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The Poisson thinning theorem, or its multinomial marking version, makes the bin counts at time $n$ independent Poisson variables with mean $n/n=1$. Independence is more than is needed for the following union bound. Fix $ϵ>0$ and put $x_n=(1+ϵ)\log n/\log \log n$. ThenSincethe logarithm of the upper bound is $-ϵ\log n+o(\log n)\to -\infty$. Thus $\boxed{\mathbb{P}(M_n\ge x_n)\to 0}$. The estimate handles the noninteger threshold directly, since the supplied Poisson tail bound holds for real $x$.

In an interval of length $t$, condition on the Poisson process count $X_t=k$. Independently marking its points gives $Y_t\mid X_t=k\sim \mathrm{Bin}(k,p)$. The joint probability generating function of kept and discarded counts isIt factors into the generating functions of independent Poisson variables of means $\lambda pt$ and $\lambda (1-p)t$. The same calculation applies to any interval, and counts and marks in disjoint intervals are independent. Hence $Y$ is a Poisson process of intensity $\boxed{\lambda p}$, and $Y_t$ and $X_t-Y_t$ are independent. This is the Poisson thinning theorem.

For a probability distribution $m$ on $\mathbb{R}$, take the canonical sample space $\Omega ={\mathbb{R}}^{\mathbb{N}}$, the product of the Borel $\sigma$-algebras, and the product measure $\mathbb{P}=m^{\otimes \mathbb{N}}$. The coordinate functions $X_j(\omega )={\omega }_j$ have joint cylinder probabilitiesThe Kolmogorov extension theorem supplies this measure from the consistent finite-dimensional distributions, and uniqueness follows from the generating cylinder sets. Thus the coordinates are independent and identically distributed; the same construction works for a given standard Borel state space.

The one-sided Bernoulli shift is $T({\omega }_1,{\omega }_2,\dots )=({\omega }_2,{\omega }_3,\dots )$. It is measurable and preserves all cylinder probabilities. The uniqueness theorem for probability measures extends this to the product $\sigma$-algebra, proving measure preservation.

To prove ergodicity, let $A$ be invariant modulo null sets and approximate it in probability by an event $C$ depending on only the first $r$ coordinates, with $\mathbb{P}(A△C)<ϵ$. Such approximations follow from the fact that finite-coordinate cylinder events generate the product $\sigma$-algebra. For $n\ge r$, independence gives $\mathbb{P}(C\cap T^{-n}C)=\mathbb{P}(C{)}^2$. Invariance and measure preservation giveAlso $|\mathbb{P}(A{)}^2-\mathbb{P}(C{)}^2|\le 2ϵ$. Letting $ϵ\downarrow 0$ proves $\mathbb{P}(A)=\mathbb{P}(A{)}^2$, so $\boxed{\mathbb{P}(A)\in \{0,1\}}$. This is exactly ergodicity.

For a Borel set $A\subset [0,1)$, the two inverse branches of the doubling map give the disjoint setsEach has half the Lebesgue measure of $A$, by translation and scaling invariance. Hence $\boxed{\mu (T^{-1}A)=\mu (A)}$. The piecewise affine map is Borel measurable, so it is a measure-preserving transformation. The printed one-dimensional “baker's map” is this doubling map.

A measure-preserving transformation is a measurable map $T:E\to E$ satisfying $\mu (T^{-1}A)=\mu (A)$ for every measurable $A$. An invariant set satisfies $T^{-1}A=A$ (or equality modulo null sets, in the measure-theoretic convention).

Preimages preserve complements and countable unions. Therefore the whole space is invariant, the complement of an invariant set is invariant, and a countable union of invariant sets is invariant. Thus the invariant sets form a $\sigma$-algebra. The same proof works modulo null sets because a countable union of null exceptional sets is null.

Use the two charts of the blowup of the affine plane at the origin. In the chart $y=xv$, the total transform isso the proper transform has equation $1-x{v}^3=0$. It misses the exceptional divisor $x=0$ and is smooth, since its derivative with respect to $x$ is $-v^3\ne 0$ there.

In the chart $x=yu$, the total transform is $y^2(u^2-y)$, so the proper transform isIts equation $u^2-y=0$ has partial derivative $-1$ with respect to $y$, so it is smooth even at its unique intersection $(u,y)=(0,0)$ with the exceptional divisor. The two chart descriptions agree on their overlap. Thus one blowup makes the cusp nonsingular, although it is tangent to the exceptional divisor there.

Write $F=xyz+yzw+zwx+wxy$. A projective hypersurface singularity must have $F=0$ and all four first partial derivatives zero. If all coordinates were nonzero, writeOn $F=0$, differentiation gives $F_x=-xyzw/x^2\ne 0$, a contradiction. Thus some coordinate is zero; by symmetry take $w=0$. The surface equation gives $xyz=0$, so a second coordinate is zero, say $z=0$. Then $F_z=xy$, so a third coordinate is zero. The remaining four projective possibilities indeed have every first derivative zero. Therefore all singular points areOver the usual characteristic-zero ground field, each is an ordinary double point: near $[1:0:0:0]$, the affine equation has nondegenerate quadratic part $yz+zw+wy$, and symmetry treats the others.

For $X\subset {\mathbb{A}}^n$ with vanishing ideal $I(X)$, the Zariski tangent space at $P$ isIt is the kernel of the Jacobian matrix of generators for the ideal, or intrinsically the dual of ${\mathfrak{m}}_P/{\mathfrak{m}}_P^2$. A point is a singular point of an algebraic variety when this tangent dimension exceeds the local dimension of the variety; otherwise it is nonsingular. For an irreducible affine variety the local dimension at a closed point equals its dimension.

Let $\alpha =\sqrt[5]{3}$ and let $\zeta$ be a primitive fifth root of unity. Then $L=\mathbb{Q}(\alpha ,\zeta )$. The field degrees $[\mathbb{Q}(\alpha ):\mathbb{Q}]=5$ and $[\mathbb{Q}(\zeta ):\mathbb{Q}]=4$ both divide $[L:\mathbb{Q}]$, while adjoining the two generators gives $[L:\mathbb{Q}]\le 20$. Consequently $[L:\mathbb{Q}]=20$.

The cyclotomic field $\mathbb{Q}(\zeta )/\mathbb{Q}$ is Galois with cyclic group of order $4$. The Fundamental theorem of Galois theory gives a normal subgroupThus $N\cong C_5$ and the chain $\boxed{G▹N▹\{1\}}$ has abelian successive quotients, proving that $G$ is a solvable group. One may refine it by inserting the inverse image of the order-two subgroup of $C_4$, giving cyclic factors $C_2,C_2,C_5$.

Label the distinct roots ${\alpha }_1,\dots ,{\alpha }_n$. Define $\tau (\sigma )$ by $\sigma ({\alpha }_i)={\alpha }_{\tau (\sigma )(i)}$. Composition respects this rule, so it is a group homomorphism. If its image permutation is the identity, $\sigma$ fixes all the roots; they generate $L$, so $\sigma$ is the identity. Thus $\tau$ is injective.

For $f=t(t^2-2)$, label the roots $0,\sqrt{2},-\sqrt{2}$. The splitting field is $\mathbb{Q}(\sqrt{2})$ and its nontrivial automorphism swaps precisely roots $2,3$. Hence $\boxed{\mathrm{im}\tau =⟨(23)⟩}$.

For $f=t^3-2$, Eisenstein criterion gives irreducibility. The real root $\alpha =\sqrt[3]{2}$ generates a degree-three real field. Adjoining a primitive cube root of unity ${\zeta }_3$ gives a further extension of degree two, since it is nonreal and satisfies $t^2+t+1=0$. Thus its splitting field has degree $6$. In characteristic zero this is a Galois extension, so its Galois group has order $6$. The injection into $S_3$ is therefore an isomorphism.

If $f$ is an irreducible polynomial, any two roots $\alpha ,\beta$ have the same minimal polynomial. Thus the map $\mathbb{Q}(\alpha )\to \mathbb{Q}(\beta )$ carrying $\alpha$ to $\beta$ is a field embedding. The extension theorem for field embeddings extends it to an embedding of the splitting field $L$ into an algebraic closure. Since $L/\mathbb{Q}$ is normal, its image is $L$, so it is an element of the Galois group.

Conversely, factor $f$ into irreducible factors over $\mathbb{Q}$. An automorphism preserves the vanishing of each factor, so it cannot send a root of one factor to a root of another. Since there are no repeated roots, different factors have disjoint root sets. Transitivity on all roots therefore forces a single irreducible factor. This proves the equivalence.

For a finite group $G$, subgroups $H,K$, and a complex group representation $V$ of $H$, the Mackey restriction formula isChoose one representative $g$ of each double coset $KgH$. The conjugate subgroup representation ${}^{g}V$ is a representation of $gH{g}^{-1}$ on the same vector space, with $(gh{g}^{-1})v=hv$. Restriction of a representation keeps the vector space and restricts the acting subgroup. induced representation is $\mathbb{C}G{\otimes }_{\mathbb{C}H}V$, and analogously for each subgroup of $K$. The displayed isomorphism is an isomorphism of $K$-representations. These definitions specify every summand and its subgroup.

Add missing cross-edges to a hypothetical non-Hamiltonian balanced bipartite graph until it is maximal with that property. The complete bipartite graph is Hamiltonian for $n\ge 2$, so a missing cross-edge $uv$ remains. Adding it creates a Hamilton cycle, and removing it gives a Hamilton pathFor each $i$, consider the conditions $u{B}_i\in E$ and $A_{i}v\in E$. Since $\mathrm{deg}u+\mathrm{deg}v>n$, they hold together for some $i$. Following the path from $u$ to $A_i$, crossing to $v$, following the remaining path backwards to $B_i$, and crossing back to $u$ gives a Hamilton cycle already in the original maximal graph. This contradiction proves the stated Hamiltonicity criterion.

For $n=2m$, two disjoint copies of $K_{m,m}$ have $n$ vertices in each class and minimum degree $m$. For $n=2m+1$, the disjoint union $K_{m,m+1}\bigsqcup K_{m+1,m}$ has $n$ vertices in each class and minimum degree $m$. Both are disconnected and hence non-Hamiltonian. Thus the sharp counterexamples have $\boxed{\delta =\lfloor n/2\rfloor }$ for every $n\ge 2$.

Let $v_1\cdots v_k$ be a longest path. Every neighbour of either endpoint lies on it. Among the $k-1$ indices $i=1,\dots ,k-1$, mark those with $v_1{v}_{i+1}$ an edge and those with $v_i{v}_k$ an edge. Their total number isso some index has both marks. The edges at that index close the path into the cycle graphThe minimum-degree hypothesis also forces the graph to be a connected graph: each component would otherwise have at least $n/2+1$ vertices. If $k<n$, an edge from this cycle to an outside vertex would give a longer path by opening the cycle at its incident vertex. Hence $k=n$, proving the Dirac theorem and Hamiltonicity.

The bound is sharp. For $n=2m$, take two disjoint copies of $K_m$, with minimum degree $m-1$. For $n=2m+1$, take two copies of $K_{m+1}$ sharing exactly one vertex. The minimum degree is $m$, but the shared vertex is a cut vertex, which no Hamiltonian graph can have. Both constructions have

A Hamilton cycle is a cycle graph which visits every vertex of a finite graph exactly once, apart from returning to its starting vertex. A graph is Hamiltonian when it contains such a cycle.

Replace the strict partial order by its reflexive closure $\le$. Partially order all partial orders on $X$ extending $\le$ by inclusion. A chain has an upper bound given by its union: transitivity and antisymmetry each concern finitely many pairs, which all lie in one member of the chain. By Zorn lemma there is a maximal extension $R$.

If $a,b$ are incomparable under $R$, defineThis adds $a{R}^{\prime }b$. It is transitive: composing an old pair with a new one stays in the added set; composing two new pairs would require $bRa$, which is impossible. It is antisymmetric: an added pair and a reverse old pair would again imply $bRa$, and two reverse added pairs would do the same. Thus $R^{\prime }$ is a larger partial order, contradicting maximality. Therefore $R$ is total, and removing its diagonal gives a total order extending the original strict order.

A well-founded relation has a minimal element in every nonempty subset; with the axiom of choice this is equivalent to having no infinite strictly descending sequence. If an extension is a well-order, its least element in any nonempty subset is minimal for the original relation, so the original order must be well-founded.

However not every total extension need be a well-order. Take the empty strict order on $\mathbb{N}$, which is well-founded. Its extension defined by $x\prec y$ if $x>y$ in the usual order is total and has no least element.

Zorn lemma states that a nonempty partially ordered set in which every chain in a partial order has an upper bound has a maximal element. Here is a proof from the axiom of choice and Hartogs theorem.

Suppose there is no maximal element. Every chain $C$ has a strict upper bound: choose an upper bound $b$ and then an element strictly greater than $b$. The family of sets of strict upper bounds is a set of nonempty subsets of the underlying set $P$. The axiom of choice is used here to choose, simultaneously for every chain $C$, one strict upper bound $b(C)$.

By Hartogs theorem there is an ordinal $\kappa$ admitting no injection into $P$. Transfinite recursion defines $x_{\alpha }$ for $\alpha <\kappa$ byAt each stage the previous elements form a chain, and $x_{\alpha }$ is strictly above every one of them. Hence $\alpha \mapsto x_{\alpha }$ is an injection $\kappa \to P$, a contradiction. Therefore a maximal element exists.