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Letbe reflection in the unit circle, and represent a Möbius transformation $g$ byA direct calculation shows that $JgJ$ is represented byThus $gJ=Jg$ exactly when $M$ and $\tau (M)$ differ by a nonzero scalar. Applying the conjugate-linear involution $\tau$ twice shows that this scalar has modulus one. Rescaling $M$ by a suitable complex scalar then makes $M=\tau (M)$. Consequently all commuting maps, and only those maps, have the formThis is the Möbius maps commuting with reflection in the unit circle classification.

For such a map,It follows that $|g(z)|<1$ whenever $|z|<1$ precisely whenThe inverse has the same property, so these and only these maps preserve the unit disc.

DefineThis is continuous and takes values on the unit sphere. It agrees at $(0,y)$ and $(1,y)$, is constant on the bottom edge, and is constant on the top edge. Thus it is constant on every $R$-class, so the universal property of the quotient topology gives a continuous map

For $0<y<1$, the third coordinate $\cos (\pi y)$ determines $y$, and the first two coordinates determine $x$ modulo one. Consequently the only equal values of $f$ in the open strip arise from $x=0$ and $x=1$. At $y=0$ the whole edge maps to the north pole, and at $y=1$ the whole edge maps to the south pole. These are exactly the identifications defining $R$, so $\overline{f}$ is injective. The spherical-coordinate formula also shows that it is surjective.

The square is compact, hence its quotient $X/R$ is compact by part (ii), while $S^2$ is Hausdorff as a subspace of ${\mathbb{R}}^3$. The compact-to-Hausdorff continuous bijection theorem now makes $\overline{f}$ a homeomorphism. Geometrically, identifying the vertical sides produces a cylinder and collapsing each boundary circle to a point produces the suspension of a topological space $S^1$, which is $S^2$. This proves the square quotient model of the two-sphere.

Let $X$ be compact and $q:X\to X/R$ the quotient map. If $(U_i{)}_{i\in I}$ is an open cover of $X/R$, then $(q^{-1}(U_i){)}_{i\in I}$ is an open cover of $X$. A finite subfamily covers $X$; because $q$ is surjective, the corresponding $U_i$ cover $X/R$. Thus every quotient of a compact space is compact.

The Hausdorff property need not survive. On the real line, defineThe quotient $\mathbb{R}/\mathbb{Q}$ has more than one point. If two nonempty open subsets of the quotient were disjoint, their inverse images would be disjoint nonempty open subsets of $\mathbb{R}$ invariant under rational translation. But any two nonempty open intervals acquire an intersection after one is translated by a suitably chosen rational number, so two such saturated open sets cannot be disjoint. Distinct quotient points cannot be separated, and the quotient is not Hausdorff. This is the non-Hausdorff quotient of the real line by rational translation.

Finally let $f:X\to Y$ be a continuous bijection, with $X$ compact and $Y$ Hausdorff. Every closed subset $C\subseteq X$ is compact. Its continuous image $f(C)$ is compact, and every compact subset of a Hausdorff space is closed. Hence $f$ is a closed map. For every closed $C\subseteq X$,is closed in $Y$, so $f^{-1}$ is continuous. Therefore $f$ is a homeomorphism. This is the compact-to-Hausdorff continuous bijection theorem.

A compact space is a topological space in which every open cover has a finite subcover. A Hausdorff space is one in which every two distinct points have disjoint open neighbourhoods. A homeomorphism is a bijection that is continuous and whose inverse is continuous.

For an equivalence relation $R$ on $X$, let $X/R$ be the set of equivalence classes and letThe quotient topology declares $U\subseteq X/R$ open exactly when $q^{-1}(U)$ is open in $X$. It follows directly from the definition that $q$ is continuous.

Suppose that the continuous map $f:X\to Y$ is constant on equivalence classes. The only possible factorisation iswhich is well-defined by the hypothesis and satisfies $F\circ q=f$. For every open $V\subseteq Y$,is open in $X$. The definition of the quotient topology therefore makes $F^{-1}(V)$ open, so $F$ is continuous. This proves the universal property of the quotient topology.

By stationarity,The two-step return probability isTherefore the requested probability is

Yes. The three nontrivial detailed balance identities areandThus the chain is a reversible Markov chain.

Let $\pi =({\pi }_A,{\pi }_B,{\pi }_C)$. Solving the stationary distribution equations $\pi P=\pi$ together with ${\pi }_A+{\pi }_B+{\pi }_C=1$ gives

Suppose ${\chi }_1$ and ${\chi }_2$ are two bound-state solutions with the same energy. Their Wronskiansatisfies $W^{\prime }=0$ because the Schrödinger equation contains no first-derivative term. Hence $W$ is constant. Both wavefunctions and their derivatives decay at infinity, so $W\to 0$ there and therefore $W\equiv 0$. The two solutions are linearly dependent. Thus every one-dimensional bound-state energy is nondegenerate.

As $|x|\to \infty$, the equation approachesIf $E>U_0$, the asymptotic solutions are oscillatory and do not decay, so there are no bound states. If $0<E<U_0$, the asymptotic solutions are exponential; one decaying branch can be selected at each end. Bound states are therefore possible, although only energies satisfying the global matching conditions actually occur.

The spatial wavefunction $\chi (x)$ is the position amplitude of a stationary quantum state, so $|\chi (x){|}^2$ is its probability density after normalization. The real function $U(x)$ is the potential energy, $E$ is the energy eigenvalue, $m$ is the particle mass, and $\hslash$ is the reduced Planck constant.

Differentiate the affine relation with respect to $x$ and $y$:Using the Cauchy-Riemann equations, the second equation becomesThusThe determinant is $-(a^2+b^2)\ne 0$, so $u_x=v_x=0$ and therefore $f^{\prime }=0$. A holomorphic function with zero derivative on a connected open set is constant. Hence $f$ is constant, as in analytic function with image in an affine real line.

Taking the complex derivative along the real direction givesTaking it along the imaginary direction and equating the two expressions gives the Cauchy-Riemann equations

Orient every edge of the feasible convex polytope in the direction of increasing objective. WriteThe possible improving simplex method paths from $O$ to $A$ areThese exhaust the directed edge graph: the ordinary cube edges remain except those incident to the removed vertex $111$; each such edge ends at one of $A,B,C$; and the cutting face contributes the triangle with edges $AB,AC,BC$.

The first two paths use three pivots, while the two paths through both $C$ and $B$ use five. Therefore the smallest and largest possible numbers of simplex steps areand the total number of distinct outcomes isThis is the simplex paths on a cube with one truncated corner calculation.

Put $y_i=1-x_i$. Feasibility gives $y_i\ge 0$ andwhile the objective isTo maximize it, the required total deficit $1/2$ must be assigned entirely to the cheapest variable $y_1$. Thusand the unique optimizer isThis is the vertex $A$ from part (b).

An extreme point of a convex set $C$ is a point $x\in C$ that cannot be written as $x=(1-t)y+tz$ with $y\ne z$ in $C$ and $0<t<1$.

Start with the eight vertices of the unit cube. The constraint $x_1+x_2+x_3\le 5/2$ removes $(1,1,1)$ and retains the other seven. The cutting plane meets the three cube edges incident to the removed vertex atThese are new vertices. No others occur: at a vertex in three dimensions, three linearly independent bounding planes are active, and choosing triples from the six cube faces and the cutting plane yields precisely the listed points. Hence the ten extreme points are

The origin belongs to $S$, so $S$ is nonempty. Each displayed inequality defines a closed half-space, and every half-space is a convex set. An intersection of convex sets is convex: if $x,y\in S$ and $0\le t\le 1$, each linear inequality remains valid for $(1-t)x+ty$. Thus $S$ is a nonempty convex polytope.

A possibly randomized test is a function $\phi (X)\in [0,1]$, interpreted as the conditional probability of rejecting $H_0$. Its power function isIt has size at most $\alpha$ whenSuch a test is a uniformly most powerful test of size $\alpha$ if, for every other test $\psi$ of size at most $\alpha$,Thus one test maximizes power simultaneously at every parameter value in the alternative.

Fix a continuous function $f$ and $\epsilon >0$. By the Weierstrass approximation theorem, choose a polynomial $p\in {\mathcal{P}}_m$ such thatFor every $n\ge m$, exactness gives $I_n(p)=I(p)$. Positivity of the quadrature weights and exactness on the constant polynomial giveConsequentlySince $\epsilon$ is arbitrary, $I_n(f)\to I(f)$. This is the convergence of positive quadrature rules.