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A colour class of a finite colouring $c:X\to \{1,\dots ,r\}$ is a fibre $c^{-1}(i)$. Each colour class is a monochromatic set, and these fibres give a finite set partition of $X$ after empty fibres are omitted.

Fix positive integers $r,k,m$ with $m\ge r$. There is an integer $n\ge m$, depending only on $r,k,m$, such that every finite coloringhas a subset $H\subseteq [n]$ of size $m$ for which the restriction of a function $c$ to $\left(\genfrac{}{}{0ex}{}{H}{r}\right)$ is a constant function. Here $[n]=\{1,\dots ,n\}$, and $\left(\genfrac{}{}{0ex}{}{S}{r}\right)$ denotes the collection of all $r$-element subsets of $S$, rather than its binomial coefficient cardinality. The notation $[n]⟶(m{)}_k^r$ abbreviates this property. Thus $H$ is a homogeneous set for a colouring; when $r>1$, the monochromatic set consists of its $r$-element subsets, not its individual elements. This is the finite assertion in Ramsey's original Theorem B, p. 267.

For $r=1$, this is the pigeonhole principle, with sharp threshold $n=k(m-1)+1$. For instance, every red-blue coloring of five objects has three of one color; four objects can split two and two. For $r=2$, $k=2$ and $m=3$, the sharp threshold is six, as shown by the Ramsey number for triangles. Coloring pairs makes this a problem about edges of a complete graph. For larger $r$, the analogous viewpoint uses a hypergraph.

The finite Ramsey theorem lets an existence argument discard most of a large finite set while retaining a prescribed amount of uniform structure. Its threshold applies to every coloring; one does not need an assumption that colors were assigned randomly. A diagonal compactness argument deduces finite existence from Ramsey's theorem, but does not by itself identify the smallest threshold. For pairs, a Ramsey number makes that optimization explicit.

A subset $A\subseteq X$ is monochromatic for a finite coloring $c:X\to C$ if there is a color $i\in C$ such thatEquivalently, the restriction of a function $c\upharpoonright A$ is a constant function, or $A$ lies in one colour class $c^{-1}(i)$. The word describes agreement of colors; it does not require $A$ to equal an entire colour class.

For example, color the integers by integer parity, using red for even and blue for odd. The set $\{2,4,8\}$ is monochromatic, whereas $\{2,3\}$ is not. If the colored objects are edges, a monochromatic triangle in a graph means that its three edges form a monochromatic set. In a coloring of $r$-element subsets of a set $S$, a homogeneous set for a colouring $H\subseteq S$ means that the collection of all $r$-element subsets of $H$ is monochromatic. For pairs, this is a complete graph on $H$ whose edges all have one color, rather than a coloring of the vertices in $H$.

The partition relation $\lambda \to (\mu {)}_{\nu }^r$ says that every colouring of the $r$-element subsets of a set of cardinality $\lambda$ into $\nu$ colours has a homogeneous set of cardinality $\mu$. Define the beth numbers over an infinite cardinal $\kappa$ by ${\beth }_0(\kappa )=\kappa$ and ${\beth }_{n+1}(\kappa )=2^{{\beth }_n(\kappa )}$. The Erdős-Rado theorem for finite arities gives the precise boundThus any fixed finite arity and any fixed infinite number of colours admit arbitrarily large infinite monochromatic sets once the starting cardinal is sufficiently large. For pairs, the especially useful case is $(2^{\kappa }{)}^{+}\to ({\kappa }^{+}{)}_{\kappa }^2$. The successor in this bound matters: a binary first-difference colouring on a suitably ordered set of binary sequences supplies the familiar obstruction at $2^{\kappa }$. Also, a theorem for every fixed finite arity is not a theorem homogenizing all arities simultaneously.

The mechanism is to build an end-homogeneous routing tree: each new vertex is routed according to its colours to earlier vertices on its branch. A long branch is end-homogeneous, meaning the colour from a fixed earlier vertex is independent of the later vertex. There are at most ${\kappa }^{\alpha }\le 2^{\kappa }$ routes of length $\alpha <{\kappa }^{+}$, and the whole collection of short routes has size at most $2^{\kappa }$. Inserting $(2^{\kappa }{)}^{+}$ vertices therefore forces a branch of length ${\kappa }^{+}$. One colour occurs at ${\kappa }^{+}$ earlier vertices on that branch, giving the pair theorem. Iterating the higher-arity version of this routing construction explains the successive powers in the Erdős-Rado theorem for finite arities.

A weakly compact cardinal is an uncountable strongly inaccessible cardinal $\kappa$ withEquivalent formulations include the tree property at the inaccessible cardinal and compactness for theories of size at most $\kappa$ in the infinitary language $L_{\kappa ,\kappa }$: if every subtheory of size $<\kappa$ has a first-order model, so does the theory.

A measurable cardinal is an uncountable $\kappa$ carrying a nonprincipal ultrafilter closed under intersections of fewer than $\kappa$ members. Such an ultrafilter contains no set of size $<\kappa$, since all singletons have complements in it and it is kappa-complete. Consequently a cofinal sequence of length $<\kappa$ would partition $\kappa$ into too few small pieces, so $\kappa$ is regular cardinal. If $\lambda <\kappa$ and $\kappa$ distinct subsets of $\lambda$ existed, choose, for each coordinate of $\lambda$, the bit selected by the ultrafilter. Kappa-completeness would make all those bits simultaneously constant on a large set of indices; only one distinct subset could have that bit pattern, a contradiction. Hence $2^{\lambda }<\kappa$: $\kappa$ is also a strong limit cardinal.

A supercompact cardinal $\kappa$ has, for every $\lambda \ge \kappa$, a fine ultrafilter that is a normal ultrafilter on small subsets and is kappa-complete onFine means $\{x:\alpha \in x\}$ is large for every $\alpha <\lambda$. Normal means that a function $f$ with $f(x)\in x$ on a large set is constant on some large set. Equivalently, for every such $\lambda$ there is an elementary embedding $j:V\to M$ with critical point $\kappa$, $j(\kappa )>\lambda$, and $M$ closed under $\lambda$-sequences from the universe.

These notions form the implication chain supercompact $\Rightarrow$ measurable $\Rightarrow$ weakly compact. For the first implication, an embedding as above gives the measure $U=\{X\subseteq \kappa :\kappa \in j(X)\}$. For the second, use a normal ultrafilter on a cardinal obtained by normalizing its measure. Indeed the ultrapower embedding gives the measure $U=\{X\subseteq \kappa :\kappa \in j(X)\}$; for a regressive $f$, $j(f)(\kappa )<\kappa$ is a fixed ordinal, yielding the required constant large fibre. Given a pair-colouring, select for each $\alpha$ the measure-one colour on the tail above $\alpha$. One selected colour occurs on a measure-one set of $\alpha$'s. Normality makes the diagonal intersection of their selected tails measure one; any two members of that intersection have the selected colour. This gives the required partition relation, and the preceding regularity and strong-limit argument gives inaccessibility.

A partial computable function is the function computed by a finite program, with an undefined value when its computation never terminates. A total computable function terminates on every input. Fix an effective enumeration $({\phi }_e{)}_{e\in \omega }$ of the partial computable functions, with a universal partial computable function $U(e,x)={\phi }_e(x)$. Composition, primitive recursion and unbounded minimization give a machine-independent description of this class. Unbounded minimization is allowed to diverge. The S-m-n theorem permits parameters in a program to be specialized effectively: an algorithm described uniformly in $e$ has an index obtained computably from $e$.

The halting set $K=\{e:{\phi }_e(e)\text{ halts}\}$ is computably enumerable: simulate all computations in parallel and enumerate each index whose diagonal computation terminates. It is not a computable set. Otherwise the program which terminates exactly when the proposed decision procedure says its own diagonal computation does not terminate gives a contradiction. There is also no effective enumeration consisting exactly of all total computable functions: from such an enumeration $(f_n)$ the total computable function $g(n)=f_n(n)+1$ differs from every listed function. Thus the effective enumeration of partial programs cannot be replaced by a decidable catalogue of total ones.

For the Rice theorem, let $C$ be a nontrivial property of partial computable functions, depending on the computed function rather than the program text. First suppose the nowhere-defined function is not in $C$, and choose a program computing some $g\in C$. Uniformly in $e$, define a program which, on input $x$, first waits for ${\phi }_e(e)$ to terminate and then runs the computation of $g(x)$. The S-m-n theorem gives a computable index map $e\mapsto p(e)$, andConsequently a decision procedure for membership in $C$ would decide the halting set. If the nowhere-defined function belongs to $C$, apply the same argument to its complement. Every nontrivial extensional property of partial computable functions has an undecidable index set. Syntactic properties of program descriptions are outside this assertion, and the theorem concerns unrestricted program indices, not merely an assumed list of terminating programs.

Here is an explicit Jockusch triple coloring computing the halting set. Choose computable increasing finite stages $K_s$ with union $K$. For $x<y<z$, putThis is a computable colouring. An infinite monochromatic set cannot have color $1$: fix its first element $x$ and choose two later elements beyond the stage at which $K_s\cap x$ stabilizes. Their triple has color $0$.

Suppose $H$ is infinite and homogeneous of color $0$. To decide whether $n\in K$ using $H$, find $x\in H$ with $x>n$, and then $y\in H$ with $y>x$. For every $z\in H$ above $y$, homogeneity gives $K_y\cap x=K_z\cap x$. Such $z$ are unbounded, so $K_y\cap x=K\cap x$. Testing $n\in K_y$ therefore decides $n\in K$. Hence every homogeneous set computes the halting set, and there is no infinite recursive monochromatic set. The infinity qualification is essential: finite homogeneous sets are recursive.

Van der Waerden theorem says that for every pair of positive integers $r,k$ there is $W(r,k)$ such that every finite colouring of $[W(r,k)]=\{1,\dots ,W(r,k)\}$ using at most $r$ colours contains a monochromatic arithmetic progression of length $k$ and positive common difference. Consequently every finite colouring of $\mathbb{N}$ contains arbitrarily long monochromatic arithmetic progressions.

Here is a colour-focusing proof of Van der Waerden theorem. Induct on $k$, simultaneously for all $r$. For $k=1$, take $W(r,1)=1$. Suppose $W(R,k)$ exists for every number of colours $R$, and fix $r$. We will construct finite bounds $F_t$, $1\le t\le r$, with the following property: every finite colouring of $[F_t]$ using at most $r$ colours contains either a monochromatic length-$(k+1)$ arithmetic progression, or $t$ length-$k$ arithmetic progressions of pairwise distinct colours, all focused at the same point $f\in [F_t]$. Specifically, their terms are“Focused” means that adjoining $f$ completes the corresponding length-$(k+1)$ arithmetic progression if $f$ has its colour.

For $t=1$, put $F_1=2W(r,k)$. Find a monochromatic length-$k$ arithmetic progression in the first $W(r,k)$ points. Its next point is at most $2W(r,k)$ and supplies the focus. For $k=1$, give the selected singleton common difference $1$.

Suppose $F_t=N$ has been constructed. SetDivide $[2NM]$ into $2M$ consecutive blocks of length $N$. Colour the first $M$ block indices by their full colour profiles, that is, the ordered list of their $N$ colours. There are at most $r^N$ profiles, so the induction hypothesis on $k$ gives equally profiled blocks with indicesWhen $k=1$, choose $D=1$. Otherwise $D>0$ and $j_0+kD\le 2M$, since the last selected block is at most $M$ and $D\le M$.

Assume the original finite colouring contains no monochromatic length-$(k+1)$ arithmetic progression. Applying the property of $F_t$ to the common profile gives $t$ focused length-$k$ arithmetic progressions at a position $f$, with common differences $d_i$ and distinct colours $q_i$. The colour $q$ of position $f$ differs from every $q_i$: otherwise the profile already contains a length-$(k+1)$ monochromatic arithmetic progression, and so does any selected block.

Put $j_{\ast }=j_0+kD$ and $F=N(j_{\ast }-1)+f$, a point in the original interval. For $1\le h\le k$, the pointlies in a selected block at profile position $f-h{d}_i$, so has colour $q_i$. Similarly $F-hND$ has colour $q$. Thus $t+1$ length-$k$ arithmetic progressions of distinct colours focus at $F$. This proves the step in $t$. At $t=r$, the focus necessarily has the colour of one of the $r$ progressions, giving a monochromatic length-$(k+1)$ arithmetic progression. Hence $W(r,k+1)$ exists, completing the mathematical induction and the proof of Van der Waerden theorem.

For the partition regularity deduction we first prove a strengthened progression lemma, the Brauer progression theorem: for positive $r,s,L$ with $L\ge 2$, there is $B(r,s,L)$ such that every finite colouring of $[B(r,s,L)]$ using at most $r$ colours contains a monochromatic setInduct on $r$. For $r=1$, take $B(1,s,L)=\max (s,L)$ and $a=d=1$. Given the result for $r-1$, putVan der Waerden theorem supplies a colour-$q$ arithmetic progression $a+j{d}_0$, $0\le j<K$, inside $[W]$. If some $sj{d}_0$, $1\le j\le N$, also has colour $q$, take $d=j{d}_0$ and use the $L$ terms $a,a+j{d}_0,\dots ,a+(L-1)j{d}_0$. They are in the progression because $(L-1)j\le K-1$.

Otherwise the integers $sj{d}_0$, $1\le j\le N$, use at most $r-1$ colours. Colour $j\in [N]$ by the colour of $sj{d}_0$ and apply the induction hypothesis. It supplies a monochromatic set $\{se,b,b+e,\dots ,b+(L-1)e\}$. Multiplying by $s{d}_0$ gives the required configuration with $a=s{d}_{0}b$ and $d=s{d}_{0}e$: its distinguished member is $s^2{d}_{0}e=sd$. All of these points are at most $sN{d}_0\le sNW$, so the finite bound is valid. This proves the Brauer progression theorem using only the just-proved Van der Waerden theorem.

A row is a partition regular matrix over the positive integers if every finite colouring of $\mathbb{N}$ admits a positive vector $x$ whose coordinates have one colour and whose product with the row is zero. Coordinates may repeat. Since the coefficients are rational numbers, multiplying them by a common positive denominator leaves the equation and every zero subset sum unchanged, so write the row as nonzero integer coefficients $b_1,\dots ,b_n$.

To prove necessity, suppose no nonempty subset of the $b_i$ sums to zero. Choose a prime $p>{\sum }_i|b_i|$ and use the last nonzero digit coloringwhere $v_p$ is the P-adic valuation. If positive $x_i$ all had the same colour $u$ and satisfied ${\sum }_i{b}_i{x}_i=0$, let $m={\min }_i{v}_p(x_i)$ and $I=\{i:v_p(x_i)=m\}$. After division by $p^m$ and reduction modulo $p$, the equation becomesAs $u\ne 0$, this says $p\mid {\sum }_{i\in I}{b}_i$. The latter sum is nonzero and has absolute value at most ${\sum }_i|b_i|<p$, a contradiction. This explicitly proves the necessary obstruction without invoking a general partition regularity criterion.

Conversely, suppose $\varnothing \ne I\subseteq \{1,\dots ,n\}$ and ${\sum }_{i\in I}{b}_i=0$. Select an index $j\in I$, and putFor any finite colouring, the Brauer progression theorem gives one-coloured $\{sd,a,a+d,\dots ,a+(L-1)d\}$. The indices $u$ and $u+q$ both lie between $0$ and $L-1$. Define positive, same-coloured coordinates byThenThis also covers $I$ equal to the whole index set and $q=0$. Reversing the clearing of denominators proves the exact criterion:

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.

Ramsey theory studies configurations that every finite coloring of a sufficiently large structure must contain. Its striking message is that some order is unavoidable even when no organized construction was used. A basic example is the Ramsey number for triangles: color every edge of a complete graph on six vertices red or blue, in any way, and some triangle in a graph has all three edges the same color. Five vertices do not suffice.

The objects receiving colors matter. In the Ramsey number for triangles, colors belong to edges, which represent all unordered pairs of distinct vertices; the vertices themselves have no colors. In the finite Ramsey theorem, colors can instead belong to all $r$-element subsets of a finite set. A monochromatic set or homogeneous set for a colouring expresses agreement among the appropriate colored objects.

Why is this useful? Ramsey theory provides existence guarantees in combinatorics that survive every possible assignment of a fixed finite number of colors. The Ramsey theorem turns this into a structural dichotomy: every sufficiently large graph contains a large clique or a large independent set. In arithmetic, the Van der Waerden theorem guarantees monochromatic arithmetic progressions. The beauty is the universal quantifier: an adversary can choose every color and still cannot prevent the promised configuration.

These guarantees concern precisely specified configurations under stated hypotheses. They do not justify selecting arbitrary patterns in empirical data after seeing the data, or attributing statistical significance to a selected pattern. A mathematical guarantee of some monochromatic configuration alone says nothing about the empirical meaning of that configuration.

The subject's name comes from F. P. Ramsey's ^[ref]. The ^[ref] records receipt on 28 November 1928 and presentation on 13 December 1928. Ramsey introduced the Ramsey theorem for r-sets and finite Ramsey theorem as tools for a problem in mathematical logic, while recognizing their independent interest in combinatorics. He also distinguished proving that a finite threshold exists from finding its smallest value, the quantitative question captured by a Ramsey number.