False with the $\mathrm{limsup}$ printed in the paper. Let the even-indexed surfaces be one fixed unit sphere and the odd-indexed surfaces one fixed ring torus. Thenbut infinitely many surfaces are tori. The two fixed surfaces lie in one ball and have uniformly bounded area. Their injectivity radii have a positive common lower bound, and compactness gives a common bound on $|\nabla K|$, hence on every radial derivative $|{\partial }_{r}K|$. Thus this alternating sequence satisfies all four additional conditions as well.

The intended statement becomes true if $\mathrm{limsup}$ is replaced byHere is the proof. Put ${\delta }_n=\max (0,-{\inf }_{S_n}{K}_n)$, so ${\delta }_n\to 0$. Choose $p_n$ maximizing distance from the centre of the fixed containing ball. The supporting-sphere curvature bound givesWrite $K_0=K_n(p_n)$ and $j(r,\theta )=\sqrt{G(r,\theta )}$ in geodesic polar coordinates. Choose a fixed sufficiently small $c>0$ and set ${\rho }_n=c{K}_0^{-1/2}$. Conditions (3) and (4) allow $c$ to be chosen so that ${\rho }_n\le {ϵ}_0$ andThe Jacobi equation $j_{rr}=-K_{n}j$, with $j(0)=0$ and $j_r(0)=1$, and the Sturm comparison theorem then give $j(r,\theta )\ge c_{1}r$ on this ball for a constant $c_1>0$ independent of $n$. ConsequentlySince $K_n\ge -{\delta }_n$ elsewhere and $\mathrm{Area}(S_n)\le M$,for all sufficiently large $n$. Gauss-Bonnet forces $\chi (S_n)>0$, and the classification theorem for surfaces makes each such $S_n$ a sphere.

The preceding estimate makes $(x_k)$ equicontinuous, andmakes it uniformly bounded. The Arzelà-Ascoli theorem therefore supplies a subsequence converging uniformly to a continuous function $x:[0,1]\to {\mathbb{R}}^n$. On every mesh interval,including an endpoint by using the adjacent interval. Consequently $y_k\to x$ uniformly along the same subsequence.

All $x_k(t)$ and $y_k(t)$ lie in one compact ball. The restriction of the continuous vector field $F$ to that ball is uniformly continuous, so $F\circ y_k\to F\circ x$ uniformly. Passing to the limit in the defining equation givesThe fundamental theorem of calculus now makes $x$ differentiable and yields $x^{\prime }(t)=F(x(t))$. This proves the required special case of the Peano existence theorem; uniqueness need not hold because $F$ was assumed continuous but not locally Lipschitz continuous.

Set $y_k(0)=x_0$ and let $M={\sup }_{z\in {\mathbb{R}}^n}\Vert F(z)\Vert <\infty$. On the first interval, $y_k$ is the known constant $x_0$, so the integral equation defines $x_k$. Its endpoint value then defines the constant $y_k$ on the next interval, and induction over the $k$ subintervals defines the pair on all of $[0,1]$. An integral of the bounded, piecewise constant function $F\circ y_k$ is continuous, so every $x_k$ is continuous. In fact,so the approximations are uniformly Lipschitz continuous.

Write the closed oriented surface of genus $g$ as ${\Sigma }_g=A\cup B$, where $B$ is a closed disk, $A$ is a slightly enlarged once-punctured surface, and $A\cap B$ is an annulus. Thus $B$ is contractible, $A\cap B$ is homotopy equivalent to the circle, and $A$ is homotopy equivalent to the wedge sum of $2g$ circles. The boundary circle represents the product of the $g$ commutators in the fundamental group of $A$, so its image in the first homology group, which is the abelianization of that group, is zero.

The relevant part of the Mayer–Vietoris sequence is consequentlyThe degree-zero part says that the connected space ${\Sigma }_g$ has $H_0({\Sigma }_g)\cong \mathbb{Z}$, while the groups above degree two vanish. Hence

Yes. Order the proper filters containing the cofinite filter by inclusion. The union of every chain is again a proper filter containing $\mathcal{F}$: any finite collection of its members already lies in one filter in the chain. Zorn lemma gives a maximal proper filter $\mathcal{U}$.

A maximal proper filter is an ultrafilter. Indeed, if $A\notin \mathcal{U}$, adjoining $A$ cannot generate another proper filter, so some $U\in \mathcal{U}$ has $U\cap A=\varnothing$. Thus $U\subseteq A^c$, and upward closure gives $A^c\in \mathcal{U}$. Finally, $\mathcal{U}$ cannot equal any principal ultrafilter ${\mathcal{U}}_n$, because it contains the cofinite set $\mathbb{N}\setminus \{n\}$ whereas ${\mathcal{U}}_n$ does not. Thus $\mathcal{U}$ is a nonprincipal ultrafilter on $\mathbb{N}$.

The cofinite filter $\mathcal{F}$ is nonempty, because $\mathbb{N}\in \mathcal{F}$; finite unions of finite complements show closure under intersections, and taking a superset can only shrink the complement. The empty set is excluded because its complement is infinite. It is not an ultrafilter: the even numbers and the odd numbers both have infinite complements, so neither belongs to $\mathcal{F}$. Finally, $\mathbb{N}\setminus \{n\}$ is cofinite but does not belong to ${\mathcal{U}}_n$, proving $\mathcal{F}⊈{\mathcal{U}}_n$ for every $n$.

The family ${\mathcal{U}}_n=\{A\subseteq \mathbb{N}:n\in A\}$ is nonempty, excludes the empty set, is closed under finite intersections, and is upward closed. It is therefore a filter. For every $A\subseteq \mathbb{N}$, either $n\in A$ or $n\in A^c$, so either $A\in {\mathcal{U}}_n$ or $A^c\in {\mathcal{U}}_n$. Hence ${\mathcal{U}}_n$ is the principal ultrafilter at $n$.

For $H=H(p)$, Hamilton's equations giveThus $P=p$ and $Q=q-t{H}^{\prime }(p)$. The Poisson bracket is $\{Q,P\}=1$, so the transformation is canonical for every $t$. A type-2 generator isIndeed $p={\partial }_q{F}_2=P$, $Q={\partial }_P{F}_2=q-t{H}^{\prime }(P)$, and the transformed Hamiltonian is $K=0$, as expected because $Q,P$ are constants of motion.

Letbe the block-saturated mean space, and letbe the fitted GLM mean space. Full column rank and the regularity assumptions give dimensions $n$ and $p$. The factorization in part (ii) makes the nuisance factor cancel from their likelihood ratio, while $\bar{Y}\stackrel{d}{=}Y$ givesAs ${\sigma }^2\to 0$, the replicate count $m=1/{\sigma }^2$ tends to infinity. Wilks theorem applied to these nested regular models makes twice the log-likelihood ratio converge in distribution to a chi-squared variable with the difference in dimensions:

Writing $B_i$ for block $i$ and using ${\sum }_{j\in B_i}{\tilde{y}}_j={\bar{y}}_i/{\sigma }^2$, the replicated likelihood factors asThe second factor is independent of $\beta$, while the first depends on the data only through $\bar{y}$. The Fisher-Neyman factorization theorem therefore makes $\bar{Y}$ a sufficient statistic for $\beta$ in the replicated experiment.

Let $m=1/{\sigma }^2$. Within block $i$, the $m$ replicated observations have dispersion $a_i$ and common natural parameter ${\theta }_i$. Evaluating their joint MGF at the blockwise argument ${\sigma }^2{t}_i$ givesMultiplying over blocks reproduces the MGF of $Y$ from part (a). Since moment-generating functions uniquely determine distributions in this family, $\bar{Y}\stackrel{d}{=}Y$.

Consecutive convergents obey the standard error estimateIf the partial quotients are unbounded, then for every $c>0$ there are infinitely many indices with $a_{n+1}>1/c$; otherwise the whole sequence would be bounded after adjoining its finite initial segment. At each such index,The denominators of the infinite continued fraction tend to infinity, so these give infinitely many distinct rational numbers of the required kind.

Suppose $a_j\le A$ for every $j$. Given $b>0$, choose the first convergent denominator $q_m\ge b$. Then $q_{m-1}<b$ and the recurrence gives $q_m=a_m{q}_{m-1}+q_{m-2}<(A+1)b$. Part (i), applied at index $m$, gives $|b\alpha -a|\ge |q_m\alpha -p_m|$.

If ${\alpha }_{m+1}=[a_{m+1};a_{m+2},\dots ]$ is the complete quotient, the exact error formula isSince ${\alpha }_{m+1}<A+1$, this is greater than $1/((A+2)q_m)$ and therefore greater than $1/((A+1)(A+2)b)$. It follows that every rational $a/b$ satisfiesThus one may take $c=((A+1)(A+2){)}^{-1}$.

The convergents satisfyso $(p_n,q_n)$ and $(p_{n-1},q_{n-1})$ form an integer basis of ${\mathbb{Z}}^2$. Put ${\delta }_j=q_j\alpha -p_j$. Consecutive errors have opposite signs and $|{\delta }_{n-1}|>|{\delta }_n|$.

For integers $p,q$ with $0<q\le q_n$, writeIf $s>0$, the denominator bound forces $r\le 0$, while if $s<0$ positivity forces $r\ge 1$. Because ${\delta }_n$ and ${\delta }_{n-1}$ have opposite signs, in either case $s\ne 0$ impliesIf $s=0$, the denominator bound gives $r=1$ unless the fraction is zero or has the same value as $p_n/q_n$. Thus $p_n/q_n$ is a best approximation of the second kind. Dividing by $q\le q_n$ gives

SetWriting $C=\cosh z$, direct differentiation givesOn the other hand,Thus $F^{\prime \prime }=1-F^2$, which is precisely $f^{\prime \prime \prime }=1-(f^{\prime }{)}^2$. The far-field condition is automatic because $F\to -1$. No slip requiressoFor $c=+\mathrm{arcosh}5$, $F$ decreases directly from $0$ to $-1$, matching flow everywhere toward the sink. For the negative root, $F$ initially becomes positive and creates a reverse-flow region next to the wall. The physically likely choice is therefore

LetUsing the Cartesian streamfunction convention $u={\psi }_y$, $v=-{\psi }_x$ givesAt fixed $y$,The terms proportional to $\eta f^{\prime }{f}^{\prime \prime }$ cancel, leavingSince $U=-Q/x$, $U{U}^{\prime }=-Q^2/x^3$, while $\nu u_{yy}=Q^2{f}^{\prime \prime \prime }/x^3$. The boundary-layer equation therefore becomesChoosing the wall value of the streamfunction as zero, imposing no slip, and matching to $U=-Q/x$ give

Along $y=0$, put $x=r$ and $U(x)=-Q/x$. If $\delta (x)$ is the layer thickness, thenThe boundary-layer thickness from an advection-diffusion balance therefore givesand henceup to a constant factor absorbed into the similarity coordinate.

Conservation of volume flux through a circular arc of radius $r$ givesso the outer flow is the sink flow in a sectorThis is a potential flow and already satisfies the no-penetration boundary condition on the radial walls. It has nonzero tangential speed $Q/r$ there, however, and therefore violates the no-slip boundary condition. Thin vorticity-bearing boundary layers are needed to reduce that speed to zero at each wall.

Let $\phi =\varphi -F(R)$, soThe new coefficient of $d{R}^2$ isIt vanishes ifOne choice that tends to zero as $R\to \infty$ and also has a regular $a\to 0$ limit isSubstitution gives the outgoing null oblate spheroidal coordinates formThere is no $d{R}^2$ term, so $g_{RR}=0$ and ${\partial }_R$ is null for every $a\ge 0$.

Set $u=t-r$ and $R=r$, soWriting now $\Sigma =R^2+a^2{\cos }^2\theta$ and $\Delta =R^2+a^2$, the Minkowski metric becomesThe squared norm of the coordinate vector is its $RR$ metric component:For $a>0$ this is nonzero away from the symmetry axis, so ${\partial }_R$ is not a null vector.