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A ring is a set with an abelian-group addition, associative multiplication and both distributive laws; here we use rings with multiplicative identity. A field is a nonzero commutative ring in which every nonzero element has a multiplicative inverse.

The positive integers form neither a ring nor a field. They do not contain an additive identity zero, and they also lack additive inverses.

After division by $z(z+2)$, the coefficients areTheir only finite singularities are $z=0$ and $z=-2$. At either point $z_0$, the products $(z-z_0)p$ and $(z-z_0{)}^{2}q$ are analytic, so both points are regular singular points. All other finite points are ordinary.

The lowest power after substituting the Frobenius series gives $\sigma (2\sigma -1)=0$. The coefficient of $z^{n+\sigma }$ then givesFor both exponents the denominator is nonzero for every $n\ge 0$, so each arbitrary nonzero $a_0$ determines a series. Unless it terminates, its coefficient ratio tends to $-1/2$, proving convergence for $|z|<2$. For the half-integer exponent choose a square-root branch. Their leading powers $1$ and $z^{1/2}$ show that the two local solutions are linearly independent.

If $\alpha =m$ is a positive integer, choose $\sigma =0$. The recurrence numerator is nonzero for $0\le n<m$ and vanishes at $n=m$. Thus $a_m\ne 0$, $a_{m+1}=0$, and all later coefficients vanish: a polynomial solution of exact degree $m$ exists. It is proportional to the Chebyshev polynomial $T_m(z+1)$, since the substitution $t=z+1$ transforms the equation into $(t^2-1)y^{\prime \prime }+t{y}^{\prime }-m^{2}y=0$.

First classify nonzero polynomial solutions of the differential equation itself. If $y$ has degree $d$ and leading coefficient $a_d\ne 0$, the coefficient of $x^{d+1}$ is $(d+\lambda )a_d$. Hence $\lambda =-d$ is necessary.

An analytic solution at zero belongs to the exponent-zero Frobenius branch. Its odd-power coefficients vanish, because $a_1=0$ and the recurrence relation preserves parity. For the even-power coefficients,All denominators are nonzero for $j\ge 0$. With $\lambda =-2m$, the coefficients through $a_{2m}$ are nonzero when $a_0\ne 0$, the next coefficient vanishes, and every subsequent coefficient vanishes. This proves sufficiency as well as the restriction to even polynomial degree:For example, setting $a_0=1$ gives $1$ when $\lambda =0$, $1+x^2$ when $\lambda =-2$, and $1+2x^2+x^4/5$ when $\lambda =-4$. A polynomial cannot instead begin with a positive integer power: its lowest power must have an indicial exponent, and the only integer indicial exponent is zero. This also proves that no odd-degree family was missed. The finite-series mechanism is a terminating Frobenius series.

If the additional endpoint conditions from part (b) are imposed on this part too, there is no nonzero polynomial solution: every nonzero polynomial just classified has $y(0)=a_0\ne 0$. The displayed family and degree answer therefore concern the differential equation without those extra endpoint conditions.

Mathematical induction consists of an induction base case $P(1)$ and an inductive step $P(n)\Rightarrow P(n+1)$ for every positive integer $n$. To derive it from the well-ordering principle, suppose the set of counterexamples were nonempty and take its least element $n$. The base case excludes $n=1$. If $n>1$, minimality gives $P(n-1)$, and the step then gives $P(n)$, a contradiction. Hence there are no counterexamples.

For the stated integer congruence, the base case is $9\equiv 2(\mathrm{mod}7)$. If $9^n\equiv 2^n(\mathrm{mod}7)$, multiplication and $9\equiv 2(\mathrm{mod}7)$ givecompleting the induction.

The purported shifted-sum proof has a correct conditional successor calculation, but the base case is missing and false: at $n=1$ it would assert $1=1+126$. An inductive step without a true induction base case cannot establish any instance.

Van der Waerden theorem. Given positive integers $r,k$, there is a finite $W(r,k)$ such that every finite colouring of $\{1,\dots ,W(r,k)\}$ with at most $r$ colours has a monochromatic arithmetic progression of length $k$ and positive common difference. In particular, every finite colouring of the positive integers contains such progressions of every finite length.

Here is a colour-focusing proof of Van der Waerden theorem. We induct on $k$, simultaneously for all numbers of colours. The assertion for $k=1$ is immediate. Suppose $W(r,k)$ exists for every $r$, and fix $r$. We construct bounds $F_t$, for $1\le t\le r$, with this property: every colouring of $[F_t]$ either has a monochromatic progression of length $k+1$, or has $t$ progressions of length $k$, of distinct colours, which share a next point $f\in [F_t]$. Write those progressions aswith positive steps $d_i$ and all displayed points in $[F_t]$.

For $t=1$, put $L=W(r,k)$ and $F_1=2L$. A monochromatic progression of length $k$ in $[L]$ has its next point in $[2L]$, since its step is at most $L$; for $k=1$ choose step $1$. This establishes the initial focusing property.

Suppose $F_t=N$ has been constructed. Put $M=W(r^N,k)$ and $F_{t+1}=2NM$. Divide the first $NM$ integers into $M$ consecutive blocks of size $N$. Colour a block by its full colour profile, an $N$-tuple having at most $r^N$ possible values. The induction hypothesis on $k$ gives block indiceswith the same colour profile. For $k=1$ use any one block and $D=1$. Suppose that the whole interval has no monochromatic progression of length $k+1$. Its repeated profile has $t$ focused length-$k$ progressions with positions $f-h{d}_i$ and distinct colours $q_i$. The colour $q$ of position $f$ differs from every $q_i$, since equality would complete a length-$(k+1)$ progression in the block.

Set $j_{\ast }=j_0+kD$ and $F=N(j_{\ast }-1)+f$. The index $j_{\ast }$ is at most $2M$, so this focus is in the full interval. For $1\le h\le k$, the pointis in one of the chosen blocks and has colour $q_i$. Also $F-hND$ is at position $f$ in that block and has colour $q$. We have produced $t+1$ focused length-$k$ progressions of distinct colours. This proves the recursive focusing property. At $t=r$, the focus itself must have one of the $r$ colours, so one of the progressions extends to length $k+1$. Thus $W(r,k+1)=F_r$ is a valid bound, completing the mathematical induction and the requested proof.

To obtain the equation criterion, we first prove the small strengthening of Van der Waerden theorem that we need. For any positive integers $r,s,L$, there is a finite bound forcing a monochromatic configurationThis is the Brauer progression theorem, and the following argument proves it from the theorem just established. Induct on $r$. One colour is immediate. Given a bound $M$ for $r-1$ colours with the same $s,L$, let $T=W(r,(L-1)M+1)$ and colour an interval at least as long as $\max (T,sMT)$. Find a monochromatic progression $a,a+d,\dots ,a+(L-1)Md$ in $[T]$, of colour $q$; take $d=1$ if this progression has one term. If some $jsd$, $1\le j\le M$, has colour $q$, its first $L$ suitably spaced terms $a,a+jd,\dots ,a+(L-1)jd$ and the point $s(jd)$ give the configuration. Otherwise the colouring of the indices $1,\dots ,M$ obtained from $j\mapsto c(jsd)$ uses at most $r-1$ colours. The induction hypothesis supplies indices $se,b,b+e,\dots ,b+(L-1)e$ of one colour. Multiplication by $sd$ gives the desired configuration with step $esd$. This completes the strengthening without assuming any Rado theorem.

Multiply the rational coefficients by a common positive denominator, so that all $a_i$ are nonzero integers; this does not alter the equation or partition regularity. By definition, the row is partition regular when every finite colouring admits positive monochromatic $x_1,\dots ,x_n$ satisfying ${\sum }_i{a}_i{x}_i=0$. The variables need not be distinct.

For necessity, choose a prime number $p>{\sum }_i|a_i|$. Write each positive $x$ uniquely as $p^{v_p(x)}u$, with $p\nmid u$, and colour $x$ by $u\mathrm{mod}p$. This is a finite colouring with $p-1$ colours. If a monochromatic solution exists, let $v={\min }_i{v}_p(x_i)$ and $I=\{i:v_p(x_i)=v\}$. Dividing the equation by $p^v$ and reducing modulo $p$ giveswhere $u_0\ne 0(\mathrm{mod}p)$ is the common colour. Thus ${\sum }_{i\in I}{a}_i\equiv 0(\mathrm{mod}p)$. Its absolute value is strictly less than $p$, so ${\sum }_{i\in I}{a}_i=0$. The index set $I$ is nonempty. This proves the leading-residue obstruction to partition regularity.

Conversely, suppose ${\sum }_{i\in I}{a}_i=0$ for a nonempty $I$. If $I$ is all indices, taking every variable equal to any one positive integer already gives a monochromatic solution. Otherwise fix $i_0\in I$, put $s=|a_{i_0}|$, $C={\sum }_{j\notin I}{a}_j$, and define integer offsetsLet $u={\min }_{i\in I}{\lambda }_i$ and choose $L=1+{\max }_{i\in I}{\lambda }_i-u$. The proved Brauer progression theorem gives a monochromatic configuration consisting of $sd$ and a length-$L$ progression starting at $b$. SetAll variables are positive and of one colour. Their weighted sum isThis explicitly gives a Brauer configuration for a one-row zero-sum equation. We concludeBoth directions have been proved independently of the general Rado theorem.

We take $\mathbb{N}=\{1,2,\dots \}$. An ultrafilter $p$ on $\mathbb{N}$ is a proper filter on a set which decides every subset: it contains $\mathbb{N}$ but not $\varnothing$, is closed under finite intersections and passage to supersets, and contains exactly one of $A$ and $\mathbb{N}\setminus A$ for every $A\subseteq \mathbb{N}$. A principal ultrafilter has the form ${\delta }_n=\{A:n\in A\}$; all other ultrafilters are nonprincipal ultrafilters.

To prove existence, start with the cofinite filter and order its proper filter extensions by inclusion. The union of a chain is again a proper filter on a set: the finitely many sets involved in an intersection all belong to one member of the chain, and no member contains the empty set. Thus Zorn's lemma supplies a maximal proper filter $p$. If $A\notin p$, adjoining $A$ must make the generated filter improper; equivalently, some $B\in p$ has $A\cap B=\varnothing$. Then $B\subseteq \mathbb{N}\setminus A$, so $\mathbb{N}\setminus A\in p$. This proves that $p$ is an ultrafilter. It contains the complement of every finite set and hence no finite set. In particular, it contains no singleton and is nonprincipal.

The Stone-Čech compactification of the natural numbers $\beta \mathbb{N}$ is the set of all ultrafilters with basic open setsWe have $\hat{A}\cap \hat{B}=\widehat{A\cap B}$ and $\beta \mathbb{N}\setminus \hat{A}=\widehat{\mathbb{N}\setminus A}$, so these basic sets define a topology and are clopen sets. If $p\ne q$, choose $A\in p\setminus q$. The disjoint basic neighbourhoods $\hat{A}$ and $\widehat{\mathbb{N}\setminus A}$ separate them. Thus the space is Hausdorff.

For compactness, it suffices to show every basic open cover has a finite subcover, since an arbitrary open cover can be refined by basic open sets. Suppose $\{\widehat{A_i}:i\in I\}$ has no finite subcover. The complements $\mathbb{N}\setminus A_i$ have the finite intersection property: if a finite intersection were empty, the corresponding finite family of $\widehat{A_i}$ would cover every ultrafilter. Their generated proper filter on a set extends to an ultrafilter, by the same Zorn's lemma argument. This ultrafilter contains every $\mathbb{N}\setminus A_i$, and therefore belongs to none of the alleged cover members, a contradiction. We have proved that $\beta \mathbb{N}$ is a compact Hausdorff space. The map $n\mapsto {\delta }_n$ embeds the discrete natural numbers densely: any nonempty $\hat{A}$ contains ${\delta }_n$ for some $n\in A$.

For addition on the Stone-Čech compactification of the natural numbers, defineand putThe map $D_q$ preserves intersections and complements, because translation does and $q$ is an ultrafilter. It also sends $\mathbb{N}$ to $\mathbb{N}$ and the empty set to itself. Consequently the displayed rule defines an ultrafilter $p+q$. On principal points it satisfies ${\delta }_n+{\delta }_m={\delta }_{n+m}$, so it extends ordinary addition.

Write $({\forall }^{p}n)\phi (n)$ to mean that $\{n:\phi (n)\}\in p$. The ultrafilter addition rule is then an iterated quantifier, with its order fixed. Expansion givesOrdinary addition is an associative operation, and the quantifiers are in the same order on both lines. Thus $\boxed{(p+q)+r=p+(q+r)}$. This argument does not interchange ultrafilter quantifiers and does not assert that this extended addition is commutative.

For fixed $q$, the translation $R_q:p\mapsto p+q$ hasa basic clopen set. Therefore addition is continuous in its left variable. This is the left-continuity convention used here: right translations are continuous, giving a left-topological semigroup. Continuity in both variables jointly is not part of the assertion.

Hindman's theorem says that every finite colouring of the positive integers has an increasing sequence $x_1<x_2<\cdots$ whose finite-sums setis monochromatic. Assume, as permitted, that the addition just defined has an idempotent ultrafilter $p$, so $p+p=p$. This ultrafilter is nonprincipal: a principal ${\delta }_n$ would satisfy ${\delta }_n+{\delta }_n={\delta }_{2n}\ne {\delta }_n$ for positive $n$. Hence every member of $p$ is infinite and every cofinite set belongs to $p$.

Exactly one cell $A$ of the finite colour partition belongs to $p$. Define its star set bySince $A\in p+p$, we have $D_p(A)\in p$ and therefore $A^{\ast }\in p$. Moreover, for each $n\in A^{\ast }$, the set $A-n$ belongs to $p$. Idempotence gives $D_p(A-n)\in p$, and translation of the defining formula showsThis proves the idempotent-ultrafilter star-set lemma rather than merely invoking it.

Choose $x_1\in A^{\ast }$. Suppose $x_1<\cdots <x_k$ have been chosen with every nonempty finite sum in $A^{\ast }$. Choose $x_{k+1}$ fromThis is a finite intersection of members of the ultrafilter $p$, and so it is nonempty. The new sums are $x_{k+1}$ and $s+x_{k+1}$; all lie in $A^{\ast }$ by the choice. Mathematical induction yields an increasing infinite sequence withThus the assumed idempotent gives the full Hindman theorem, with distinct increasing generators and all their nonempty finite sums of one colour.

Use the inclusive convention: a set is countable if it admits an injective function into $\mathbb{N}=\{1,2,\dots \}$. This includes every finite set and the empty set. If $j:X\to \mathbb{N}$ is injective and $Y\subseteq X$, its restriction to $Y$ is still injective, proving directly that every subset of a countable set is countable.

Enumerate pairs by diagonals. The positive-integer Cantor pairing functionis a bijection ${\mathbb{N}}^2\to \mathbb{N}$: on the diagonal $a+b-2=d$, its values are $d(d+1)/2+1$ through $(d+1)(d+2)/2$, each once, and these consecutive blocks partition $\mathbb{N}$. This proves countability without assuming any product theorem.

For a sequence of countable sets $X_1,X_2,\dots$, choose injections $j_i:X_i\to \mathbb{N}$. For $x$ in their union, let $i(x)$ be the least index containing $x$. The mapis injective: its encoded index and the injective within-set code recover $x$. Thus a countable union of countable sets is countable. Selecting the injections for an arbitrary given family uses the usual axiom of countable choice; when the encodings are supplied, the displayed construction is entirely explicit.

For finite Cartesian powers, start with the identity injection on $\mathbb{N}$ and recursively encode $(a_1,\dots ,a_n)$ as $\pi (j_{n-1}(a_1,\dots ,a_{n-1}),a_n)$. This proves that ${\mathbb{N}}^n$ is countable for every positive integer $n$ and is the constructive proof of the finite Cartesian power of a countable set property.

For each positive integer $m$, let $P_m$ consist of the functions for which $m$ is a period. Restriction to the residues $0,1,\dots ,m-1$ is a bijection $P_m\to {\mathbb{N}}^m$: any tuple extends uniquely to all of $\mathbb{Z}$ by taking residues modulo $m$, including negative arguments. Hence $P_m$ is countable. Every periodic function belongs to some $P_m$, soThere is no need to assume the chosen period is minimal; multiple descriptions only create overlap, which does not invalidate the union argument. This is an instance of the cardinality of integer-periodic function spaces principle.

For integers $m,n$ not both zero, their greatest common divisor is the greatest positive integer dividing both. For the degenerate pair $(0,0)$ use the conventional value zero. A number $d$ divides both $m,n$ exactly when it divides both $m+kn,n$: one direction uses closure of divisibility under sums; the reverse subtracts $kn$. The positive common-divisor sets are therefore identical, provingThis also holds for $(0,0)$ under the stated convention and does not depend on the signs of the integers.

Bezout identity states that there exist integers $a,b$ such that $am+bn=\gcd (m,n)$. If a prime $p$ divides $mn$ but does not divide $m$, then $\gcd (p,m)=1$, so choose $a,b$ with $ap+bm=1$. Multiplying by $n$ gives $apn+bmn=n$. Both terms on the left are divisible by $p$, so $p\mid n$. Thuswhich proves the Euclid lemma from the requested identity. The separately numbered entries use these arithmetic facts for the Fibonacci numbers.

For a finite alphabet $K$, a combinatorial line in $K^N$ has a nonempty active set $J\subseteq \{1,\dots ,N\}$, fixed letters outside $J$, and the same variable letter at every coordinate in $J$. The Hales-Jewett theorem states that for every alphabet size $k\ge 1$ and number of colours $r\ge 1$, some $N$ makes every r-colouring of $\{1,\dots ,k{\}}^N$ contain a monochromatic combinatorial line.

We first prove the alphabet insensitivity lemma with a concrete bound. Fix two distinct letters $a,b$, a desired dimension $d$, and a colouring $\chi :K^N\to \{1,\dots ,r\}$. A combinatorial subspace of dimension $d$ has disjoint nonempty active blocks $J_1,\dots ,J_d$ and fixed other coordinates. It is $(a,b)$-insensitive when exchanging $a,b$ in any one variable leaves the colour unchanged, whatever the other variable letters are.

Choose consecutive blocks of lengths $L_1,\dots ,L_d$ recursively byProcess these blocks from right to left. When block $i$ is reached, the later blocks already have one variable each; the earlier $P_i$ coordinates are still entirely unrestricted. For each of the $L_i+1$ words $a^t{b}^{L_i-t}$ in block $i$, form its complete colour profile over all $k^{P_i}$ earlier words and all $k^{d-i}$ assignments to the later variables. There are at most $r^{k^{P_i+d-i}}=L_i$ profiles. The pigeonhole principle gives $s<t$ with equal profiles. Fix the first $s$ coordinates of this block to $a$, the coordinates after $t$ to $b$, and make the interval from $s+1$ to $t$ variable. Substitution of $b$ gives the chain word indexed by $s$, and substitution of $a$ gives the one indexed by $t$; they have identical profiles. Thus this variable is insensitive to $a,b$ for every remaining context. Previously constructed later variables remain insensitive, because their colour equalities held for every possible earlier word. After all $d$ blocks, we have the required insensitive combinatorial subspace. This proves the explicit block bound for alphabet insensitivity.

Now induct on $k$, simultaneously for every $r$. The case $k=1$ is immediate with $N=1$. For $k\ge 2$, assume the Hales-Jewett theorem for the alphabet of size $k-1$, and take $d$ large enough for that theorem with $r$ colours. Apply the proved alphabet insensitivity lemma to the letters $k-1,k$, obtaining a d-dimensional combinatorial subspace $\varphi :K^d\to K^N$. On $\{1,\dots ,k-1{\}}^d$, the induced colouring $\chi \circ \varphi$ has a monochromatic combinatorial line. Its value at letter $k$ has the same colour as its value at $k-1$: change the active variables one at a time and use insensitivity. The union of their nonempty active blocks is the active set of a combinatorial line in the original word space. All its k points have one colour, completing the induction and the proof.

To deduce the Van der Waerden theorem, take $K=\{0,\dots ,m-1\}$ and pull a colouring of the positive integers back throughA combinatorial line with active set $J$ maps toHence every finite colouring has a monochromatic arithmetic progression of every prescribed finite length. In finite form, if $H(k,r)$ is a word-length bound for the Hales-Jewett theorem, then $W(r,m)\le 1+(m-1)H(m,r)$ is a valid interval bound.

The Strengthened Van der Waerden theorem also puts the common difference in the colour class:Here is a finite proof by induction on the number $r$ of colours. For $m=1$ choose $a=d=1$. For $r=1$ choose $a=d=1$ in $[m]$. For $m\ge 2,r\ge 2$, let $M$ be a bound for the strengthened theorem with $r-1$ colours, put $L=(m-1)M+1$, and let $W=W(r,L)$ be an ordinary Van der Waerden theorem bound. Work in $[MW]$. Its first $W$ points contain a monochromatic progressionof colour $c$. Every $jd$, $1\le j\le M$, lies in $[MW]$. If some $jd$ has colour $c$, then $a,a+jd,\dots ,a+(m-1)jd$, together with its difference $jd$, gives the answer. Otherwise the colouring of $\{d,2d,\dots ,Md\}$ uses only $r-1$ colours. Pull it back to $[M]$ and apply the induction hypothesis to get a monochromatic set $\{e,b,b+e,\dots ,b+(m-1)e\}$. Multiplication by $d$ gives the required configuration in the original colouring. This proves the strengthened theorem, including finite bounds.

The Gallai theorem for an integer lattice states that every finite colouring of ${\mathbb{Z}}^h$ contains a monochromatic homothetic copy $u+tF$ of every prescribed finite nonempty pattern $F\subseteq {\mathbb{Z}}^h$, where $t$ is a positive integer. The positive-lattice version holds for finite $F\subseteq {\mathbb{N}}^h$ as well. Write $F=\{v_1,\dots ,v_k\}$ and colour $w\in \{1,\dots ,k{\}}^N$ by the colour ofA monochromatic combinatorial line maps to $u+|J|F$, with $u=b+{\sum }_{j\notin J}{v}_{w_j}$. Thus the Hales-Jewett theorem supplies the copy, with nonzero dilation $|J|$. Choosing $b$ large and positive places all these word sums in the positive lattice if required. This sum map from words to homothetic copies also proves the real-vector-space version: replace ${\mathbb{Z}}^h$ by ${\mathbb{R}}^h$ and allow an arbitrary finite real pattern, while retaining positive integer dilations.

For a rational matrix $A$ with columns $a_1,\dots ,a_n$, partition regularity means that every finite colouring of the positive integers admits a positive monochromatic vector $x$ with $Ax=0$. Repeated coordinates are allowed. The columns condition is an ordered partition $\{1,\dots ,n\}=B_1\bigsqcup \cdots \bigsqcup B_s$ into nonempty blocks such thatRado's theorem says that $A$ is partition regular if and only if it satisfies the columns condition. We prove both directions.

For necessity, clear denominators so the columns are integer vectors. For every pair of disjoint index sets $I,B$ with $B\ne \varnothing$ for which ${\sum }_{i\in B}{a}_i$ is not in the rational linear span of the I-columns, choose an integer linear functional ${\ell }_{I,B}$ that vanishes on those I-columns but not on that sum. Such a functional exists by finite-dimensional linear algebra; clear its rational denominators. There are only finitely many pairs, so choose a prime number $p$ dividing none of the nonzero integersIf there are no offending pairs, any prime number suffices.

Colour a positive integer $x=p^{v_p(x)}u$, $p\nmid u$, by $u\mathrm{mod}p$. This is the last nonzero digit coloring, using $p-1$ colours. Partition regularity gives a positive monochromatic solution. Group its coordinate indices by their distinct P-adic valuations $e_1<\cdots <e_s$, writing $B_j=\{i:v_p(x_i)=e_j\}$. All units $x_i/p^{e_j}$ have the same nonzero residue $\alpha$ modulo $p$.

For each $j$, let $I=B_1\cup \cdots \cup B_{j-1}$. If its block sum failed the required span condition, apply the chosen ${\ell }_{I,B_j}$ to ${\sum }_i{a}_i{x}_i=0$. The earlier terms vanish exactly, not just modulo a power of $p$. Divide by $p^{e_j}$; terms with later valuations are still divisible by $p$. Reduction modulo $p$ then yieldscontrary to the choice of $p$. Thus every required span relation holds, and with $I=\varnothing$ the first block sum is zero. This is the finite separating-functional proof of the columns condition; it avoids an unjustified passage from congruences to rational equality.

For sufficiency, assume the columns condition. Choose rational coefficients ${\lambda }_{ij}$, for $j\ge 2$ and $i\in B_1\cup \cdots \cup B_{j-1}$, such thatChoose a positive integer $c$ clearing every denominator, and a positive integer $p\ge \max |c{\lambda }_{ij}|$, taking $p=1$ if the list is empty. By the allowed monochromatic m-p-c set theorem, there are positive generators $z_1,\dots ,z_s$ such that all numbersare positive and have one colour. This is a full m-p-c set; positivity is part of its definition, not a rule discarding negative expressions. For $i\in B_h$, putThese coordinates all belong to that monochromatic m-p-c set. Collecting coefficients of $z_j$ givesThus the Rado solution inside an m-p-c set proves partition regularity and completes the theorem.

To obtain the Finite sums theorem from Rado's theorem, introduce a positive variable $y_F$ for every nonempty $F\subseteq \{1,\dots ,k\}$ and imposePartition the columns of this system into $B_j=\{F:\min F=j\}$. The $B_1$-column sum is zero: in a row indexed by $F$ containing $1$, the contributions of $y_F$ and $y_{\{1\}}$ cancel, and other rows receive neither. For $j>1$, the $B_j$-column sum iswhere $a_F$ is the column of $y_F$. Those columns lie in earlier blocks. Hence the columns condition holds. A monochromatic positive solution gives $x_i=y_{\{i\}}$ and $y_F={\sum }_{i\in F}{x}_i$, provingFor $k=1$ the system has no rows and the conclusion is immediate. The construction is also visible directly in the sufficiency proof: a monochromatic $(k,1,1)$-set contains every nonempty sum of its generators, classified by the smallest chosen index.

Write $[\mathbb{N}{]}^{\omega }$ for the space of infinite subsets of the natural numbers, with each subset identified with its increasing enumeration. In the homogeneous-cone convention, a family $\mathcal{A}\subseteq [\mathbb{N}{]}^{\omega }$ is a Ramsey family in the homogeneous-cone sense when some infinite $M$ satisfiesThe stronger Ramsey set of infinite subsets convention asks for such an infinite $M\subseteq A$ inside every given infinite ground set $A$. We will prove the open-set assertion in this stronger form, and our Ramsey-but-not-completely-Ramsey example will satisfy it as well. Thus the distinction in conventions does not affect any of the conclusions below.

For a non-Ramsey example, declare $X\sim Y$ if their symmetric difference is finite, and choose a representative $R$ from each equivalence class using the axiom of choice. Setfor the representative of the class of $X$. Removing one point reverses this colour, because its symmetric difference with $R$ changes by exactly one point. Consequently every cone $[M{]}^{\omega }$ contains the oppositely coloured sets $M$ and $M\setminus \{\min M\}$. Either colour class of this finite-symmetric-difference parity colouring is not a Ramsey family in the homogeneous-cone sense. No definability is claimed for the representative selection.

The ordinary topology on infinite subsets, denoted $\tau$, has basic cylinders $[s]=\{X:s⊏X\}$, where $s$ is a finite stem of the increasing enumeration. Let $O$ be a $\tau$-open family and fix an arbitrary infinite ground set $A$. We give a complete fusion proof of a homogeneous cone inside $A$.

Use the convention $\max \varnothing =0$. For finite $s$ and an infinite reservoir $B$ above $\max s$, say that $B$ accepts $s$ if $[s,B]\subseteq O$, whereSay that $B$ rejects $s$ if no infinite subset of $B$ accepts $s$. There is always a refinement deciding $s$: choose an accepting subreservoir if one exists, and otherwise retain $B$, which rejects. Both decisions persist under further thinning.

A first fusion produces an infinite $H\subseteq A$ deciding every finite $s\subseteq H$ on its tail $H/s=\{h\in H:h>\max s\}$. First decide the empty stem. Choose the first point, and thin the remaining reservoir successively to decide every subset of the selected prefix. Repeat after choosing each new point. There are finitely many subsets to decide at each stage. The diagonal sequence of chosen points has the required property, since its tail after the largest point of any stem lies in the reservoir where that stem was decided. The empty-stem decision persists too. This is deciding all finite stems by fusion.

If $H$ accepts the empty stem, $[H{]}^{\omega }\subseteq O$ and we are done. Suppose it rejects. For a rejected finite $s\subseteq H$, only finitely many $h\in H/s$ can have $H/h$ accepting $s\cup \{h\}$. Otherwise let $D$ be their infinite set. Every infinite $X\subseteq D$ starts with some such $h$, and its remaining tail lies in $H/h$, so $s\cup X\in O$. Then $D$ accepts $s$, contradicting rejection. This proves finitely many accepting extensions of a rejected stem.

A second fusion constructs $K\subseteq H$ whose every finite subset is rejected. If a finite prefix of chosen points has been selected, all its subsets are rejected by induction. Avoid the union of their finitely many accepting successor sets when choosing the next point. The first fusion already decides every successor, so every new subset is rejected. An infinite reservoir remains after each finite exclusion.

If some $X\in [K{]}^{\omega }$ belonged to $O$, openness would give an initial segment $s⊏X$ with $[s]\subseteq O$. Its reservoir $H/s$ would then accept $s$, contradicting the second fusion. Therefore $[K{]}^{\omega }\cap O=\varnothing$. In either case we have a homogeneous cone inside the arbitrary ground set $A$. Every $\tau$-open family is a Ramsey set of infinite subsets, under both conventions.

A completely Ramsey set requires more: for every finite $s$ and infinite reservoir $A$ above $\max s$, there is an infinite $B\subseteq A$ such thatThe finite stem must be kept fixed. These $[s,A]$ are the basic neighbourhoods of the Ellentuck topology, also called the star topology.

Let $E$ be the even positive integers, and construct the finite-symmetric-difference parity colouring $\epsilon$ on $[E{]}^{\omega }$. DefineFor any infinite ground set $A$, the set $A\setminus \{1\}$ is still infinite and its whole cone misses $\mathcal{R}$. Thus $\mathcal{R}$ is a Ramsey set of infinite subsets even in the stronger convention. But inside $[\{1\},E]$, every infinite refinement $B\subseteq E$ gives the two sets $\{1\}\cup B$ and $\{1\}\cup (B\setminus \{\min B\})$ of opposite colours. Hence no such refinement is homogeneous. The family $\mathcal{R}$ is Ramsey but not completely Ramsey. The stem-supported Ramsey family need not be completely Ramsey construction explains this example: forgetting the stem $\{1\}$ is precisely what loses the obstruction.

Finally take $D=[E{]}^{\omega }$. It is nonempty and star-open, since it equals $[\varnothing ,E]$. It is $\tau$-closed: if an infinite set contains an odd integer, an initial segment witnessing that integer has a cylinder disjoint from $D$. Its $\tau$-interior is empty: after any finite stem of even integers, append an arbitrarily large odd integer and then infinitely many further points. ThusThis is the cone on an infinite coinfinite ground set. Choose its subset $Y=\{X\in [E{]}^{\omega }:\epsilon (X)=0\}$. Since ${\overline{Y}}^{\tau }\subseteq D$, $Y$ is also $\tau$-nowhere dense. Nevertheless every $[\varnothing ,B]$ with infinite $B\subseteq E$ contains both $B$ and $B\setminus \{\min B\}$, of opposite colours. Hence $Y$ is not completely Ramsey. We have proved explicitly that ordinarily nowhere-dense sets need not be completely Ramsey; the topology in the nowhere-dense hypothesis cannot be silently changed to the star topology.

The given relation impliesTake isogeny degrees and cancel the positive integer $\mathrm{deg}\pi$. Multiplicativity of isogeny degree givesAs above, $\mathrm{deg}(1-{\varphi }_i)=\#E_i({\mathbb{F}}_q)$ and $\mathrm{deg}{\varphi }_i=q$. The degree parallelogram law on $\mathrm{End}(E_2)$ now yieldsSubstitution provesEquivalently, their Frobenius traces are negatives of one another. The isogeny degree argument works over the algebraic closure, so it does not wrongly assume that $\pi$ is defined over ${\mathbb{F}}_q$; the stated quadratic-extension hypothesis is compatible with the anticommutation relation.

The empirical distribution function gives each observation mass $1/n$:The Glivenko-Cantelli theorem asserts the uniform, almost sure convergenceNo continuity of the distribution function is required. Here is a proof that also handles its jumps. Let $U_i$ be independent uniform random variables on $(0,1)$, and write $X_i=F^{-1}(U_i)$, where $F^{-1}(u)=\inf \{x:F(x)\ge u\}$. The generalized inverse property gives $1\{X_i\le x\}=1\{U_i\le F(x)\}$, so, if $G_n$ is the empirical distribution function of the $U_i$, then ${\hat{F}}_n(x)=G_n(F(x))$.

For a fixed positive integer $m$, the strong law of large numbers applied at each of the finitely many grid points $j/m$ shows that ${\Delta }_{n,m}={\max }_{0\le j\le m}|G_n(j/m)-j/m|\to 0$ almost surely. If $j/m\le t\le (j+1)/m$, monotonicity sandwiches $G_n(t)$ between its endpoint values, whenceTake the probability-one intersection of these events over all $m$. On that event the limit superior is at most $1/m$ for every $m$, hence is zero. Therefore ${\sup }_x|G_n(F(x))-F(x)|\to 0$. The coupled sequence has the same joint law as any independent sample from $F$, which proves the asserted almost sure conclusion for the original sample.

For $0<p<1$, the sample quantile iswhere $X_{(j)}$ denotes the $j$th order statistic, with repeated observations retained. Suppose the population median $m=F^{-1}(1/2)$ is unique and $F$ is continuously differentiable in a neighbourhood of $m$, with probability density function $f(m)>0$. ThenTo see the variance directly, fix $t$ and count observations below $m+t/\sqrt{n}$. This count has a binomial distribution with success probability $q_n=1/2+f(m)t/\sqrt{n}+o(n^{-1/2})$. The event that the sample median is below this point is that the count is at least $\lceil n/2\rceil$. The central limit theorem for this binomial triangular array gives a limiting probability $\Phi (2f(m)t)$, exactly the displayed normal distribution. Centering is necessary for stating the limiting distribution; subtracting the deterministic population median has no effect on the asymptotic variance requested in the comparisons.

For a nondegenerate distribution function $G$, membership $F\in D(G)$ in its maximum domain of attraction means that there exist $a_n>0,b_n\in \mathbb{R}$ such thatat every continuity point of $G$. A max-stable distribution satisfies: for each positive integer $k$, there exist $c_k>0,d_k\in \mathbb{R}$ such that $G(c_{k}x+d_k{)}^k=G(x)$. Thus a maximum of $k$ independent copies has the same probability distribution after a change of location and scale.

Suppose first that $F\in D(G)$. For fixed $k$, the distribution functions $H_n=F^{kn}$ have the two nondegenerate limitsThe given convergence-of-types result supplies constants $a>0,b$ such that $G(x{)}^k=G(ax+b)$. Equality extends from continuity points to all points by right continuity. Substitute $x=(z-b)/a$ to obtain $G((z-b)/a{)}^k=G(z)$, which proves max-stability with $c_k=1/a,d_k=-b/a$. Conversely, if $G$ is max-stable, take $F=G$ and its max-stability constants for $k=n$. Then $F(c_{n}x+d_n{)}^n=G(x)$ exactly for every $n$, proving $G\in D(G)$. Therefore the maximum domain of attraction is nonempty exactly when $G$ is max-stable.

For the particular logarithmic tail, complete the distribution function by taking $F(x)=0$ for $x\le x_0$. Let $a_n>1$ be the unique solution of $a_n\log a_n=n$, and take $b_n=0$. Equivalently $F(a_n)=1-1/n$, including $n=1$ with $a_1=x_0$. For each fixed $x>0$, eventually $a_{n}x>x_0$, andHere $a_n\to \infty$ follows directly from its defining equation. Since $\log (1-z)=-z+O(z^2)$ as $z\to 0$, it follows that $n\log F(a_{n}x)\to -1/x$. For $x\le 0$, the probability is zero. ConsequentlyThis is the shape-one Fréchet distribution, and the calculation proves the limit at every real $x$ without needing a regular variation theorem.