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Suppose the union $W$ has a separation $W=A\cup B$ into disjoint, nonempty, relatively open subsets. Each connected $U_j$ must lie entirely in one of them: otherwise its intersections with $A$ and $B$ would separate $U_j$ in its subspace topology. In particular $U_1$ lies on one side, say $A$. Every other $U_j$ intersects $U_1$, so it cannot lie in $B$ and must also lie in $A$. Then $W\subseteq A$, contradicting $B\ne \varnothing$. The union is connected. This is the connected union with a connected hub argument.

For path-connectedness, take any $x\in U_i$ and $y\in U_j$. Choose $a\in U_i\cap U_1$ and $b\in U_j\cap U_1$. There is a path from $x$ to $a$ inside $U_i$, a path from $a$ to $b$ inside $U_1$, and a path from $b$ to $y$ inside $U_j$. Concatenate the three paths on successive thirds of $[0,1]$. They agree at the joining points, so the resulting map is continuous and remains in $W$. The union is path-connected. There need not be a single point common to all the subsets.

A function $f:I\to \mathbb{R}$ is uniformly continuous ifThe same $\delta$ must work at every point of $I$.

Every uniformly continuous function on $(0,1)$ is bounded. Choose the tolerance $\delta$ for $\epsilon =1$ and an integer $N$ with $1/N<\delta$. The finitely many centers $c_j=(j+1/2)/N$, $0\le j<N$, cover the interval by cells of radius $1/(2N)$. For each $x$, one center satisfies $|f(x)-f(c_j)|<1$. Thus $|f(x)|\le 1+{\max }_j|f(c_j)|$. This is the uniformly continuous function on a totally bounded set is bounded argument; closedness of the interval is unnecessary.

The reciprocal function is not uniformly continuous on $(0,1)$. It is unbounded, already contradicting the preceding result. Directly, $x_n=1/(n+1)$ and $y_n=1/(n+2)$ have distance tending to zero while $|1/x_n-1/y_n|=1$.

The oscillating function $\sin (1/x)$ is not uniformly continuous either, despite being bounded. TakeTheir distance tends to zero, but their function values are $1$ and $-1$. The fixed output gap of two contradicts the uniform-continuity condition, for example with $\epsilon =1$.

A group automorphism is a bijective group homomorphism from $G$ to itself. The multiplication in $\mathrm{Aut}(G)$ is composition: $(\alpha \beta )(g)=\alpha (\beta (g))$. Composition is associative, the identity map is the identity element, and each inverse bijection is again a homomorphism, so these maps form the automorphism group.

For conjugation by $h$, multiplication is preserved becauseConjugation by $h^{-1}$ is its inverse, so $\psi (h)$ is an group automorphism. MoreoverHence $\psi :G\to \mathrm{Aut}(G)$ is a homomorphism, with image the inner automorphisms. Its group homomorphism kernel is the group center of $G$, since $\psi (h)$ is the identity precisely when $hg=gh$ for every $g$.

The transition row sums to one, since $p{\sum }_{k=0}^i{q}^k+q^{i+1}=1$. From any state the Markov chain can reach zero, and from zero consecutive upward moves reach every state; hence it is irreducible. Its self-transition at zero has probability $q>0$, so it is aperiodic. It is the reflected chain with geometric downward jumps.

Put $h_0=1$ for the hitting problem. The first-step equations for $i\ge 1$ areSubtracting $q$ times the equation at $i-1$ gives the linear recurrence relation, valid for $i\ge 2$,The separate boundary equation at $i=1$ is $h_1=q^2+pq{h}_1+p{h}_2$. It must not be replaced by the linear recurrence relation with $h_0=1$: the absorbing hitting convention at zero is different from applying a first-step return equation there.

For $p\ne q$, the linear recurrence relation has roots $1$ and $r=q/p$, so $h_i=A+B{r}^i$ for $i\ge 1$. The boundary equation gives $B=r(1-A)$. If $p<1/2$, then $r>1$ and boundedness of probabilities forces $B=0$, hence $A=1$. If $p=1/2$, the repeated-root solution is $A+Bi$; boundedness again gives $B=0$ and the boundary equation gives $A=1$.

For $p>1/2$, $r<1$ and the nonnegative harmonic candidate ${\tilde{h}}_i=r^{i+1}$, together with ${\tilde{h}}_0=1$, satisfies every first-step equation. Hitting-by-time-$n$ probabilities start from zero off the target and iterate those equations, so by induction they are bounded above by this candidate. Taking their increasing limit gives $h_i\le r^{i+1}$. But the general bounded solution has limit $A\ge 0$ as $i\to \infty$, forcing $A=0$ under that bound. ThusIn particular the transient exponent is $i+1$, fixed by the original boundary equation.

The probability of a subsequent return to zero when starting there is $f_{00}=q+p{h}_1$. It is one for $p\le 1/2$, so the Markov chain is recurrent there. For $p>1/2$ it equals $q+p(q/p{)}^2=q/p<1$, so the irreducible Markov chain is transient.

For a stationary probability vector $\pi$, balance at zero and one gives ${\pi }_1=(p/q){\pi }_0$. Subtracting adjacent balance equations givesso this initial ratio propagates and ${\pi }_j={\pi }_0(p/q{)}^j$. It can be normalized only when $p<q$. In that caseDirect substitution verifies all balance equations, including ${\pi }_0=q{\sum }_i{\pi }_i{q}^i$. An irreducible Markov chain admitting this stationary probability law is positive recurrent, with mean return time $1/{\pi }_0=q/(q-p)$. At $p=q$ the linear recurrence relation and boundary relation force every ${\pi }_j$ equal; no such nonzero sequence is summable. The recurrent chain is therefore null recurrent. In summary,

Direct matrix multiplication gives the multiplication tableTogether with $I$ acting as the identity, this shows that the product of any two real linear combinations is again a real linear combination. Thus $\Omega$ is closed under multiplication. It is the complex matrix representation of quaternions.

Its general element has the formIf this matrix is zero, real and imaginary parts of its first row give $a=b=c=d=0$. Hence $I,J,K,L$ are linearly independent over $\mathbb{R}$ and form a basis:Put $v=bJ+cK+dL$. Since distinct imaginary units anticommute, the mixed terms in $v^2$ cancel, leaving $v^2=-(b^2+c^2+d^2)I$. Scalars commute with $v$, soThe reversed product is the same. The coefficient is strictly positive for every nonzero element. ConsequentlyThis also identifies $\Omega$ as a real noncommutative division algebra, not as the full algebra of complex two-by-two matrices.

For observed data $x$, let $F_x$ be the posterior cumulative distribution function. Assuming finite posterior first moment, the posterior risk under asymmetric absolute-error loss isAt points where the posterior has a density, differentiation gives $R_x^{\prime }(a)=(\gamma +\delta )F_x(a)-\gamma$. Risk is convex, so a global minimum is a posterior quantile:More generally any $a$ with $F_x(a^{-})\le \gamma /(\gamma +\delta )\le F_x(a)$ minimizes risk. This is the Bayes quantile under asymmetric absolute-error loss; a larger cost for underestimation moves the chosen quantile upward.

For the uniform sampling model set $M={\max }_i{X}_i$. For a valid positive sample the likelihood is ${\theta }^{-n}{1}_{\{\theta >M\}}$. Multiplying by the prior cancels its power of $\theta$ and gives posterior density proportional to $e^{-\theta }{1}_{\{\theta >M\}}$. Normalization yieldsSolving the quantile equation gives the explicit Bayes estimateThe posterior is the sample maximum plus a unit-rate exponential distribution displacement; it has finite first moment, so the preceding risk minimization is justified here.

In Cartesian coordinates the magnetic field is $(bx,by,-2bz)$. Its divergence is $b+b-2b=0$, and all components of its curl vanish. Thus it satisfies the static vacuum magnetic equations $\nabla \cdot \mathbf{B}=0$ and $\nabla \times \mathbf{B}=0$, with no electric current in the region.

Fix clockwise orientation as viewed from above, looking down the positive $z$-axis. Its positive area normal is $-{\mathbf{e}}_z$. The loop's external magnetic flux is thereforeWith the opposite normal its flux has the opposite sign. On the circle $\mathbf{B}=ba{\mathbf{e}}_r-2bz{\mathbf{e}}_z$ and clockwise line element is $d\mathbf{\ell }=-a{\mathbf{e}}_{\phi }d\phi$. The Lorentz force element isThe radial terms cancel around the circle, soThis fixes the current/flux/force signs together.

Neglect self-inductance initially, so Faraday's law and Ohm's law give $RI=-K\dot{z}$. Including gravity then givesFor $R>0$ and $K\ne 0$, the terminal solution is $z=z_0-vt$ withThe general velocity relaxes exponentially to $-v$ on time scale $mR/K^2$. In the terminal state $I=Kv/R$, and the Joule-heating rate isThus all the lost gravitational potential energy becomes resistive heat when kinetic energy is constant.

Within the zero-inductance approximation $v\to 0$ as $R\to 0$; finite-speed motion would require an unbounded current. This limit is singular, so it is important to describe the physical ideal-conductor limit separately. With finite self-inductance $\mathcal{L}$, the circuit equation is $\mathcal{L}\dot{I}+RI=-K\dot{z}$. At $R=0$, total linked flux $\mathcal{L}I+Kz$ is conserved and there is no Joule heating. If $I=0$ initially at $z_0$, thenThis lossless limit of magnetic loop braking has undamped oscillations about $z_0-mg\mathcal{L}/K^2$, with magnetic energy storing and returning mechanical energy. It has no generic dissipative terminal descent. Thus the vanishing terminal speed is the formal resistive-model limit, not a justification for dropping inductance in a perfectly conducting moving loop.

On a suitable dense domain of smooth rapidly decreasing wave functions, the canonical commutator is $[x,p]=i\hslash$. Expanding the two ladder operators givesTheir sum and difference giveDirect multiplication, retaining the operator order, givesUsing ${\beta }^2=m\omega /\hslash$ identifies the Hamiltonian operator:For any admissible wave function, adjointness givesFor the conventional normalized state $\Vert \Psi \Vert =1$, this is the stated lower bound $E\ge \hslash \omega /2$. For an unnormalized function the norm factor must be retained; rescaling a wave function cannot leave its unnormalized energy integral bounded below by a fixed positive number.

Equality holds exactly when $a{\Psi }_0=0$. In position representation this equation is $(\beta x+{\beta }^{-1}{\partial }_x){\Psi }_0=0$, whose normalized square-integrable solution isThis construction establishes existence of a state attaining the bound, and hence proves that it is the ground state.

The commutator $[a,a^{†}]=1$ implies $[H,a^{†}]=\hslash \omega a^{†}$. Repeatedly commuting through the product givesThese vectors are nonzero: the identity $a(a^{†}{)}^n{\Psi }_0=n(a^{†}{)}^{n-1}{\Psi }_0$ gives their norm squared $n!$ by induction. ThereforeFor time-dependent stationary states multiply each spatial eigenfunction by $e^{-i{E}_{n}t/\hslash }$.

Fixed endpoint variations yield the Euler–Lagrange equationWhen $F$ is independent of $z$, differentiation givesThis is the Beltrami identity. Denote the constant by $1/k$ when nonzero; for a general Lagrangian a zero constant is also possible.

The film area is $S=2\pi {\int }_{-H}^{H}r\sqrt{1+r^{\prime 2}}dz$. Suppressing its harmless positive factor $2\pi$, take $F=r\sqrt{1+r^{\prime 2}}$. The Euler–Lagrange equation reduces toThe first integral is $r/\sqrt{1+r^{\prime 2}}=1/k$ with $k>0$. Hence $1+r^{\prime 2}=k^2{r}^2$ and, using the differential equation without dividing by $r^{\prime }$, $r^{\prime \prime }=k^{2}r$. The positive solution obeying that first integral is $r=k^{-1}\cosh (k(z-z_0))$. Equal endpoint radii force $z_0=0$, givingThus every smooth positive stationary film in this axisymmetric class is a catenoid.

Put $u=kH>0$. The boundary equation is $R/H=f(u)=\cosh u/u$. The function tends to infinity at both endpoints of $(0,\infty )$, andIts minimum occurs when $u=\coth u$. The function $u-\coth u$ is strictly increasing, with derivative $1+{\mathrm{csch}}^{2}u>0$, and has limits $-\infty$ and $+\infty$. It therefore has exactly one positive zero $A$. At this point $1/A=\tanh A$, so $f(A)=\sinh A$. This proves the catenoid existence threshold for equal rings:At equality there is one stationary solution and above it two. The equation describes candidates for an area minimum, not a guarantee that both branches minimize. The thinner-neck branch is unstable to suitable axisymmetric variations; topology-changing collapse can also compete with connected films. In particular two separate planar disks have area $2\pi R^2$. The derived threshold concerns existence of the smooth annular stationary film; it does not exclude such disconnected competitors.

Close the real-line contour by a positively oriented upper semicircle of radius $R$, beyond every pole. The rational function is $O(R^{-2})$ uniformly on that arc, and $|e^{iz}|=e^{-\Im z}\le 1$. Its arc integral is therefore $O(R^{-1})$ and vanishes. The real integral is absolutely convergent, while no pole lies on it. The residue theorem givesOnly actual poles contribute; any removable zero of the denominator has zero residue. The same contour proof covers poles of any order.

For the particular denominator, the upper poles are $a=1+i$ and $b=-1+i$. They are simple, with residues $e^{ia}/(4a^3)$ and $e^{ib}/(4b^3)$. Sincetheir sum isMultiplication by $2\pi i$ gives the real answer. The sine part of the real-line integral is zero by oddness, so

For a Riemannian metric $g$, write $s(t)=\sqrt{g_{\gamma (t)}(\dot{\gamma }(t),\dot{\gamma }(t))}$. Use the energy of a curve normalization compatible with the requested inequality:The Cauchy-Schwarz inequality here has the direct variance formEquality holds exactly when $s$ is constant almost everywhere, hence everywhere for a smooth curve. With the alternative energy of a curve convention containing a factor $1/2$, the same assertion reads $L^2\le 2E$.

In the upper half-plane write $\gamma =u+iv$, $v>0$, and use $d{s}^2=(d{u}^2+d{v}^2)/v^2$. Let the endpoints be $P=ip$, $Q=iq$, $p,q>0$. ThenThe first equality requires $\dot{u}=0$ throughout. Endpoint values force $u=0$. The second equality requires the derivative of $\log v$ to have one sign, so $v$ is monotone. Conversely every smooth monotone positive $v$ with these endpoints attains this bound. Consequently the absolute length minimizers are precisely the vertical curves with monotone height, including a constant curve when $P=Q$, and the distance is $|\log (q/p)|$.

For curves stationary for the energy of a curve, fixed-endpoint variations give the Euler–Lagrange equations for $F=({\dot{u}}^2+{\dot{v}}^2)/v^2$:Thus $\dot{u}=c{v}^2$ for a constant $c$. Since $u(1)-u(0)=0$, integration gives $c{\int }_0^1{v}^{2}dt=0$, hence $c=0$. The second equation then becomes $(\dot{v}/v{)}^{\prime }=0$. ThereforeIt has constant hyperbolic speed, is one of the absolute length minimizers, and also has minimum energy of a curve $E=[\log (q/p){]}^2$.

Let $F=\mathbb{Z}{e}_1\oplus \mathbb{Z}{e}_2$ and $N=\mathbb{Z}(6e_1+9e_2)$. The precise presented group is $A=F/N$, with $x=e_1+N$ and $y=e_2+N$. It is abelian and has exactly the imposed additive relation and its integer multiples.

Change the integral basis to $u=2e_1+3e_2$, $v=e_1+2e_2$. The change-of-basis determinant is $2\cdot 2-3\cdot 1=1$, so these are a basis of the free abelian group. Explicitly, $e_1=2u-3v$ and $e_2=-u+2v$. The relation generator becomes $6e_1+9e_2=3u$. ThereforeThe image of $u$ is nonzero, since $(2,3)$ is not an integer multiple of $(6,9)$, but three times it vanishes. Thus $A$ has an element of exact order three. A free abelian group has no nonzero finite-order element, so $A$ is not free abelian.

Let $\dim V=n$ and $\dim W=k$. Extend a basis $w_1,\dots ,w_k$ of the vector subspace $W$ to $w_1,\dots ,w_n$ of $V$, and let $f_1,\dots ,f_n$ be its dual basis. Every linear functional is uniquely ${\sum }_i{c}_i{f}_i$. It vanishes on $W$ exactly when $c_1=\cdots =c_k=0$. Thus the annihilator of a vector subspace has basis $f_{k+1},\dots ,f_n$, and

To apply this to a matrix, regard its rows as elements of $({\mathbb{R}}^n{)}^{\ast }$ and let $R$ be their linear span. Elementary invertible row and column operations reduce a rank-$r$ matrix to a block identity $\mathrm{diag}(I_r,0)$, showing that $\dim R=r$. The equations require that every functional in $R$ vanish on $x$. The canonical identification ${\mathbb{R}}^n\cong (({\mathbb{R}}^n{)}^{\ast }{)}^{\ast }$ sends $x$ to evaluation $f\mapsto f(x)$: it is injective because coordinate functionals separate points, and is bijective by equal dimensions. Consequently the solution vector space is the annihilator of $R$ under this identification. The dimension formula just proved givesA basis of this kernel consists of exactly $n-r$ linearly independent solutions, and every solution is their linear combination.

Associate free real dual variables $u_i$ and $v_j$ with the row and column equality constraints. The Lagrangian for the transportation problem isIts infimum over $x_{ij}\ge 0$ is finite exactly when $u_i+v_j\le c_{ij}$ for every pair. The linear programming duality therefore givesThe dual variables have no sign restrictions because the primal constraints are equalities.

A feasible shipment $x$ is optimal if and only if there exist feasible transportation dual potentials satisfying complementary slackness:Thus every occupied route must have $u_i+v_j=c_{ij}$. To see sufficiency explicitly, the primal minus dual objective isand complementary slackness makes it zero. This is the weak duality certificate. Necessity follows from strong duality for the feasible, bounded linear program. If the total supply $M$ is positive, $x_{ij}=a_i{b}_j/M$ gives feasibility; if $M=0$, $x=0$ does. The feasible shipments form a closed bounded set, so the primal optimum is attained. This also covers zero supplies and demands. The one redundant equality merely leaves the dual gauge freedom $(u,v)\mapsto (u+t,v-t)$.