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An Lp norm measures the size of a measurable function by averaging a power of its absolute value with respect to a specified measure. For $1\le p<\infty$,when the integral is finite. The exponent $p$ controls how strongly large values influence the result. The Lp space identifies functions equal almost everywhere, so this becomes a norm on the corresponding equivalence classes. For $p=\infty$, use the essential supremum of $|f|$, as defined by the L-infinity norm. ^[ref].

For a two-coordinate vector $(3,4)$, treat the coordinates as a function on two inputs with counting measure. Then $\Vert (3,4){\Vert }_1=3+4=7$, $\Vert (3,4){\Vert }_2=\sqrt{9+16}=5$, and $\Vert (3,4){\Vert }_{\infty }=4$. The middle case is the Euclidean norm; the first adds the coordinate magnitudes, and the last takes their largest value. These are different answers to different questions about size, all satisfying the triangle inequality.

If instead each input has probability $1/2$, the same values give $\Vert f{\Vert }_1=3.5$ and $\Vert f{\Vert }_2=\sqrt{12.5}$, while $\Vert f{\Vert }_{\infty }=4$ remains unchanged. Thus the measure is not an optional label. On a probability space, the finite Lp norms are nondecreasing in $p$ whenever they are finite, by Hölder's inequality; the unnormalized two-coordinate example above decreases instead. Statements comparing different exponents must specify their measure convention.

Why are these sizes useful? If $e$ is an error function, its L1 norm measures aggregate absolute error, its L2 norm measures the square root of aggregate quadratic loss, and its L-infinity norm bounds error almost everywhere. Concentrating a large error on a set of small positive measure can give a modest finite Lp norm but a large L-infinity norm. The three-point example and diagram in L-infinity norm show this explicitly, including why a null set behaves differently from a rare event.

The odd part of a positive integer $n$ is the unique odd positive integer $d$ obtained by repeatedly dividing $n$ by two. The process terminates because the positive integers decrease. If $2^{r}d=2^{s}e$ with $d,e$ odd, then $r>s$ would make $e$ even and $s>r$ would make $d$ even; hence $r=s$ and $d=e$. Two positive integers have the same odd part exactly when their ratio is an integral power of two. The odd positive integers therefore give representatives for the resulting equivalence classes.

Both functions represent the same $L^1$ equivalence class. Fubini's theorem shows that the interval-mass function has derivative $Du$, so its difference from $u$ is constant. Continuity of a finite measure gives left continuity for $u_l$ and right continuity for $u_r$. Their opposite one-sided limits differ by the measure atom $Du(\{t\})$. There are only countably many nonzero measure atoms, since every set of measure atoms above a fixed magnitude threshold is finite.

For a group action of a group $G$ on a set $X$, the orbit of a group action of $x\in X$ is the subset of all positions reachable from $x$:Two points belong to the same group orbit exactly when one can be moved to the other by a group element. This is an equivalence relation: the identity element gives $x\sim x$, the inverse element of a transformation reverses a move, and a product combines two moves. Hence the group orbits form a partition of a set of $X$ into equivalence classes. A transitive group action on a nonempty set has just one group orbit. See Ackerman's algebra notes, lemma 0.5.1.3, for the orbit partition.

For the square and the eight-element dihedral group $D_4$ defined under group action, direct enumeration gives the following separate group orbits:

These counts depend on the chosen group action, not just the group. An orbit of a group action describes all configurations equivalent under the permitted transformations. For ordered pairs under the symmetric group, equality of the two entries is preserved, separating the two group orbits in equality-pattern orbits of ordered pairs; the orbit-stabilizer theorem computes their sizes when the symmetric group is $S_3$.

A partition of a set $X$ is a family of pairwise disjoint nonempty subsets whose union is $X$. Every equivalence relation defines a partition into equivalence classes, and conversely belonging to the same cell of a partition defines an equivalence relation. The nonempty colour classes of a finite colouring form a finite partition.

Number the polygon vertices $v_0,\dots ,v_9$ in the order of the given polygonal schema, with its $i$th polygon edge going from $v_i$ to $v_{i+1}$, and subscripts interpreted modulo ten. Equal labels identify equal arrow endpoints. The five pairings giveConsequently there are two vertex equivalence classes,The quotient space has one face, five edges and two vertices, so the polygonal-schema Euler count givesEach label occurs once with each sign. Hence every pairing reverses the direction induced by traversing the boundary of the polygon, so an orientation of a surface on its interior extends consistently across every glued edge. Thus $X$ is orientable. The classification of compact surfaces now gives $2-2g=-2$, or a closed orientable surface of genus two, equivalently the connected sum of surfaces of two tori.

Write ${\theta }_i$ for the internal angle at $v_i$, using the numbering in part (b), and use the hyperbolic plane with Gaussian curvature $-1$. The necessary and sufficient conditions are equal lengths for each pair of equally labelled polygon sides, and total angle $2\pi$ at each vertex equivalence class. Explicitly, the two angle conditions areNecessity follows because the identification must preserve arc length along each seam, and a smooth interior point of a Riemannian surface has total angle $2\pi$. For sufficiency, equally long hyperbolic geodesic segments can be identified by isometries. Across their interiors, two half-disc neighbourhoods join to a hyperbolic disc. At an identified vertex, the incident sectors form a cyclic disc with total angle $2\pi$, so no cone singularity of a surface metric remains. These local isometries therefore extend the Riemannian metric smoothly everywhere. This is the smooth hyperbolic polygon gluing criterion.

The ten internal angles sum to $4\pi$. The area of a hyperbolic geodesic polygon consequently givesEquivalently, the Gauss-Bonnet theorem gives $-\mathrm{Area}(X)=2\pi \chi (X)=-4\pi$.

Use Lebesgue measure restricted to the Lebesgue measurable set $E$. For $1\le p<\infty$, start with the measurable real-valued functions $f$ on $E$ satisfying ${\int }_E|f{|}^{p}dm<\infty$. Define an equivalence relation by $f\sim g$ when $f=g$ almost everywhere, and let $L^p(E)$ consist of their equivalence classes. Addition and scalar multiplication are defined using representatives. The Lp norm isFor $p=\infty$, use measurable functions bounded almost everywhere, apply the same identification, and define the L-infinity norm byThese expressions are independent of the representative. For finite $p$, zero norm is equivalent to $f=0$ almost everywhere: each set $\{|f|>1/n\}$ must have measure zero, and their union is $\{f\ne 0\}$. The same implication follows from the essential supremum for $p=\infty$. Absolute homogeneity is immediate; Minkowski inequality proves the triangle inequality and closure under addition. Thus these are normed vector spaces. Using complex-valued representatives gives their complex versions by the same construction.

We use the finite-structure convention of Section B. Express that $<$ is a strict linear order by irreflexivity, transitivity and comparability of distinct points. Express that $E$ is an equivalence relation by reflexivity, symmetry and transitivity. All these assertions use at most the three variable names $x,y,z$.

In a finite linearly ordered structure, every equivalence class has a unique least element. The two-variable formulaselects exactly those least elements. Apply part (b) to this formula, using $k=3$, to obtain ${\mathrm{Rep}}_n\in L^3$ saying that at least $n$ representatives exist. The required sentence isThis gives the least representatives of finite equivalence classes construction. Under the two structural axioms, counting representatives is exactly counting equivalence classes.

Finiteness is essential to the assertion as written in this section. If it were interpreted as applying to all infinite structures, the general claim would fail, not merely this choice of representatives. Partition the rational order into three dense classes, or into four dense classes, and let $E$ mean belonging to the same class. In the three-pebble version of the Ehrenfeucht-Fraïssé game, the duplicator can play indefinitely between these structures: after one pebble is moved, at most two old classes remain represented; a new point can be placed in the required interval in its corresponding old class, or in a class different from both. Density makes every required placement possible in either structure. Thus the structures agree on all $L^3$ sentences although one has three classes and the other four. The finite reading of the printed request is therefore necessary for $n=4$.

For nonzero complex numbers, $x\overline{y}=(x/y)|y{|}^2$. Since $|y{|}^2$ is positive real, the condition is equivalent to $x/y\in \mathbb{R}\setminus \{0\}$. This is an equivalence relation. The ratio $x/x=1$ gives reflexivity; the reciprocal of a nonzero real ratio shows that the relation is symmetric; and $(x/y)(y/z)=x/z$ gives transitivity. An equivalence class is a real line through zero with zero removed, or equivalently all nonzero complex numbers with the same argument modulo $\pi$.

This is an equivalence relation. The difference $x-x=0$ is an even integer, proving reflexivity. If $x-y=2k$ then $y-x=-2k$, proving that the relation is symmetric. If $x-y=2k$ and $y-z=2\ell$ then $x-z=2(k+\ell )$, proving transitivity. Its equivalence classes are the translates $x+2\mathbb{Z}$; two real numbers need not be integers themselves to belong to the same class.

Work in the frame $A$, writing $a\to x$ for Heyting implication. A nucleus on a frame is an inflationary, idempotent map preserving finite meets. The open nucleus and closed nucleus areTheir meet in the frame of nuclei on a frame is pointwise. The distributivity of a Heyting algebra and the defining adjunction for Heyting implication giveIf a nucleus on a frame $l$ bounds both, then $l(a)=1$ and $a\le l(0)$. Consequently $1=l(a)\le l(l(0))=l(0)$, so $l$ is the top nucleus. Thus $o(a)$ and $c(a)$ are complements in $N(A)$.

For any nucleus on a frame $j$, put $t_a=o(a)\wedge c(j(a))$. At an arbitrary $x$,since $j(a)\wedge (a\to x)\le j(a)\wedge j(a\to x)=j(a\wedge (a\to x))\le j(x)$. Moreover $t_a(a)=j(a)$. ThereforeIndeed the join is at most $j$, and its value at each $a$ is at least $j(a)$. This proves generation by the open nuclei and closed nuclei, without assuming that joins of nuclei on a frame are computed pointwise.

For the given frame homomorphism $h$, restrict the displayed join to $a=h(b)$ and denote the resulting nucleus on a frame by $r(j)$. The same inequalities show $r(j)\le j$ and $r(j)(h(b))=j(h(b))$. If $k$ has the same values on $h(B)$, its restricted formula is identical, so $r(j)\le k$. Hence $r(j)$ is the least representative of the equivalence class. We will also use the consequence

The least representatives form a subframe. For a family $j_i=r(j_i)$, let $J={\bigvee }_i{j}_i$ in $N(A)$. Restricted monotonicity gives $j_i\le r(J)$ for every $i$, hence $J=r(J)$. For two least representatives, distribution in $N(A)$ expresses their meet as the join, over $b,d\in B$, ofHere $o(hb)\wedge o(hd)=o(h(b\vee d))$ and $c(u)\wedge c(v)=c(u\wedge v)$. Sinceeach summand is below $r(j\wedge k)$. The opposite inequality always holds, proving closure under binary meets. The top is $c(1)$ and is a least representative. Also $r(c(a))=c(a)$, because the summand at $b=0$ is $c(a)$; and $r(o(hd))=o(hd)$, because the summand at $b=d$ is $o(hd)$. It follows from the restricted normal form that $N_B(A)$ is exactly the subframe generated by $c(A)$ and $o(h(B))$.

The map $c:A\to N_B(A)$ preserves finite meets by the formula above. A nucleus on a frame $l$ bounds every $c(a_i)$ exactly when $a_i\le l(0)$ for every $i$, equivalently ${\bigvee }_i{a}_i\le l(0)$. The least such nucleus is $c({\bigvee }_i{a}_i)$, so $c$ preserves arbitrary joins too, including $0$. Evaluation at $0$ recovers $a$, so this frame homomorphism is injective.

For the pushout, we prove the universal property explicitly. Let $u:A\to C$ be a frame homomorphism for which every $u(hb)$ has a complement $d_b$ in $C$. Define, for $j\in N_B(A)$,To prove preservation of arbitrary joins, let $u_{\ast }$ be the right adjoint of $u$ and set $k_z(a)=u_{\ast }(z\vee u(a))$ for $z\in C$. This is a nucleus on a frame: inflationarity follows from the adjunction, finite-meet preservation follows from distribution and meet preservation by $u_{\ast }$, and $u{u}_{\ast }\le 1$ gives $k_z{k}_z\le k_z$; inflationarity gives the reverse inequality. Complementation and the adjunction now giveThe last equivalence uses leastness of $j$. Thus $F$ is a left adjoint, so preserves arbitrary joins. For binary meets, expand $F(j)\wedge F(k)$. A term indexed by $b,d$ isThis proves $F(j)\wedge F(k)\le F(j\wedge k)$; monotonicity proves the reverse inequality. The term at $b=0$ proves $F(1)=1$. Directly,For the second identity the term at $b=d$ gives $d_d$, while every term is disjoint from $u(hd)$, because $hd\wedge (hd\to hb)\le hb$.

Apply this construction first with $A=B$, $h=1_B$, target $N_B(A)$, $u=c_{A}h$ and complements $d_b=o_A(hb)$. It gives a frame homomorphism $v_0:N(B)\to N_B(A)$ satisfying $v_0{c}_B=c_{A}h$ and $v_0{o}_B=o_{A}h$. Given any compatible pair $u:A\to C$, $v:N(B)\to C$, take $d_b=v(o_B(b))$; compatibility supplies the required complements. The constructed $F$ satisfies $F{c}_A=u$ and $F{v}_0=v$, since the latter equality holds on the open and closed generators of $N(B)$. Uniqueness follows because $c(A)$ and $o(h(B))$ generate $N_B(A)$. Hence the square with edges $h,c_B,c_A,v_0$ is a pushout in $\mathbf{Frm}$.

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.

Identify the space of infinite subsets of the natural numbers with their increasing enumerations, with the ordinary topology on infinite subsets inherited from the discrete product ${\omega }^{\omega }$. This is the sequence convention in the Open Ramsey theorem. For a general infinite ground set, first choose a countably infinite subset and use its enumeration. The theorem cannot mean unrestricted sequences with repetitions: the clopen partition according as the first two entries agree or differ has both colours over every infinite ground set.

For a finite finite stem $s$ and an infinite tail $B$ above $\max s$, writewhere the enumeration of $s$ precedes the tail. Let $O$ be open. The tail $B$ accepts $s$ if $[s,B]\subseteq O$; it rejects $s$ if no infinite subtail accepts it. Both properties persist on infinite subtails, and any stem can be decided by thinning: choose an accepting subtail if one exists; otherwise the current tail already rejects it.

Use deciding all finite stems by fusion. At stage $n$, choose a new element $b_n$ from the current tail, then thin the remaining tail to decide each of the finitely many subsets of $\{b_0,\dots ,b_n\}$. First decide the empty stem as well. The final selected set $B$ decides every finite stem from $B$, because all subsequent selections lie in the tail that decided that stem. If $B$ accepts the empty stem, $[B{]}^{\omega }\subseteq O$ and we are done.

Otherwise the empty stem is rejected. If a stem $s\subset B$ is rejected, only finitely many $a\in B$ above $\max s$ can have the tail $B/a$ accepting $s\cup \{a\}$. Indeed, if infinitely many such $a$ formed $D$, then every $X\in [D{]}^{\omega }$ starts with one of them and has its remaining tail in $B/a$. It follows that $[s,D]\subseteq O$, contradicting rejection of $s$. All other one-point extensions are rejected, since the first fusion has already decided them.

Build an infinite $C\subseteq B$ recursively. Maintain that every subset of the finite selected prefix is rejected. To select its next element, avoid the finite union of the finitely many accepting-extension obstructions for these rejected stems. Such an element exists, and all the new stems remain rejected. Hence every finite stem from $C$ is rejected. If some $X\in [C{]}^{\omega }$ were in $O$, openness would supply a finite initial segment $s$ of $X$ whose entire ordinary cylinder is contained in $O$. The tail $C/\max s$ would accept $s$, a contradiction. ThusIn particular every open payoff has an infinite homogeneous set. A finite partition into open colour classes can be handled by successive application of this two-colour assertion.

For an arbitrary partition use finite-symmetric-difference parity colouring. On the infinite subsets of the ground set, declare $X\sim Y$ when $X△Y$ is finite, and choose a representative $R$ from each equivalence class. Colour $X$ byFor any infinite $H$, removing one member gives an infinite $H^{\prime }$ in the same class and reverses the parity. Thus $H$ and $H^{\prime }$ have opposite colours, so neither colour contains all infinite subsets of any infinite ground set. This two-colouring has no infinite homogeneous set. Representative selection uses choice; the construction is deliberately unrestricted by topological regularity.

An ultraproduct converts coordinatewise finite information into a single first-order structure. Let $(M_i{)}_{i\in I}$ be nonempty structures in a common first-order language, and let $U$ be an ultrafilter on $I$. In ${\prod }_i{M}_i$, setThe ultraproduct ${\prod }_i{M}_i/U$ has these equivalence classes as its elements. Functions are interpreted coordinatewise; a relation holds of classes exactly when it holds on a $U$-large set of indices. Intersecting finitely many large agreement sets shows that these interpretations are independent of representatives.

The Łoś theorem, which explains the construction's usefulness, saysHere is the proof. Induction on terms establishes the coordinatewise interpretation of atomic formulas. Finite intersections handle conjunction, and the ultrafilter property handles negation, since precisely one of a set and its complement is large. For an existential quantification, a witness class gives coordinate witnesses on a large set. Conversely, when the coordinate existential truth set is large, select a witness in each of those structures and a default element elsewhere. Its class is a witness in the ultraproduct. This uses choice. Structural induction on formulas now proves the theorem, with universal quantification obtained by negation.

Compactness. If every finite subset of a first-order theory $T$ has a model, take $I$ to be the finite subsets $\Gamma \subseteq T$ and choose $M_{\Gamma }\models \Gamma$. The cones $\{\Gamma :\Delta \subseteq \Gamma \}$ for finite $\Delta$ have the finite intersection property. Extend their filter to an ultrafilter by the ultrafilter lemma. For each $\phi \in T$, its cone is large, so the Łoś theorem makes the ultraproduct satisfy $\phi$. This proves the compactness theorem, including theories of arbitrary set size.

Elementary extensions and nonstandard elements. An ultrapower uses a fixed structure at every coordinate. The diagonal map sending $a$ to the class of the constant function $a$ is an elementary embedding by the Łoś theorem. For a nonprincipal ultrafilter on $\omega$, the class of $i\mapsto i$ in an ultrapower of the natural numbers exceeds every standard integer, because each corresponding tail is large. In an ultrapower of the real ordered field, the reciprocal of this element is positive and smaller than every positive standard reciprocal integer: an infinitesimal. These examples implement ordinary finite statements in a larger structure without claiming that every external subset or second-order assertion transfers.

Saturation. Such an ultraproduct over $\omega$ is countably saturated. To see the actual diagonal argument, list a finitely satisfiable type ${\phi }_0(x),{\phi }_1(x),\dots$ with parameters in the ultraproduct and fix representatives for those parameters. For each $n$, the set $E_n$ of coordinates admitting a simultaneous witness to its first $n$ formulas belongs to $U$ by the Łoś theorem. Arrange that the $E_n$ decrease and put $D_n=E_n\cap \{i:i\ge n\}$. The $D_n$ decrease, belong to $U$, and have empty intersection. At coordinate $i$, choose a witness to the first $m(i)$ formulas, where $m(i)$ is the largest $n$ with $i\in D_n$, or choose a default element if there is no such $n$. For each fixed $n$, $m(i)\ge n$ on the large set $D_n$. Therefore the resulting class realizes every formula. This proves countable saturation of an ultraproduct over omega; it does not assert saturation at all larger cardinalities.

Algebraic limits. Take finite fields ${\mathbb{F}}_p$ indexed by the primes and a nonprincipal ultrafilter. Their ultraproduct is a field of characteristic zero: for every fixed positive integer $n$, only finitely many primes divide $n$, so $n\cdot 1\ne 0$ on a large set. Every first-order sentence true in all finite fields holds in this infinite field, and more generally so does any sentence true for all but finitely many of the factors. This is a way to create an infinite field retaining uniform first-order information from finite fields. It does not make the field algebraically closed: the first-order sentence that every element is a square already fails in infinitely many of the factors.

Large cardinals. A countably complete ultrafilter gives a well-founded membership ultrapower of the universe. If it had an infinite descending chain represented by $f_n$, each setwould be large. Countable completeness makes ${\bigcap }_n{A}_n$ nonempty, giving an actual infinite membership descent at any index in the intersection, contrary to foundation. The Mostowski collapse therefore gives a transitive target and an ultrapower embedding $j:V\to M$.

For a measure on a measurable cardinal $\kappa$, every function $\kappa \to \alpha$ with $\alpha <\kappa$ is constant on a large set: otherwise intersect the fewer than $\kappa$ complements of all its fibres. Consequently $j$ fixes the ordinals below $\kappa$. The class of the identity function is below $j(\kappa )$ and above every constant class below $\kappa$, since all tails are large. Thus $j(\kappa )>\kappa$ and its critical point is $\kappa$. Fine ultrafilters on $P_{\kappa }(\lambda )$ that are normal ultrafilters on small subsets and kappa-complete yield the stronger embeddings witnessing supercompact cardinals.

These applications share a single principle: an ultraproduct preserves precisely the first-order information seen by its ultrafilter. Principal ultrafilters merely select a factor; nonprincipal ones can produce new sizes, elements and saturation. Existence of the needed ultrafilters and selection of coordinate witnesses are set-theoretic assumptions, and changing the ultrafilter may change the resulting model.

An equivalence relation is reflexive, symmetric and transitive: for all $a,b,c\in S$, one has $aRa$; $aRb$ implies $bRa$; and $aRb$, $bRc$ imply $aRc$. Define the equivalence class $[a]=\{b\in S:aRb\}$. Reflexivity puts $a$ in its class, so these classes cover $S$ and are nonempty. If $[a]$ and $[b]$ share $c$, symmetry and transitivity give $aRb$; the same properties show that every element of either class is in the other. Thus two classes are equal or disjoint. The distinct equivalence classes form a partition of the set.

For the relation defined by a subgroup $H\le G$, $a^{-1}a=1\in H$ proves reflexivity. If $a^{-1}b\in H$, its inverse $b^{-1}a$ belongs to $H$, proving symmetry. If also $b^{-1}c\in H$, their product is $a^{-1}c\in H$, proving transitivity. Finally,These are exactly the left cosets. The map $H\to gH$, $h\mapsto gh$, is a bijection, with inverse $x\mapsto g^{-1}x$. If $G$ is finite and has $r$ such cosets, their disjoint union gives $|G|=r|H|$. This proves the Lagrange theorem:For $g\in G$, finitely many possible powers imply $g^a=g^b$ for some $a<b$, so a positive power is the identity element. If its least positive exponent is $n$, the powers $1,g,\dots ,g^{n-1}$ are distinct: equality of two would produce a smaller positive exponent. They are closed under products and inverses by reducing exponents modulo $n$, hence form the cyclic subgroup $⟨g⟩$ of cardinality $n$. The Lagrange theorem gives $n\mid |G|$, and consequentlyNow take a positive modulus $m$. On residue classes modulo $m$, consider the classes with representatives coprime to $m$. The following integer facts are used: multiplication is associative; reducing representatives respects products; a prime dividing a product divides a factor; and the Bezout identity gives $ra+sm=1$ when $\gcd (a,m)=1$. If $a,b$ are coprime to $m$, no prime divisor of $m$ divides $ab$, so their product class is again in the set. The class of $1$ is its identity element. The Bézout equation gives $ra\equiv 1(\mathrm{mod}m)$, and $r$ is also coprime to $m$ because any common divisor of $r,m$ would divide $1$. Thus every class has an inverse, and these classes form the group of units modulo an integer.

Representing classes in $\{1,\dots ,m\}$ gives precisely the group in the question; for $m=1$ this is the one-element group. Its cardinality is the Euler totient function $\phi (m)$. Applying the finite-group exponent result yields the Fermat-Euler theorem:For $m=1$ the congruence is automatic.

The Brauer group of a field consists of equivalence classes of central simple algebras, with addition induced by tensor product, zero represented by matrix algebras, and inverse represented by the opposite algebra. Localization sends a global algebra to its scalar extensions at the completions. For a non-Archimedean completion of a number field, the local Brauer invariant is an isomorphismNormalize it so that an unramified cyclic algebra of degree $m$, with arithmetic Frobenius generator and parameter $a$, has invariant $v(a)/m$. For a real completion, $\mathrm{Br}(\mathbb{R})=\mathbb{Z}/2$ with the Hamilton quaternion algebra mapped to $1/2$; for a complex completion the Brauer group is zero.

The global computation is the exact sequence in the Albert-Brauer-Hasse-Noether theorem:Thus a global Brauer class is determined uniquely by a finite-support family of local invariants. Such a family occurs if and only if its sum is zero; at real places the only allowed entries are $0,1/2$, and at complex places the entry is zero. Injectivity is a local-global splitting principle; the middle exactness is the reciprocity constraint; surjectivity of the sum follows already from any one finite local summand. These are separate assertions, not merely different descriptions of the sum formula.

For an explicit group description, choose one finite place $v_0$. Assign arbitrary finite-support invariants at all other finite places and arbitrary allowed invariants at the real places, then set the invariant at $v_0$ equal to minus their sum. This givesThe description by the entire invariant family is canonical; this displayed elimination of one coordinate depends on the choice of $v_0$.

The connection with Artin reciprocity is especially transparent for cyclic algebras. Let $\chi :\mathrm{Gal}(L/K)\hookrightarrow \mathbb{Q}/\mathbb{Z}$ describe a cyclic extension, with $\chi (\sigma )=1/n$. Write $(\chi ,a)$ for the class of $(L/K,\sigma ,a)$. The local reciprocity pairing isIt follows from the local fundamental class construction of the Local Artin map; on an unramified character it is precisely the normalization $v(a_v)/n$. The group kernel in the parameter is the local norm group, and restriction/corestriction give the norm compatibility of this pairing.

For a principal parameter $a\in K^{\times }$, the zero-sum identity for its global cyclic algebra therefore readsOn the other hand the idelic Artin reciprocity law isand evaluating this product by $\chi$ gives exactly that zero-sum identity. Conversely, testing every cyclic character of a finite abelian Galois group separates its elements, so the zero-sum identities for cyclic algebras imply this principal-idele reciprocity law for every finite abelian extension. At almost all places a unit and an unramified character pair trivially, so the sums and products are finite.

The full idelic statement identifies the group kernel as $K^{\times }{N}_{L/K}{J}_L$ and induces $J_K/(K^{\times }{N}_{L/K}{J}_L)\cong \mathrm{Gal}(L/K)$. It explains why the local norm-residue pairings must glue globally. The global Brauer theorem supplies, in addition, the injectivity and realization statements needed to compute all algebra classes; the principal-product identity alone would not prove those statements. For cyclic algebras its injectivity yields the Hasse norm theorem, as in Question 3. In degree two, an invariant is either zero or $1/2$, so the zero-sum law becomes the Hilbert reciprocity law ${\prod }_v(a,b{)}_v=1$. For example, a quaternion algebra over $\mathbb{Q}$ can be ramified at precisely a prescribed finite set of places, including possibly the real place, if and only if that set has even cardinality; the invariants uniquely determine its Brauer class. This illustrates how arithmetic reciprocity becomes a concrete constraint on global algebra structure.

Use the usual positive-definite convention for binary quadratic forms with negative discriminant. Reduction by integral changes of variables of determinant one produces a form $a{x}^2+bxy+c{y}^2$ with $|b|\le a\le c$, and with $b\ge 0$ on the boundary. For completeness, replacing $x$ by $x+ny$ makes $|b|\le a$; if then $c<a$, interchange the variables, reversing one sign to keep determinant one. The new positive leading coefficient is smaller. Repeating terminates and gives a reduced form.

For discriminant $-7$, $4ac-b^2=7$ and the reduction inequalities give $3a^2\le 7$. Thus $a=1$, $b$ is odd and $|b|\le 1$, and $c=2$. The boundary convention chooses $b=1$. Hence there is exactly one positive-definite equivalence class, represented by $x^2+xy+2y^2$.

For an odd prime $p\ne 7$ represented by this form, the identity $4p=(2x+y{)}^2+7y^2$ shows that $-7$ is a square modulo $p$. Here $p\nmid y$, because otherwise $p\mid x$ and $p^2$ would divide the represented number. Conversely, if $-7$ is a square modulo $p$, choose odd $b$ with $b^2\equiv -7(\mathrm{mod}p)$. Then $(p,b,(b^2+7)/(4p))$ is an integral positive-definite form of discriminant $-7$. By the preceding reduction it is equivalent to the given form, so its value $p$ at $(1,0)$ is represented by that form. Quadratic reciprocity gives $(-7/p)=(p/7)$, whose value is one precisely for $p\equiv 1,2,4(\mathrm{mod}7)$. The exceptional primes are represented: $2=Q(0,1)$ and $7=Q(1,-2)$. ThereforeWithout the positive-definite convention, the first printed assertion is false: $-x^2-xy-2y^2$ also has discriminant $-7$ but cannot be equivalent to a positive-definite form, because an invertible change of variables preserves its sign.

Write $[M{]}^{\omega }$ for the set of all infinite subsets of $M$. A family $Y\subseteq [\mathbb{N}{]}^{\omega }$ is a Ramsey family in the homogeneous-cone sense if there is an infinite $M$ such that either $[M{]}^{\omega }\subseteq Y$ or $[M{]}^{\omega }\cap Y=\varnothing$. Thus membership in $Y$ is constant on an infinite homogeneous cone. We will prove the stronger version for open families: such a cone exists inside every prescribed infinite ground set, as in the definition of a Ramsey set of infinite subsets.

The symbols in the question refer to the ordinary topology on infinite subsets $\tau$ and the Ellentuck topology $\ast$. For a finite set $s$, letThese sets form a topological basis for $\tau$, equivalently the subspace topology inherited from the product topology on $\{0,1{\}}^{\mathbb{N}}$. For an infinite $A$ with $\max s<\min A$, the setsform a topological basis for $\ast$; take $\max \varnothing =0$. In particular $[s]=[s,\{n:n>\max s\}]$, so $\ast$ is finer than $\tau$.

For a non-Ramsey example use the finite-symmetric-difference parity colouring. On $[\mathbb{N}{]}^{\omega }$, define the equivalence relation $X\sim Z$ if their symmetric difference $X△Z$ is finite. By the axiom of choice, select a representative $R$ in every equivalence class. Give $X$ the colourwhere $R_X$ is its class representative. Deleting one point stays in the same class and toggles the parity. Thus for every infinite $M$, the sets $M$ and $M\setminus \{\min M\}$ are both in $[M{]}^{\omega }$ but have opposite colours. Either colour class, regarded as a family $Y$, is therefore not Ramsey.

Now let $U$ be $\tau$-open. We prove that for every infinite $A_0$ there is an infinite $C\subseteq A_0$ with $[C{]}^{\omega }\subseteq U$ or $[C{]}^{\omega }\cap U=\varnothing$. The proof uses acceptance and rejection of finite stems, which we define explicitly. For a finite stem $s$ and an infinite tail $A$ above $\max s$, say that $A$ accepts $s$ if $[s,A]\subseteq U$, and rejects $s$ if no infinite subset of $A$ accepts $s$. Both properties persist on taking infinite subsets. Every tail has an infinite subset that decides $s$: take an accepting subset if one exists, and otherwise it already rejects $s$.

First thin $A_0$ to decide the empty stem. Then construct $b_1<b_2<\cdots$ and nested infinite tails. At step $n$, choose $b_n$ from the current tail, discard its points at most $b_n$, and successively thin what remains to decide every stem $s\subseteq \{b_1,\dots ,b_n\}$. There are only finitely many such stems, so an infinite tail remains. Let $B=\{b_1,b_2,\dots \}$ and write $B/s=\{b\in B:b>\max s\}$. For every finite $s\subseteq B$, the tail $B/s$ decides $s$, since it lies in the tail obtained at the stage when the largest point of $s$ was selected. The empty stem is also decided. This is deciding all finite stems by fusion.

If $B$ accepts the empty stem, then $[B{]}^{\omega }\subseteq U$, as required. Otherwise $B$ rejects it. We need the finitely many accepting extensions of a rejected stem observation. Suppose a tail $B/s$ rejects $s$. Only finitely many $a\in B/s$ can have $B/a$ accepting $s\cup \{a\}$. Indeed, if infinitely many such $a$ formed a set $D$, every member of $[s,D]$ would have some first new point $a$ and all later points in $B/a$, hence would lie in $U$. This would make $D$ an accepting subset of $B/s$, contradicting rejection.

Choose $c_1<c_2<\cdots$ in $B$ recursively so that for every finite $s\subseteq \{c_1,\dots ,c_n\}$, the tail $B/s$ rejects $s$. The empty stem satisfies this initially. At the next step, each of the finitely many previous stems has only finitely many accepting one-point extensions by the observation. Choose $c_{n+1}$ larger than all previous points and outside all those finite exceptional sets. The first fusion ensured that the tail for every new stem decides it; since it does not accept it, it rejects it. This maintains the recursion.

Put $C=\{c_1,c_2,\dots \}$. If some $X\in [C{]}^{\omega }$ belonged to $U$, ordinary openness would give a finite initial segment $s$ of $X$ with $[s]\subseteq U$. In particular $[s,C/s]\subseteq U$. But $C/s$ is an infinite subset of $B/s$, which rejects $s$, a contradiction. Hence $[C{]}^{\omega }\cap U=\varnothing$. We have proved

The equivalence class of $x\in X$ is $[x]=\{y\in X:y\sim x\}$. By the reflexivity of the equivalence relation, $x\in [x]$, so every equivalence class is nonempty and their union is $X$.

Suppose $[x]\cap [z]\ne \varnothing$, and choose $w$ in this intersection. We have $w\sim x$ and $w\sim z$. Symmetry and transitivity give $x\sim z$. If $y\in [x]$, then $y\sim x\sim z$, so $y\in [z]$; interchanging $x,z$ proves the reverse inclusion. Thus two equivalence classes are either identical or disjoint. The distinct equivalence classes therefore form a set partition of $X$. This also covers $X=\varnothing$, whose family of equivalence classes is empty.