}^{-1}=H\}$. It is a subgroup: the identity normalizes $H$, products of normalizing elements normalize it, and conjugating the equality by $g^{-1}$ proves inverse closure. The mapis well-defined and onto. Two images are equal exactly when $g_2^{-1}{g}_1\in N_G(H)$, exactly the condition that their cosets coincide. Thus the number of conjugates is $\boxed{[G:N_G(H)]}$.

By the permitted conjugacy assertion for Sylow subgroups, the number $n_p$ of Sylow $p$-subgroups equals $[G:N_G(P)]$ for one such subgroup $P$. Lagrange's theorem makes this divide $|G|$, and $P\subseteq N_G(P)$ also makes it divide $|G|/|P|$.

For order $72=8\cdot 9$, a Sylow $3$-subgroup has order nine, so $n_3$ divides eight. To obtain the required congruence without assuming an additional Sylow counting theorem, let $P$ act by conjugation on the set of Sylow $3$-subgroups. Its orbit sizes are powers of three. A fixed subgroup $Q$ is normalized by $P$, so $PQ$ is a subgroup and its order $|P||Q|/|P\cap Q|$ is a power of three. Since nine is the largest power of three dividing $72$, this forces $PQ=P=Q$. There is exactly one fixed point, namely $P$ itself. All other orbits have sizes divisible by three, giving $n_3\equiv 1(\mathrm{mod}3)$. Among $1,2,4,8$, only one and four satisfy this. Therefore $\boxed{n_3\in \{1,4\}}$.

The directed positive-transition graph gives the communicating classes $\{1,3,6\}$, $\{2,4\}$ and $\{5\}$. The first and last classes are closed; $\{2,4\}$ is open. In particular the PDF's fourth row has all six entries equal to $1/6$. The converted TeX incorrectly drops the last entry; that incomplete row is not a transition distribution.

The finite irreducible closed class $C=\{1,3,6\}$ reaches state six almost surely. For example, from every state in $C$ there is a uniformly positive probability of hitting six within a bounded number of steps, so the probability of avoiding it for successive blocks tends to zero. Put $h_i={\mathrm{Pr}}_i(T_6<\infty )$, with $h_1=h_3=h_6=1$ and $h_5=0$. First-step conditioning at states two and four givesSolving $4h_2-h_4=2$ and $5h_4-h_2=3$ yields $\boxed{h_2=13/19}$, with $h_4=14/19$.

Starting in state three, the chain remains in $C$. In the order $(1,3,6)$ its transition matrix isIts characteristic polynomial is $(\lambda -1)(\lambda +1/4)(\lambda +1/6)$. Thus, by the Cayley-Hamilton theorem, $q_n=(M^n{)}_{2,3}$ is a linear combination of $1,(-1/4{)}^n,(-1/6{)}^n$. The initial values are $q_0=0$, $q_1=1/3$, and $q_2=(1/3)(1/2+1/3+1/4)=13/36$. Solving for the three coefficients givesThe limiting value $12/35$ agrees with the state-six probability of the stationary distribution $(8/35,3/7,12/35)$ of this closed class.

The Rao-Blackwell theorem states that if an estimator $D$ has finite second moment and $T$ is a sufficient statistic, then $D^{\ast }(T)=\mathbb{E}[D\mid T]$ can be chosen without knowing the parameter. It has the same expectation as $D$ and no larger variance. In particular an unbiased estimator remains unbiased and its squared-error risk cannot increase. Sufficiency makes the conditional distribution, hence the function of $T$ used here, parameter-independent.

The tower property gives $\mathbb{E}{D}^{\ast }=\mathbb{E}D$. The law of total variance givesso the variance decreases, with equality exactly when $D$ is already a function of $T$ almost surely. Since the bias is unchanged, the same identity proves the squared-error risk claim. More generally the corresponding convex-loss inequality follows from conditional Jensen's inequality.

For the given geometric distribution, the joint mass is $(1-p{)}^n{p}^{\sum x_j-n}$ on positive-integer sample vectors. Thus $T={\sum }_j{X}_j$ is a one-dimensional sufficient statistic. Directly, conditional on $T=t$, every positive-integer composition of $t$ into $n$ parts has the same probability; there are $\left(\genfrac{}{}{0ex}{}{t-1}{n-1}\right)$ such compositions, independently of $p$.

The simple estimator $D=1_{\{X_1>1\}}$ is unbiased for $p$, since $\mathrm{Pr}(X_1>1)=p$. For $n\ge 2$, compositions with $X_1>1$ are counted by subtracting one from their first part, giving $\left(\genfrac{}{}{0ex}{}{t-2}{n-1}\right)$ when $t>n$. Consequently the Rao-Blackwell estimator of a geometric failure probability isAt $T=n$ its value is zero, as it must be when all observations are one. For the possible one-observation case, use $\hat{p}=1_{\{T>1\}}$; the displayed fraction would otherwise be undefined at $T=1$. The conditional-expectation argument proves unbiasedness without needing to sum the negative-binomial mass explicitly.

In the inviscid incompressible irrotational model, write the total velocity as $(U+{\varphi }_x,{\varphi }_y)$. The perturbation potential obeys Laplace's equation in $-h<y<\eta (x,t)$. The complete boundary conditions, neglecting surface tension and taking the air pressure constant, areThe first surface condition expresses that the free surface is material; the second is the Bernoulli equation with atmospheric pressure and the constant background kinetic energy absorbed into the potential's time gauge.

At first order, evaluation is on $y=0$ and products of perturbations vanish. HenceFor a mode $e^{i(\omega t-kx)}$, the bottom condition selects $\varphi =C\cosh [k(y+h)]e^{i(\omega t-kx)}$. Let $\Omega =\omega -kU$. The linear surface conditions areEliminating $C$ gives the dispersion relationIt is the intrinsic gravity-wave frequency squared, with the uniform current supplying the Doppler shift. For either sign of real $k$, $k\tanh (kh)$ is nonnegative.

Expand $e_{r+1}$ in the given spanning set:Some $b_j$ must be nonzero, since otherwise $e_{r+1}$ would lie in the span of the earlier $e_i$, contradicting linear independence. Reorder the remaining $f_j$ so $b_{r+1}\ne 0$. Solving this equation for $f_{r+1}$ expresses it in terms of $e_1,\dots ,e_{r+1},f_{r+2},\dots ,f_m$. Every member of the old spanning set is therefore in the new span, so the replacement set still spans $V$. This proves the needed step of the Steinitz exchange lemma directly.

Starting with $f_1,\dots ,f_m$, apply that replacement successively to $e_1,e_2,\dots$. After $r\le m$ steps the spanning list is $e_1,\dots ,e_r$ together with $m-r$ unreplaced $f$'s. If $n>m$, after $m$ replacements the list $e_1,\dots ,e_m$ spans $V$, forcing $e_{m+1}$ to be a linear combination of it. That contradicts linear independence. Hence $\boxed{n\le m}$.

The angular momentum commutation relations are $[L_i,L_j]=i\hslash {ϵ}_{ijk}{L}_k$. The commutator product rule givessince the bracketed expression is symmetric in $j,k$ and ${ϵ}_{ijk}$ is antisymmetric. Alsoproving $\boxed{L^2=L_{-}{L}_{+}+L_3^2+\hslash L_3}$.

For orbital angular momentum, $L_i=-i\hslash {ϵ}_{ijk}{x}_j{\partial }_k$. A differentiable radial function has ${\partial }_{k}f(r)=f^{\prime }(r)x_k/r$, so its contraction with the antisymmetric symbol vanishes: $Lf(r)=0$. Put $s=x_1+i{x}_2$. The coordinate differential operators give $L_{3}s=\hslash s$, whileannihilates every power of $s$, because $({\partial }_1+i{\partial }_2)s=0$. Since all $L_i$ annihilate the radial factor, the product rule yieldsSubstitution into the expression for $L^2$ gives the highest-weight complex-coordinate orbital wavefunction resultAs $[L^2,L_{-}]=0$, lowering preserves this eigenvalue whenever the lowered function is nonzero. Explicitly,For $n=0$ the lowered function is zero, so it is not an eigenfunction. For $n\ge 0$ the original nonzero function has the usual regular angular dependence. The PDF's “any integer” differential identities also hold locally for negative $n$ where $s\ne 0$, but those negative powers are singular on the axis and are not square-integrable regular angular eigenstates. The converted TeX incorrectly writes $L_4$ in place of the PDF's $L_{+}$.