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Take $X$ to be the set $\mathbb{Q}$, equipped with the discrete metricEvery Cauchy sequence is eventually constant: its terms must eventually have mutual distance in this metric less than $1/2$. Consequently $X$ is a complete metric space. Its topology is the discrete topology, since every singleton is a ball of radius $1/2$ and every subset is a union of such balls.

For the codomain use the usual subspace topology on $\mathbb{Q}$. The inverse image of any open set there is a subset of the discrete $X$, hence is open. Therefore $f$ is continuous and plainly surjective. A continuous image of a complete metric space need not be complete. The two occurrences of $\mathbb{Q}$ have different domain and codomain metrics.

Let $K$ be a compact set of a Hausdorff space $X$. For any $x\notin K$ and each $y\in K$, the Hausdorff property gives disjoint open sets $U_y\ni y$ and $V_y\ni x$. The sets $U_y$ cover $K$; compactness supplies a finite subcover $U_{y_1},\dots ,U_{y_r}$. The open set $V={\bigcap }_i{V}_{y_i}$ contains $x$ and misses every $U_{y_i}$, hence misses $K$. Thus $X\setminus K$ is open, and every compact subset of a Hausdorff space is closed. The empty-set case is immediate.

Since $\mathbb{R}$ with its usual topology is Hausdorff, its compact sets are closed. The open cover $\{(-n,n):n\ge 1\}$ of $\mathbb{R}$ has a finite subcover of any compact set $K$. Taking the largest index in that subcover gives $K\subset (-N,N)$, so $K$ is a bounded set. This proves the required direction of the Heine-Borel theorem without assuming its converse.

The function $g(t)=|t{|}^{3/2}$ is differentiable everywhere: for $t\ne 0$, its derivative is $(3/2)\mathrm{sgn}(t)\sqrt{|t|}$, while at zero the difference quotient is $\mathrm{sgn}(t)\sqrt{|t|}\to 0$.

The function $k(t)=|t{|}^{1/2}$ is differentiable for $t\ne 0$, with derivative $\mathrm{sgn}(t)/(2\sqrt{|t|})$. At zero its difference quotient is $\mathrm{sgn}(t)/\sqrt{|t|}$, which has no finite limit. Applying the coordinatewise-sum argument givesAt those points,where the first coefficient is zero at $x=0$. At any point with $y=0$, the restriction to the vertical coordinate line already fails to be differentiable.

First suppose each $g_j$ is differentiable at $x_j$. Write $g_j(x_j+t)=g_j(x_j)+v_{j}t+r_j(t)$, where $v_j\in {\mathbb{R}}^m$ and $r_j(t)=o(|t|)$. Define the linear map $Lh={\sum }_j{v}_j{h}_j$. Given $\epsilon >0$, choose a common sufficiently small neighbourhood so that $\Vert r_j(h_j)\Vert \le \epsilon |h_j|$ for all $j$, with $r_j(0)=0$. ThenThis proves differentiability, withConversely, suppose $f$ is differentiable with Fréchet derivative $L$. Along the $j$th coordinate direction $e_j$, the other summands cancel exactly:Thus $g_j$ is differentiable at $x_j$, with $g_j^{\prime }(x_j)=L{e}_j$. The equivalence works because each summand varies in only one coordinate; it is stronger than merely having partial derivatives.

Differentiability here is understood in the sense of the Fréchet derivative: there is a linear map $L:{\mathbb{R}}^n\to {\mathbb{R}}^m$ such thatThe map $L=Df(x)$ is the Fréchet derivative. In finite dimensions a linear map is bounded, so $\Vert Lh\Vert \le C\Vert h\Vert$ for some $C$. The defining estimate therefore impliesThis is precisely continuity at $x$: differentiability implies continuity.

Let $C$ be an open communicating class, and fix $i\in C$. There is a path of positive probability from $i$ to a state outside $C$. Choose a shortest such path, so it does not revisit $i$ before exiting. No state reached outside $C$ can have a path back to $C$: combined with the exit path and communication within $C$, such a path would put it in the same communicating class. Thus with positive probability the chain never returns to $i$. Its return probability is less than one, making $i$ a transient state. This proves transience throughout $C$.

Conversely, suppose a finite communicating class were both closed and transient. Starting inside it, a path stays there forever. Since it has finitely many states, at least one state must be visited infinitely often. But a transient state is visited only finitely often almost surely, and the union of finitely many exceptional null events still has probability zero. This is a contradiction. Every finite transient communicating class is therefore open.

For the infinite counterexample, take the biased random walk on $\mathbb{Z}$ with independent increments $+1$ of probability $3/4$ and $-1$ of probability $1/4$. All states communicate, so $\mathbb{Z}$ is one infinite closed communicating class. The strong law of large numbers gives $X_n/n\to 1/2$ almost surely. Consequently every fixed state is visited only finitely often, and the class is transient. Finiteness is exactly what fails in the preceding argument.

A QR decomposition of an $m\times n$ matrix, $m\ge n$, is $A=QR$, where $Q$ is an $m\times m$ orthogonal matrix and $R$ is $m\times n$ and upper trapezoidal: $R_{ij}=0$ for $i>j$. In particular its bottom $m-n$ rows vanish.

Successive left Givens rotations annihilate the entries below the diagonal, column by column. When working on column $j$, rotate rows $j$ and $i>j$; their earlier columns are already zero, so the earlier eliminations persist. If $G_k\cdots G_{1}A=R$, thenA product of orthogonal matrices is orthogonal. The construction remains valid for a rank-deficient matrix, although uniqueness is not asserted. This is QR decomposition by Givens rotations.

A Givens rotation is the identity outside rows and columns $p,q$, with the $2\times 2$ blockThe block's transpose times itself is the identity, so the whole matrix satisfies $\boxed{{\Omega }^T\Omega =I}$ and is an orthogonal matrix. In particular it preserves the Euclidean norm. To annihilate the lower entry of a column pair $(a,b{)}^T$, take $c=a/r$, $s=b/r$, $r=\sqrt{a^2+b^2}$; the rotated pair is $(r,0{)}^T$. If both entries are zero, take the identity rotation.

For $E_0\ne 0$, put $\beta =v/c$. Part (c) givesThe vanishing of their cross product is equivalent toFor $0<\theta <\pi /2$ the two roots are $\tan (\theta /2)$ and $\cot (\theta /2)$. Only the first has $|\beta |<1$. The continuous endpoint $\theta =0$ gives $\beta =0$. Thus the physical boost isSubstitution gives ${\mathbf{E}}^{\prime }=c{\mathbf{B}}^{\prime }$, so the fields point in the same direction. At $\theta =\pi /2$ the formal root is $v=c$, which is not an inertial observer's velocity. The nonzero perpendicular equal-strength fields then have both electromagnetic field invariants zero. If they were parallel in a finite frame, those invariants would force both fields to vanish, impossible under an invertible Lorentz transformation. Hence no finite inertial boost makes them parallel in this case. If $E_0=0$, both fields are zero in every frame and no velocity is singled out.

Transform the electromagnetic field tensor by $F^{\prime \mu \nu }={\Lambda }^{\mu }{}_{\alpha }{\Lambda }^{\nu }{}_{\beta }{F}^{\alpha \beta }$. For example,Reading off the other components in the same way yields the transverse field transformationThe longitudinal components remain $E_x^{\prime }=B_x^{\prime }=0$. These signs correspond to the stated boost in which $S^{\prime }$ moves with velocity $v$ relative to $S$.

Contractions of Lorentz tensors have no free indices and are Lorentz scalars. The antisymmetry and the components in part (a) giveThus $\mathbf{E}\cdot \mathbf{B}$ and $|\mathbf{B}{|}^2-|\mathbf{E}{|}^2/c^2$ are the two electromagnetic field invariants under proper Lorentz transformations. The second contraction uses the orientation-dependent epsilon tensor: strictly, $\mathbf{E}\cdot \mathbf{B}$ is a pseudoscalar, changing sign under an improper transformation such as spatial parity. Under the boosts in this question it is invariant.

Use SI units and the Minkowski metric $\eta =\mathrm{diag}(1,-1,-1,-1)$. Write $x^{\mu }=(ct,x,y,z)$, ${\partial }_{\mu }=(c^{-1}{\partial }_t,\nabla )$ and the four-current $J^{\mu }=(c\rho ,\mathbf{J})$, with Greek indices from $0$ to $3$. The electromagnetic field tensor is the antisymmetric tensor whose components areIndices are raised and lowered with $\eta$. Define the dual electromagnetic field tensor by ${\tilde{F}}^{\mu \nu }=\frac{1}{2}{ϵ}^{\mu \nu \rho \sigma }{F}_{\rho \sigma }$, with ${ϵ}^{0123}=+1$. Then the covariant Maxwell equations areHere $\rho$ and $\mathbf{J}$ are charge and current density, ${\mu }_0$ is the vacuum magnetic permeability, and ${ϵ}_0{\mu }_0{c}^2=1$. For example the $\nu =0$ equation gives $\nabla \cdot \mathbf{E}=\rho /{ϵ}_0$, and the spatial equations give $\nabla \times \mathbf{B}-c^{-2}{\partial }_t\mathbf{E}={\mu }_0\mathbf{J}$. With the stated dual convention, ${\tilde{F}}^{0i}=-B_i$ and ${\tilde{F}}^{ij}={ϵ}_{ijk}{E}_k/c$; the second tensor equation gives the two homogeneous Maxwell equations.

Start with a nonzero eigenfunction of energy $E$ and apply the lowering ladder operator repeatedly. Whenever the resulting wavefunction is nonzero, part (b) makes it an eigenfunction of energy $E-k\hslash \omega$. This cannot remain nonzero for every $k$, since part (e) rules out negative energies. Let $\psi ={\hat{A}}^n\varphi$ be the last nonzero member. Then $\hat{A}\psi =0$, and part (d) forces its energy to be $\hslash \omega /2$. Therefore every energy eigenvalue has the formThese levels actually occur: solving $\hat{A}{\psi }_0=0$ with $\hat{p}=-i\hslash d/dx$ gives the normalizable Gaussian ${\psi }_0\propto \exp (-m\omega x^2/(2\hslash ))$. The raising operator $\hat{B}$ increases the energy by $\hslash \omega$, and $\Vert \hat{B}\psi {\Vert }^2=2m(E+\hslash \omega /2)\Vert \psi {\Vert }^2>0$. Repeated raising therefore produces a nonzero normalizable state at every listed level.

The Hamiltonian of the quantum harmonic oscillator is the sum of two positive quadratic forms. For a nonzero eigenfunction in its domain,Thus $\boxed{E\ge 0}$. Part (d) supplies the stronger bound $E\ge \hslash \omega /2$.

Apply the adjoint operator identity of part (c) with $\chi =\hat{A}\varphi$ and use part (a):For a nonzero eigenfunction, positivity of the norm in particular forces $E\ge \hslash \omega /2$. Hence the requested norm isIt vanishes exactly at the ground-state energy.