True. Write $C={\bigcup }_{n\ge 1}{C}_n$ with every $C_n$ nowhere dense in $\mathbb{R}$. ThenThe closure of $C_n\times \mathbb{R}$ is
$\overline{C_n}\times \mathbb{R}$, which has empty interior because $\overline{C_n}$ has empty interior. Each product is therefore nowhere dense, so $A$ is meagre in ${\mathbb{R}}^2$.

False. Fix $y_0\in \mathbb{R}$ and takeThis closed set has empty interior in ${\mathbb{R}}^2$, so it is nowhere dense and hence meagre. Its section at $y_0$ is $C_{y_0}=\mathbb{R}$, which is not meagre in itself by the Baire category theorem.

A set is of first Baire category, or meagre, when it is a countable union of nowhere dense sets. The statement is true: a countable union of countable unions of nowhere dense sets is again a countable union of nowhere dense sets.

The relevant Baire category theorem says that a nonempty complete metric space is not meagre in itself.

No. Every estimator in parts (b) and (c) has $a\ne 1$, so its risk contains the nonconstant quadratic term
$((a-1)\theta +b{)}^2$ and has infinite supremum over $\theta \in \mathbb{R}$. The estimator ${\overline{X}}_n$ has constant risk $1/n$, so those estimators cannot be minimax.

The risk of ${\overline{X}}_n$ is constantly $1/n$. Suppose an estimator $\delta$ dominates it, and put
$m(\theta )={\mathbb{E}}_{\theta }\delta$ and $b(\theta )=m(\theta )-\theta$. The biased Cramér-Rao bound, using total information $n$, givesDomination would implyfor every real $\theta$. The standard differential-inequality argument shows that the only globally defined differentiable solution is $b\equiv 0$: a nonzero value forces $b^{\prime }$ to retain a sign and magnitude that makes $b$ leave the permitted bounded interval in one time direction.

Thus $m(\theta )=\theta$. The Cramer--Rao inequality now gives
${\mathrm{Var}}_{\theta }\delta \ge 1/n$, so domination forces equality everywhere. Equality in Cramer--Rao makes the centred estimator proportional to the Gaussian score:and hence $\delta ={\overline{X}}_n$ almost surely. No strict improvement exists, so ${\overline{X}}_n$ is admissible.

Put ${\theta }_0=-b/(a-1)$ and compare with the constant estimator
$\delta ={\theta }_0$. Its risk is $(\theta -{\theta }_0{)}^2$. For $a<0$,for every $\theta$, since $|a-1|>1$. Hence the original estimator is inadmissible.

If $a>1$, compare withIts bias is the negative of the original bias, so the squared-bias terms agree, whileIts risk is therefore strictly smaller for every $\theta$. Thus
${\hat{\theta }}_{a,b}$ is not an admissible estimator.

Since ${\overline{X}}_n\sim N(\theta ,1/n)$, the estimator has variance
$a^2/n$ and bias $(a-1)\theta +b$. Its quadratic risk function is

Write $d_G(x)$ for the degree of $x$ in $G$ and $d_A(x)$ for the number of its neighbours lying in $A$. Each trial succeeds with probability
$d_A(x)/d_G(x)$. The stated geometric-sum fact shows that the total holding time at $x$ is exponential with rate $d_A(x)/d_G(x)$. Conditional on success, each neighbour in $A$ is chosen uniformly. Hence

LetFor every edge $xy$ of $A$,Thus detailed balance holds, so $\pi$ is invariant. Connectedness of $A$ makes the chain irreducible and this invariant distribution unique.

The first condition is that $X$ is Markov with transition semigroup
$P(t)=e^{tQ}$. The second is that, for every function
$f:S\to \mathbb{R}$,is a martingale.

The first condition implies the second by conditioning over a short time interval, using
$P(h)f=f+hQf+o(h)$, summing increments, and passing to the limit. This is Dynkin's formula.

Conversely assume the martingale problem for a continuous-time Markov chain. Fix $T>0$ and a function $f$, and putThe backward equation gives ${\partial }_{t}u+Qu=0$. Applying the martingale identity to this time-dependent function shows that $u(t,X_t)$ is a martingale. Therefore, for $s\le T$,Taking indicator functions $f$ proves both the Markov property and the transition semigroup $e^{tQ}$.

Since $f_n\uparrow f$, the functionsare nonnegative and increase to $g=f-f_1$. The monotone convergence theorem gives
$\mu (g_n)\uparrow \mu (g)$. Because $f_1$ is integrable, adding the finite number
$\mu (f_1)$ yields

Integrability of $f_1$ is essential. On $\mathbb{R}$ with Lebesgue measure, letThen $f_n\uparrow 0$ pointwise, but every integral is $-\infty$, so
$\mu (f_n)$ does not increase to $\mu (0)=0$.

The monotone convergence theorem states that if
$0\le f_n\uparrow f$ pointwise, thenallowing the value $+\infty$.

Monotonicity of the integral shows that the increasing limit $L$ of the left side is at most $\int f$. Conversely, let $s$ be a nonnegative simple function with $s\le f$, and fix $0<c<1$. The setsincrease to the support of $s$. HenceLet $c\uparrow 1$ and take the supremum over all simple $s\le f$. This gives
$L\ge \int f$, proving equality.

Lift $f$ to a doubly periodic meromorphic function on $\mathbb{C}$. Integrate it around a fundamental parallelogram whose boundary avoids all poles. Integrals over opposite edges cancel by periodicity, while the residue theorem gives

Thus the principal-part map from part (b) takes values in the codimension-one hyperplane on which the sum of residue coefficients is zero. The Mittag--Leffler existence criterion on a compact Riemann surface says that prescribed principal parts occur precisely when their residues pair trivially with every holomorphic one-form. On a complex torus the holomorphic one-forms are the scalar multiples of $dz$, so the single condition is exactly the displayed residue sum. Hence the image has dimension
${\sum }_i{m}_i-1$, and the kernel of constants has dimension one. For $n\ge 1$,Equivalently, this is the genus-one case of the Riemann-Roch theorem for a positive divisor.

Choose a local coordinate $z_i$ at each $p_i$ and map $f\in V$ to all coefficients of the negative-power terms in its Laurent expansions:The kernel consists of globally holomorphic functions on the compact Riemann surface $R$. By the maximum modulus principle, every such function is constant, so the kernel has dimension one. Rank--nullity givesFor $n=0$, this simply says $\dim V=1$.

The sum of two meromorphic functions has no poles away from the union of their pole sets, and at each $p_i$ its pole order is at most the larger of the two orders, hence at most $m_i$. Scalar multiplication cannot increase a pole order. The zero function belongs to the set, so $V$ is a complex vector space.

For $x={\sum }_n{x}_n{e}_n$, Cauchy-Schwarz inequality givesThus every $T\in H(V,V)$ is bounded and
$\Vert T\Vert \le \Vert T{\Vert }_{\ast }$. The truncationshave finite rank and satisfyHence every such Hilbert-Schmidt operator is compact.

DefineThis is an inner product inducing $\Vert \cdot {\Vert }_{\ast }$. If $(T_j)$ is Cauchy in this norm, then each $T_j{e}_n$ converges to some $y_n$, and the usual Fatou/tail argument gives
${\sum }_n\Vert y_n{\Vert }^2<\infty$ and
${\sum }_n\Vert T_j{e}_n-y_n{\Vert }^2\to 0$. Defining
$Tx={\sum }_n{x}_n{y}_n$ produces a bounded operator by the preceding estimate and gives
$T_j\to T$ in $\Vert \cdot {\Vert }_{\ast }$. Thus $(H(V,V),\Vert \cdot {\Vert }_{\ast })$ is a Hilbert space.

The norms are not equivalent in infinite dimension. The rank-$N$ orthogonal projection hasAlthough $\Vert T\Vert \le \Vert T{\Vert }_{\ast }$, no uniform reverse inequality can hold.

A compact operator maps every bounded sequence to a sequence with a norm-convergent subsequence. A Hilbertian basis is a complete orthonormal sequence.

If $T$ is compact and ${\lambda }_n$ does not tend to zero, some subsequence has
$|{\lambda }_{n_j}|\ge \epsilon$. The vectors
$T{e}_{n_j}={\lambda }_{n_j}{e}_{n_j}$ are pairwise separated by at least $\sqrt{2}\epsilon$, so they have no convergent subsequence, a contradiction.

Conversely, if ${\lambda }_n\to 0$, define the finite-rank operatorThenAn operator-norm limit of compact operators is compact, so $T$ is compact.

Let $(T_m)$ be a Cauchy sequence in the operator norm. For each $v\in V$,so $(T_{m}v)$ is a Cauchy sequence. Completeness of the Banach space $V$ permits the definition
$Tv={\lim }_m{T}_{m}v$. Pointwise passage to the limit proves that $T$ is a linear. A norm-Cauchy sequence is a bounded sequence, say $\Vert T_m\Vert \le C$, and therefore
$\Vert Tv\Vert \le C\Vert v\Vert$, so $T$ is bounded. Finally, letting $m\to \infty$ in
$\Vert (T_n-T_m)v\Vert \le \epsilon \Vert v\Vert$ gives
$\Vert T_n-T\Vert \le \epsilon$. Thus $T_n\to T$ in operator norm and
$\mathcal{B}(V,V)$ is Banach.

The Dedekind factorization theorem says that, when a rational prime $p$ does not divide the index of an order $\mathbb{Z}[\theta ]\subseteq {\mathcal{O}}_K$, a factorizationof the minimal polynomial modulo $p$ gives

Let $K=\mathbb{Q}(\sqrt{d})$ with square-free $d$, and let $p$ be odd. The resulting splitting of rational primes in a quadratic field is:

Finally suppose $d>0$ and
$\alpha =A/C+(B/C)\sqrt{d}$ has norm $-1$, with integers
$A,B,C$ chosen coprime. ThenIf an odd $p\equiv 3(\mathrm{mod}4)$ ramified, then $p\mid d$. Reduction modulo $p$ gives
$A^2\equiv -C^2(\mathrm{mod}p)$. Since $-1$ is not a square modulo such a prime, $p\mid A,C$. Dividing the equation's divisibility shows $p\mid B$ as well (use that square-free $d$ has $p$-adic valuation one), contradicting coprimality. Hence no prime $p\equiv 3(\mathrm{mod}4)$ ramifies.

Distinct prime-power ideals are pairwise comaximal. For comaximal ideals, product equals intersection, so induction givesThe Chinese remainder theorem therefore gives the ring isomorphism

The ideal approximation theorem, proved by applying this Chinese-remainder map one prime power deeper, suppliesEquivalently, $v_{P_i}(\alpha )=m_i$ at every prime dividing $I$.

Since $(\alpha )\subseteq I$, the fractional idealis an integral ideal and $I{I}^{\prime }=(\alpha )$ is a principal ideal. At every $P_i\mid I$,
$v_{P_i}(I^{\prime })=v_{P_i}(\alpha )-m_i=0$, so $I+I^{\prime }={\mathcal{O}}_K$.

Choose $\beta \in I$ and $\gamma \in I^{\prime }$ with $\beta +\gamma =1$. Then
$\alpha ,\beta \in I$. For any $x\in I$,Thusso every ideal of ${\mathcal{O}}_K$ has a generating set of two elements.