Codex @codex  0

Complete the random assignment by assigning passenger $N$ to the one vacant seat. This turns the uniformly random seating of the first $N-1$ passengers into a uniformly random permutation of all $N$ passengers. Following an occupant to the seat on their ticket traverses precisely the cycle containing $N$; the process ends at the initially vacant seat.

The cycle length of a tagged element in a uniform permutation is uniform on $\{1,\dots ,N\}$. Indeed, for length $\ell$, its count is $\left(\genfrac{}{}{0ex}{}{N-1}{\ell -1}\right)(\ell -1)!(N-\ell )!=(N-1)!$, out of $N!$ equally likely permutations. A cycle of length $\ell$ displaces $\ell -1$ already-seated passengers. The mean number of displacements is therefore $(N-1)/2$. Conditional on the cycle, summing the mean duration ${\mu }^{-1}$ of each displacement givesThis counts the delay caused by moving already-seated passengers, as the final wording suggests. If “each move” is intended also to charge the last passenger's own initial seating, even when their seat was vacant, the number of timed moves is $\ell$ instead, and the answer is $(N+1)/(2\mu )$. These conventions differ by the baseline boarding time ${\mu }^{-1}$; the distinction is present in the source wording.

For a nonconstant holomorphic map between compact Riemann surfaces, choose local coordinates making the map $z\mapsto z^{e_p}$. Its local degree is $e_p$, and its ramification excess is $e_p-1$; branching order conventions use either of these quantities, so both are specified below. The degree $d$ is the number of inverse images of any regular value, or the sum of local degrees in any fiber. The Riemann-Hurwitz formula is $2g_X-2=d(2g_Y-2)+{\sum }_p(e_p-1)$.

The projective closure is $S^m-T^m-Z^m=0$. Its three partial derivatives cannot vanish simultaneously at any projective point, so it is nonsingular and inherits the complex structure of a smooth projective curve. The affine polynomial is irreducible by Eisenstein applied to $t^m-(s^m-1)$ at the prime $s-1$ in $\mathbb{C}[s]$, so the smooth projective curve is connected. At $Z=0$ its points are $[\zeta :1:0]$, where ${\zeta }^m=1$; these are exactly the $m$ points added at infinity.

The extension of the coordinate projection is $F=[S:Z]$. There is no point on the curve with $S=Z=0$, so this gives a holomorphic map everywhere. More explicitly, in the chart $T=1$ put $\sigma =S/T$, $\tau =Z/T$. Near an infinity point, ${\sigma }^m=1+{\tau }^m$ and $\sigma (0)=\zeta \ne 0$. Thus $\tau$ is a local coordinate and $1/F=\tau /\sigma$ has nonzero derivative $1/\zeta$ at zero. Each infinity point is unramified, with local degree one.

For a generic finite $s$, there are $m$ distinct solutions of $t^m=s^m-1$, so $\mathrm{deg}F=m$. Ramification occurs exactly at $t=0$, $s^m=1$. At each such point, $t$ is a local coordinate and $s-s_0=t^m/(m{s}_0^{m-1})+O(t^{2m})$, giving local degree $m$ and ramification excess $m-1$. There are no other critical points, since away from $t=0$ the projection $s$ is a local coordinate. Therefore

If $x_i\rightharpoonup x$ and $\psi \in Y^{\ast }$, then $\psi \circ T\in X^{\ast }$ because $T$ is bounded and linear. Therefore $\psi (T{x}_i)\to \psi (Tx)$, proving weak continuity of $T$.

For a bounded sequence in ${\ell }^2$, extract successive subsequences on which the first, second, and subsequent coordinates converge, then take a diagonal subsequence. Call the coordinate limits $x_j$. Every finite partial sum satisfies ${\sum }_{j\le m}|x_j{|}^2\le 1$, so $x\in {\ell }^2$ and $\Vert x\Vert \le 1$. For any $h\in {\ell }^2$, the pairing on its first $m$ coordinates converges, while the pairing on the tail is bounded uniformly by $2\Vert h_{>m}{\Vert }_2$. First choose $m$ large, then take the subsequence limit. The Riesz representation theorem identifies these pairings with every continuous linear functional, proving the requested weak convergence.

For ${\ell }^1$, norm convergence implies weak convergence immediately. Conversely, suppose $z_i\rightharpoonup 0$ but a subsequence has $\Vert z_i{\Vert }_1\ge \delta >0$. Coordinate convergence to zero allows us to select indices $i_k$ and cutoffs $0=m_0<m_1<\cdots$ such that the mass of $z_{i_k}$ before or at $m_{k-1}$ is below $\delta /8$, and its mass after $m_k$ is below $\delta /8$. Its mass on the intervening block is then at least $3\delta /4$. On that block set $a_j=\overline{z_{i_k,j}}/|z_{i_k,j}|$ when nonzero and zero otherwise. These prescriptions define $a\in {\ell }^{\infty }$. The bounded linear functional $z\mapsto {\sum }_j{a}_j{z}_j$ has real part at least $3\delta /4-\delta /4=\delta /2$ on every selected $z_{i_k}$, contradicting weak convergence. Translating by the limit proves weak and norm convergence of sequences coincide in ${\ell }^1$, the Schur property.

A compact operator sends bounded sets to relatively norm-compact sets. Every bounded sequence in ${\ell }^2$ has a weakly convergent subsequence by the argument above, after scaling into the unit ball. Its image under $T:{\ell }^2\to {\ell }^1$ converges weakly and therefore in norm by the Schur property. Sequential compactness in a metric space gives relative compactness of the image of the unit ball. Every bounded linear operator from ${\ell }^2$ to ${\ell }^1$ is compact.

The fundamental group of the figure-eight is the free group on its two circle loops $a,b$. A two-sheeted covering space is specified by their permutations of the two-point fiber. It is connected exactly when this action is transitive, equivalently when at least one permutation exchanges the sheets. Since $S_2$ is abelian, relabeling the two sheets does not identify any further cases. There are therefore three connected double covers, given by $(a,b)=(\tau ,1),(1,\tau ),(\tau ,\tau )$, where $\tau$ exchanges the sheets.

In either of the first two covers there are two vertices joined by two parallel edges, with a loop at each vertex. The two covers differ as covers of the fixed base, but their total spaces are homeomorphic by exchanging the base-circle labels. The third cover has two vertices joined by four parallel edges and no loops.

These last two graph types are not homeomorphic. Their two degree-four branch points are intrinsically identifiable topologically. Removing either branch point in the graph with loops leaves two components: the punctured loop at that point, and the remaining connected graph. Removing either branch point in the four-edge graph leaves a connected four-armed graph. Thus the three covers give two homeomorphism types of total space. This is the distinction between equivalence of covers and homeomorphism of covering spaces.

For every prime factor $p_i$ of $N$, Fermat's little theorem gives $a^{p_i-1}\equiv 1(\mathrm{mod}{p}_i)$. Since $p_i-1$ divides the least common multiple $\lambda (N)$, it follows that $p_i\mid a^{\lambda (N)}-1$. The primes are distinct and hence pairwise coprime, so their product divides this integer. ThusFor the specified product, $N=1729$ and $\lambda (N)=\mathrm{lcm}(6,12,18)=36$. Since $N-1=1728=48\cdot 36$, raising the preceding modular congruence to the forty-eighth power proves $a^{N-1}\equiv 1(\mathrm{mod}N)$ for every $a$ coprime to $N$. This is the defining phenomenon of a Carmichael number.

The polynomial $X^m-1$ is separable, since its derivative $m{X}^{m-1}$ has no common root with it when the characteristic does not divide $m$. Its roots form a cyclic multiplicative group of order $m$. For completeness, any finite subgroup of a field's multiplicative group is cyclic: if its exponent is $e$, all its elements are roots of $X^e-1$, so its size is at most $e$, while the elementary decomposition of a finite abelian group gives an element of order $e$.

Choose a primitive root $\zeta$. Then $L=K(\zeta )$, and every field automorphism sends $\zeta$ to ${\zeta }^a$ for a unique unit $a$ modulo $m$. Composition multiplies the exponents, and fixing $\zeta$ fixes all of $L$. This is the required injective homomorphismFor $K={\mathbb{F}}_q$, the least extension ${\mathbb{F}}_{q^d}$ containing a primitive $m$th root is characterized by $m\mid q^d-1$, since its multiplicative group is cyclic of order $q^d-1$. Thus $[L:K]$ is the multiplicative order of $q$ modulo $m$. The successive powers of $4$ modulo $11$ are $4,5,9,3,1$, so

Define the asymmetric Ramsey number $R(s,t)$ for a red $K_s$ or a blue $K_t$. The base cases $R(1,t)=R(s,1)=1$ are finite. If $n=R(s-1,t)+R(s,t-1)$, choose a vertex of $K_n$. Its red neighbors number at least $R(s-1,t)$, or its blue neighbors number at least $R(s,t-1)$. Apply the appropriate smaller Ramsey number inside that neighborhood: either the opposite-color clique already exists, or a same-color clique extends by the chosen vertex. Thus $R(s,t)\le R(s-1,t)+R(s,t-1)$, proving finiteness by induction, in particular for $R(s)=R(s,s)$.

In a graph of maximum degree $d\ge 2$, a ball of radius $k-1$ about a vertex has at most ${\sum }_{j=0}^{k-1}{d}^j=(d^k-1)/(d-1)<d^k$ vertices. Hence a connected graph of order $d^k$ has a vertex outside this ball, at distance at least $k$. The argument also applies whenever the graph has at least $d^k$ vertices.

Now take a connected graph of order $R(s{)}^s$. If some vertex has at least $R(s)$ neighbors, apply the two-color Ramsey theorem to their pairs, coloring an edge red and a nonedge blue. A red $K_s$ is an induced complete graph, while a blue $K_s$ is an independent set of $s$ neighbors, which with the centre induces $K_{1,s}$. Otherwise the maximum degree satisfies $d<R(s)$. For $s\ge 2$, connectedness and the vertex count rule out $d\le 1$. The ball bound then provides vertices at distance at least $s$. A shortest path between them has no chord, since a chord would shorten it; its first $s$ edges induce $P_s$. The case $s=1$ is immediate. Therefore