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PutThe initial values are $S_1=S_2=1$. Using Pascal's identity and the convention that an out-of-range binomial coefficient is zero,Thus $(S_n)$ has the same initial values and linear recurrence relation as the Fibonacci numbers. Mathematical induction gives

The Laplace-transform formula for a semigroup resolvent is locally uniformly convergent in the half-plane $\mathrm{Re}z>\omega$, so it may be differentiated under the Bochner integral:The resolvent identity gives $R^{\prime }(z)=-R(z{)}^2$ and, by mathematical induction,ThereforeFor real $\lambda >\omega$, the integral triangle inequality and the gamma integral giveConsequentlyThe exponent $-n+1$ printed in the question is a typographical error: already at $n=0$ it contradicts the first-resolvent bound.

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.

We prove the Hilbert cube theorem directly by mathematical induction on its dimension. Dimension zero is immediate. Suppose every finite coloring contains a monochromatic Hilbert $n$-cube, and let $c:\mathbb{N}\to [k]$ be a $k$-coloring. Refine it to the finite coloringBy the induction hypothesis, some Hilbert $n$-cubeis monochromatic for $C$. Consequently, for every $j\in \{0,\dots ,k\}$ the translate $H+j$ is monochromatic for $c$; denote its color by ${\gamma }_j$. By the pigeonhole principle, two of the $k+1$ colors ${\gamma }_0,\dots ,{\gamma }_k$ agree, say ${\gamma }_s={\gamma }_t$ with $s<t$. Thenis a monochromatic Hilbert $(n+1)$-cube. This proves the result without using the Finite sums theorem or the Van der Waerden theorem.

Let $[m]$ be an alphabet. A combinatorial line in $[m{]}^N$ is obtained by choosing a nonempty set of active coordinates, fixing every other coordinate, and putting the same variable letter in all active coordinates. The Hales-Jewett theorem states that, for every pair of positive integers $m,k$, some $N$ makes every $k$-coloring of $[m{]}^N$ contain a monochromatic combinatorial line.

We prove it by mathematical induction on $m$. The case $m=1$ is immediate. Suppose the result is known for the alphabet $[m-1]$, with any finite number of colors. Order each line by its variable letter, and call its last point its focus. Say that $r$ lines are color-focused when they share a focus, each line with its focus removed is monochromatic, and those $r$ punctured lines have different colors.

We claim that for every $1\le r\le k$ there is $N_r$ such that every $k$-coloring of $[m{]}^{N_r}$ contains either a monochromatic line or $r$ color-focused lines. For $r=1$, restrict a coloring to $[m-1{]}^{N_1}$ with $N_1=HJ(m-1,k)$. A monochromatic line there becomes a punctured line over $[m]$ after adjoining the point obtained by putting $m$ in every active coordinate. It is either already monochromatic or is one color-focused line.

Assume $n=N_r$ works for $r$, and putView $[m{]}^{n+N}$ as $[m{]}^n\times [m{]}^N$. For $b\in [m-1{]}^N$, record the entire color patternThis is a coloring with at most $k^{m^n}$ colors, so the induction hypothesis on the alphabet supplies a line on which this pattern is constant. Adjoin its focus $b^{+}$, whose active coordinates contain the letter $m$, and write $L$ for the resulting line over $[m]$. Thus $c(a,b)$ is independent of $b\in L\setminus \{b^{+}\}$ for every $a$. It defines a $k$-coloring $c^{\prime }(a)$ of $[m{]}^n$.

If $c^{\prime }$ has a monochromatic line, fixing any $b\in L\setminus \{b^{+}\}$ gives one for $c$. Otherwise there are $r$ color-focused lines for $c^{\prime }$, with common focus $f$. Couple the active coordinates of each of those lines with the active coordinates of $L$; this produces $r$ punctured lines focused at $(f,b^{+})$. The line obtained by fixing the first component at $f$ and varying along $L$ supplies one more. Its punctured color differs from the preceding $r$ colors, since equality with one of them would make the corresponding $c^{\prime }$-line monochromatic. Hence there are $r+1$ color-focused lines. The claim follows by induction on $r$.

Take $r=k$. The common focus has one of the $k$ colors, so it completes the punctured line of that color to a monochromatic line. This proves the Hales-Jewett theorem.

A $d$-dimensional combinatorial subspace has $d$ disjoint nonempty active coordinate sets, one for each independent variable. The Extended Hales-Jewett theorem says that, given $m,k,d$, every $k$-coloring of $[m{]}^n$ contains such a monochromatic subspace when $n$ is large enough. To deduce it, chooseand identifyA monochromatic combinatorial line over the alphabet $[m{]}^d$ becomes a monochromatic $d$-parameter set over $[m]$: within every active block, its first coordinates form the first variable set, its second coordinates form the second, and so on. Thus $n=dN$ suffices.

Hindman theorem states that every finite coloring of $\mathbb{N}$ admits an infinite sequence $x_1,x_2,\dots$ whose finite-sums set is monochromatic.

Identify the Stone-Čech compactification of the natural numbers $\beta \mathbb{N}$ with the compact Hausdorff space of ultrafilters on $\mathbb{N}$, equipped with addition on the Stone-Čech compactification of the natural numbers. This makes $\beta \mathbb{N}$ a compact Hausdorff left-topological semigroup. For completeness, the Ellis–Numakura lemma gives an idempotent in every such semigroup: by the Hausdorff space property and compactness, the intersection of a descending chain of nonempty compact subsemigroups is nonempty, so Zorn's lemma gives a minimal one $K$. For $a\in K$, the compact subsemigroup $K+a$ equals $K$. Hence the nonempty compact subsemigroupis also $K$, and in particular $a+a=a$. Choose the resulting idempotent ultrafilter on the natural numbers $\mathcal{U}$.

One color class $A$ belongs to $\mathcal{U}$. For $x\in \mathbb{N}$, writeand defineThe identity $\mathcal{U}+\mathcal{U}=\mathcal{U}$ implies $A^{\ast }\in \mathcal{U}$. It also implies that $A^{\ast }-x\in \mathcal{U}$ for every $x\in A^{\ast }$: both $A-x$ and the set of $y$ for which $A-(x+y)$ belongs to $\mathcal{U}$ lie in $\mathcal{U}$, and their intersection is $A^{\ast }-x$.

Choose $x_1\in A^{\ast }$. Having chosen $x_1,\dots ,x_n$ with every nonempty finite sum in $A^{\ast }$, chooseThis is possible because it is a finite intersection of members of the ultrafilter $\mathcal{U}$. Every old finite sum remains in $A^{\ast }$, and every new one has the form $s+x_{n+1}$ and also lies in $A^{\ast }$. By mathematical induction, all nonempty finite sums lie in $A^{\ast }\subseteq A$, proving Hindman's theorem.

Now putfor each $A\in \mathcal{U}$. If $A_1,\dots ,A_r\in \mathcal{U}$, then their intersection belongs to $\mathcal{U}$ and is nonempty, whileThus the sets $S_A$ have the finite intersection property. Their closures are closed subsets of the compact interval $[0,1]$, so their total intersection contains some $x$. Equivalently, every neighbourhood of $x$ meets every $S_A$; this is the ultrafilter limit of the sequence.

The point is unique. If distinct points $x,y$ both had this property, choose disjoint neighborhoods $U,V$. The index set $I_U=\{n:x_n\in U\}$ must belong to $\mathcal{U}$, because otherwise its complement would belong to $\mathcal{U}$ and the associated $S_A$ would miss $U$. Similarly $I_V\in \mathcal{U}$. Their intersection is empty, contradicting the definition of an ultrafilter.

Let $X=\{v_1,\dots ,v_m\}$ be the vertices of a regular polygon, indexed cyclically so that $v_i\mapsto v_{i+1}$ is an isometry. We first prove the product lemma needed for the cyclic-symmetry argument.

Suppose $A\subseteq X$ has the property that, for every number of colors, a finite witness forces a copy of $X$ on which the vertices corresponding to $A$ are monochromatic. We claim that for every $N,k$ there is a finite witness forcing an $A$-invariant copy of $X^N$. Proceed by mathematical induction on $N$. The case $N=1$ is the assumption. For the step, choose a finite set $S$ that forces an $A$-invariant $X^{N-1}$ for a coloring with $k^{|X|-|A|+1}$ colors, and choose $T$ that forces an $A$-monochromatic $X$ for a coloring with $k^{|S|}$ colors.

Given a $k$-coloring $c$ of $S\times T$, color $t\in T$ by its complete vector $(c(s,t){)}_{s\in S}$. In the resulting copy of $X$ in $T$, that vector is the same for every member of $A$. Each $s\in S$ can therefore be colored by the vector consisting of $c(s,a)$ for one $a\in A$ and the colors $c(s,x)$ for $x\in X\setminus A$. Applying the choice of $S$ gives an $A$-invariant copy of $X^{N-1}$, and its product with the chosen copy of $X$ in $T$ is the desired $A$-invariant copy of $X^N$. This proves the lemma rather than assuming it.

We now prove by induction on $r$ that a finite witness forces a copy of $X$ whose first $r$ vertices have one color. The case $r=1$ is trivial. For $r=2$, use that a line segment is a Euclidean Ramsey set; each occurrence of a segment congruent to $v_1{v}_2$ can be extended to a copy of the regular polygon, and only finitely many extensions are needed for a finite witness.

Assume the assertion for $r$, and put $A=\{v_1,\dots ,v_r\}$. The product lemma lets us work inside an $A$-invariantly colored copy of $X^N$, with $N$ as large as required. For an increasing $(m-1)$-element subset $H=\{a_1<\cdots <a_{m-1}\}$ of $[N]$ and $1\le i\le r$, define a word $w_i(H)$ by putting $v_{i+j}$ in coordinate $a_j$, with subscripts read modulo $m$, and putting $v_1$ in every other coordinate. Color $H$ by the $r$-tupleBy the Finite Ramsey theorem, for sufficiently large $N$ there are coordinates $b_1<\cdots <b_m$ on which all $(m-1)$-subsets have the same tuple color.

For $1\le i\le r+1$, let $z_i$ be the word whose entries in coordinates $b_1,\dots ,b_m$ arecyclically, and whose other entries are $v_1$. For $i\le r$, the word $z_i$ differs from $w_i(\{b_2,\dots ,b_m\})$ only in coordinate $b_1$, where the two entries $v_i,v_1$ lie in $A$. Similarly $z_{i+1}$ differs from $w_i(\{b_1,\dots ,b_{m-1}\})$ only in coordinate $b_m$, again by two members of $A$. The $A$-invariance and the homogeneous choice of the $b_j$ therefore giveHence $z_1,\dots ,z_{r+1}$ have one color.

The cyclic words $z_1,\dots ,z_m$ form an isometric copy of $\sqrt{m}X$: in each of the $m$ varying coordinates the cyclic shift is an isometry, so every squared distance is multiplied by $m$. Rescaling the finite witness by $1/\sqrt{m}$ gives a copy of $X$. The induction reaches $r=m$, proving that every regular polygon is a Euclidean Ramsey set.

For three consecutive vertices $p_0,p_1,p_2$ of a regular $M$-gon scaled to have side length one,Since the polygon is Euclidean Ramsey, choosing $M$ large enough proves that $\{0,1,2\}$ is an approximately Euclidean Ramsey set.

In fact every finite $X\subseteq {\mathbb{R}}^d$ is approximately Ramsey. First perturb each coordinate of each point onto a sufficiently fine lattice $\delta \mathbb{Z}$, changing all pairwise distances by less than $\epsilon /2$. For each coordinate, map the finitely many required integers to consecutive vertices of a very large regular polygon, scaled so that one angular step has arc length $\delta$. If the step angle is $\theta$, the chord replacing a difference of $q$ lattice steps has lengthas $\theta \to 0$, uniformly over the finitely many differences involved. Taking the orthogonal Cartesian product of $d$ such polygons therefore embeds the perturbed points with a further distance error below $\epsilon /2$. Each polygon is Euclidean Ramsey, and their product is Euclidean Ramsey by the product theorem for Euclidean Ramsey sets. A monochromatic copy of that product contains the corresponding approximate copy of $X$, completing the proof.

This is the Baumslag-Solitar group $G=\mathrm{BS}(1,2)$. Its defining relation gives, by mathematical induction,In the cyclic subgroup $H=⟨a⟩$ with generator $a$,whereas in $G$, with generators $a,b$,If inclusion were a $(\lambda ,\epsilon )$-quasi-isometric embedding, its lower bound would implyfor every $n$, which is impossible because an exponential function eventually dominates every linear function. Hence $H$ is not quasi-isometrically embedded; it is an exponentially distorted subgroup.

We use mathematical induction on $n$. The identity is immediate for the empty partition. Add a removable corner $(r,s)$ to a partition of $n-1$, and put $c=s-r$. Only the new hook and the hooks to its left in row $r$ or above it in column $s$ change. Each old affected hook length increases by one. If $S$ is the sum of those old hook lengths, direct substitution in the hook formula givesThere are $r+s-2$ affected old hooks, so the increase in the sum of squared hook lengths isThe increase in $n^2+\sum c_{i,j}^2$ is likewise $n^2-(n-1{)}^2+c^2=2n-1+c^2$. The induction closes and proves

Set $\overline{\mathbf{y}}=({\mathbf{y}}_n+{\mathbf{y}}_{n+1})/2$. The implicit midpoint rule readsTaking the inner product with ${\mathbf{y}}_{n+1}+{\mathbf{y}}_n=2\overline{\mathbf{y}}$ givesThe last equality again follows from skew-symmetry. Mathematical induction therefore yields

Proceed by mathematical induction on $n$. The cases $n=0,1$ are immediate. Given a symmetric chain decomposition of a Boolean lattice $\mathcal{P}([n-1])$, consider one of its chainsIt produces the two chainsand, when nonempty,The first runs from rank $r$ to rank $n-r$, and the second from rank $r+1$ to rank $n-1-r$; both endpoint ranks sum to $n$. They are disjoint and together contain the old chain both without and with $n$. Doing this for every old chain partitions $\mathcal{P}([n])$ into symmetric chains.

Every positive integer $m$ has a unique binomial representationDefineThe Kruskal-Katona theorem says that every $r$-uniform family $\mathcal{A}$ of size $m$ hasand the initial segment of length $m$ in colexicographic order attains equality.

We prove the inequality by simultaneous mathematical induction on the ground-set size and on $r$. We use the elementary binomial-shadow arithmetic lemmaTo verify the lemma, greedily remove the largest term $\left(\genfrac{}{}{0ex}{}{q}{r}\right)$ from the binomial representation of $a+b$. If $a$ reaches that term first, apply the induction hypothesis to the remainders; otherwise transfer the excess to $b$. In the crossing case Pascal's identitygives exactly the two terms on the right. This induction also proves uniqueness of the greedy representation.

Choose the largest ground-set element $x$ and splitThe members of the lower shadow that avoid $x$ form $\partial {\mathcal{A}}_0\cup {\mathcal{A}}_1$, while those containing $x$ are the sets $B\cup \{x\}$ with $B\in \partial {\mathcal{A}}_1$. Hence, writing $a=|{\mathcal{A}}_0|$ and $b=|{\mathcal{A}}_1|$,This completes the induction.

Finally, the initial colex segment first contains all $r$-subsets of $[a_r]$, followed recursively by sets containing $a_r+1$ whose remaining elements form the appropriate initial segment at level $r-1$. Its shadow has the same recursive decomposition, and therefore has exactly $f_r(m)$ members. This proves sharpness.

Ramsey's theorem for $r$-sets says that every finite coloring of $[\mathbb{N}{]}^{(r)}$ has an infinite monochromatic set. We prove it by mathematical induction on $r$. The case $r=1$ is the infinite pigeonhole principle. Suppose the result holds for $r-1$, and let $c:[\mathbb{N}{]}^{(r)}\to [k]$. Choose $a_1$, then use the induction hypothesis on the coloring $F\mapsto c(F\cup \{a_1\})$ to obtain an infinite set $M_1$ on which this color is constant, say $d_1$. Inductively chooseso that $c(F\cup \{a_j\})=d_j$ for every $F\in [M_j{]}^{(r-1)}$. Some color $d$ occurs for infinitely many $d_j$. If $j_1<j_2<\cdots$ are the corresponding indices, every $r$-set from $\{a_{j_1},a_{j_2},\dots \}$ has color $d$: take its least-indexed element $a_{j_s}$, after which its other $r-1$ elements lie in $M_{j_s}$. This proves the theorem.

Apply bipartite superadditivity of quantum relative entropy first to system $1$ and systems $2,\dots ,n$, and then repeat on the remaining joint state. Mathematical induction givesThis is multipartite superadditivity of quantum relative entropy. Repeated application of bipartite additivity similarly gives

Rewrite the recurrence relation asSince $a_0=1>0$ and $c>0$, a positive $a_n$ gives a positive $a_{n+1}$. Mathematical induction therefore proves $\boxed{a_n>0\text{ for all }n\ge 0}$, and also shows that every division in the recurrence relation is defined.

Put $u_n=11^n\log n/(2^{2n+1}n!)$ for $n\ge 2$. The factorial givesThe logarithm identityfollows from continuity of the logarithm at one and $\log n\to +\infty$. The product rule for limits of sequences therefore makes the displayed ratio tend to zero. In particular, some $N$ satisfies $0<u_{n+1}\le u_n/2$ for all $n\ge N$. By mathematical induction, $0<u_{N+j}\le u_N{2}^{-j}$. A geometric sequence with ratio in $(0,1)$ tends to zero, so the squeeze theorem proves

The construction is the partial middle-third construction of the Cantor function. After $n$ rounds of assigning middle thirds, every remaining open interval has endpoint valuesfor some $k=0,\dots ,2^n-1$, and every such consecutive pair occurs. This follows by mathematical induction. Initially the endpoint values are $0,1$. An interval with those consecutive endpoint values receives the valueon its closed middle third. Its two remaining open intervals therefore have consecutive endpoint values at denominator $2^{n+1}$. All previously assigned endpoint values persist. Hence all even numerators at the next denominator were already present, and all odd numerators have just been assigned. Including the initial values zero and one provesThese are precisely the dyadic rational values between zero and one.

If $A$ is a strictly upper triangular matrix, then $A_{ij}=0$ unless $j-i\ge 1$. We prove by mathematical induction thatThe assertion for $k=1$ is the definition. In the matrix multiplication formulaa nonzero summand would require $\ell -i\ge k$ and $j-\ell \ge 1$, hence $j-i\ge k+1$. This proves the induction step. Since $j-i\le n-1$ for all entries, $\boxed{A^n=0}$. In particular, every strictly upper triangular matrix is a nilpotent matrix.

The real spectral theorem states that a symmetric matrix has an orthonormal basis of real eigenvectors, and hence is a diagonalisable matrix. Here is a proof in this setting.

For $n\ge 1$, the quadratic form $q(x)=x^{T}Ax$ attains a maximum on the unit sphere, by compactness and the extreme value theorem. Choose a unit maximizer $u$ and put $\lambda =u^{T}Au$. For any $w\perp u$, the Rayleigh quotient along $u+tw$ has a maximum at $t=0$, soThe last equality uses $A=A^T$ and $u^{T}w=0$. Thus $Au$ is orthogonal to every $w\perp u$, so $Au=\lambda u$.

The orthogonal complement $u^{\perp }$ is an invariant subspace, because for $x\perp u$,The restriction of $A$ to $u^{\perp }$ is again symmetric with respect to the induced inner product. By mathematical induction on dimension, it has an orthonormal basis of eigenvectors; adjoining $u$ gives one for ${\mathbb{R}}^n$. If $Q$ has those basis vectors as columns, then $Q^{T}AQ$ is a real diagonal matrix and $Q^{-1}=Q^T$.

Put $L={\bigcap }_{i=1}^k{H}_i$. Repeated application of part (iv), or mathematical induction on $k$, givesPart (ii) therefore implies that $L$ is an infinite set. If $G\setminus U$ is nonempty, choose $g\notin U$. We claim its left coset $gL$ is wholly outside $U$. If $g\ell \in H_i$ for some $\ell \in L$, then $\ell \in H_i$ and closure under products and inverses would give $g=(g\ell ){\ell }^{-1}\in H_i$, a contradiction.

Thus $gL\subseteq G\setminus U$. Part (i) makes $gL$ infinite, so the complement is infinite. If there is no such $g$, the complement is empty. This proves the complement of a finite union of finite-index subgroups property: