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A finite union is a union of finitely many sets. The indexing set is finite; its member sets need not be finite sets. The union of an empty indexed family is the empty set.

A function collision is a pair of distinct inputs $x\ne x^{\prime }$ with $f(x)=f(x^{\prime })$. An injective function has no collisions. For a two-to-one function every nonempty fiber has two elements and hence one unordered collision pair. Finding such a pair through an oracle is the task of quantum collision finding.

In the finite example under image and preimage of a function, the pair $\{a,b\}$ is a function collision: both inputs give zero. The function retains their shared output but loses the distinction between them. If a finite set has more inputs than the number of available outputs, the pigeonhole principle forces a function collision. Equal source and target cardinalities alone do not prevent one.

For a finite function, let $m_y$ be the cardinality of its fiber of a function over $y$. The number of unordered function collisions is ${\sum }_y\left(\genfrac{}{}{0ex}{}{m_y}{2}\right)$, because each pair must lie in one common fiber. The example's nonempty fiber sizes are $2,1,1$, giving exactly one collision pair.

Draw a $k\times D$ random matrix $G$ whose entries are independent random variables with the standard normal distribution, and setThe factor $1/\sqrt{k}$ makes the expected value of the squared L2 norm equal to the original squared L2 norm. This Gaussian Johnson–Lindenstrauss transform is a linear map, generally not an orthogonal projection: its rows are not required to be orthogonal vectors. ^[ref]

For the fixed finite set $S$, let $0<\delta <1$ be an allowed failure probability, and defineWith $\log$ denoting the natural logarithm, a sufficient integer target dimension isThen one draw of this specified random matrix satisfies all the squared Euclidean distance bounds simultaneously with probability at least $1-\delta$. Since $c(\epsilon )\ge {\epsilon }^2/12$, the simpler sufficient bound $k\ge \lceil 12{\epsilon }^{-2}\log (n(n-1)/\delta )\rceil$ also works. Both are sufficient bounds for this argument; they do not claim the smallest possible target dimension.

This is a probability statement about a map drawn independently of the fixed finite set. It implies existence because the success probability is positive, but permits failed draws. The set cannot be chosen after inspecting the map with this same guarantee. In particular, no single draw preserves every possible finite set in the ambient Euclidean space when $k<D$.

A graph coloring assigns labels called colours to elements of a graph, subject to specified constraints. Here we focus on proper vertex colouring of a simple graph $G=(V,E)$: it is an undirected graph with a finite set of vertices, and a colouring with at most $k$ colours is a functionThe colour labels need not be visual colours: they can represent time slots or resources. Unused labels are allowed. Properness concerns graph adjacency, so moving vertices in a drawing does not change whether the graph coloring is valid.^[ref]

Each colour class $c^{-1}(i)$ is an independent set: no edge has both endpoints in it. Conversely, a set partition of $V$ into independent sets gives a proper graph coloring by assigning one colour to each part. Thus graph coloring groups mutually compatible objects, and its chromatic number asks for the fewest such groups.

For an illustrative scheduling model, let each vertex represent an activity of equal duration and each edge represent a pair that cannot occur in the same time slot. A proper graph coloring is then a schedule respecting those pairwise conflicts. If the conflicts form a cycle graph $C_4$ with edges $12,23,34,41$, activities $1,3$ can share one slot and $2,4$ another. This model assumes interchangeable slots and includes only the pairwise conflicts encoded by the graph.^[ref]

The word "colouring" also occurs in edge coloring, which requires different colours on edges sharing an endpoint. The chromatic number discussed here always refers to proper vertex colouring.

The Hausdorff dimension is the critical exponent at which Hausdorff measure changes from infinite to zero:with the standard endpoint convention for an empty supremum. This measures fine-scale geometric size rather than the number of coordinates used to describe the set. A line segment has Hausdorff dimension one; a square has dimension two; the middle-third Cantor set has dimension $\log 2/\log 3\approx 0.63093$.

Why can a dimension be fractional? At a scale reduced by a factor of three, the Cantor set needs twice as many construction intervals. The balance $2(1/3{)}^d=1$ gives $d=\log 2/\log 3$, between the growth rates of finitely many points and a whole interval. Making this argument rigorous requires a cover upper bound and the mass distribution principle lower bound, as in Hausdorff dimension of the Cantor set.

At the critical exponent itself, Hausdorff measure can be zero, finite and positive, or infinite; dimension does not specify that endpoint value. For example, finite sets and countably infinite sets both have dimension zero, although their counting measures differ. Dimension is also different from a finite-resolution estimate: box-counting dimension is another asymptotic definition, and fitting a scaling slope over a handful of observed scales does not establish either dimension.

The metric matters. Multiplying every distance by a fixed positive constant multiplies ${\mathcal{H}}^s$ by that constant to the power $s$ and preserves Hausdorff dimension. A Lipschitz map cannot increase it, because mapped cover diameters increase by at most a fixed factor. A bi-Lipschitz map therefore preserves it by applying this fact in both directions. These properties make Hausdorff dimension useful for comparing geometric models while controlling how their measurements change.

For finite sets $A_1,\dots ,A_n$, the cardinality of their union isAdd the individual cardinalities, subtract pairwise intersection counts, add triple intersection counts, and continue alternating. This is useful whenever one object can satisfy several conditions: a search result can match several filters, or a permutation can have several forbidden fixed points.

Among the integers from 1 to 30, let $A$ contain the multiples of 2 and $B$ the multiples of 3. Then $|A|=15$, $|B|=10$, and $|A\cap B|=5$, because their common multiples are the multiples of 6. The naive sum $25$ counts those five twice; the correct union size is $15+10-5=20$. The regions contain ten in $A$ alone, five in both, five in $B$ alone, and ten in neither.

Add a third set $C$ containing the multiples of 5. Its size is 6; the three pairwise intersections have sizes $5,3,2$; and the triple intersection contains only 30. The inclusion-exclusion principle gives $15+10+6-5-3-2+1=22$ integers divisible by at least one of 2, 3, and 5.

The alternating corrections have a local explanation. An object belonging to exactly $r\ge 1$ of the sets is counted $\left(\genfrac{}{}{0ex}{}{r}{1}\right)-\left(\genfrac{}{}{0ex}{}{r}{2}\right)+\cdots +(-1{)}^{r+1}(\genfrac{}{}{0ex}{}{r}{r})=1$ times, by the binomial theorem applied to $(1-1{)}^r$. An object in none contributes zero. Thus overlapping descriptions are repaired without requiring the sets themselves to be disjoint. For probability, the same calculation applies to event probabilities in place of cardinalities, giving the inclusion-exclusion principle for probabilities.

For an infinite subset $A\subseteq \mathbb{N}$, repeatedly selecting the least element not yet used gives a bijection $f:\mathbb{N}\to A$. The well-order ensures each minimum exists, infinitude ensures the remaining set is nonempty, and strict increase gives injectivity. If $a\in A$ were omitted forever, all selected values would be at most $a$, impossible inside the finite set $\{0,\dots ,a\}$.

An independence system consists of a finite set $E$ and a nonempty family $\mathcal{I}$ of subsets such that $I\in \mathcal{I}$ and $J\subseteq I$ imply $J\in \mathcal{I}$. This is the hereditary condition: deleting chosen elements keeps a selection admissible. In particular, the empty set belongs to $\mathcal{I}$. An independence system becomes a matroid when it also satisfies the matroid independence axioms for exchange. The hereditary nonmatroid greedy obstruction shows why deletion alone does not justify a greedy algorithm.

The infinite pigeonhole principle says that if an infinite set has a partition of a set into finitely many parts, at least one part is an infinite set. Equivalently, assigning infinitely many objects to a finite number of boxes forces some box to contain infinitely many objects. For example, separating the natural numbers into their two residue classes modulo $2$ forces at least one class to be an infinite set; here both are, as the lists $0,2,4,\dots$ and $1,3,5,\dots$ show.

Proof outline. A union of finitely many finite sets is a finite set. If all the parts were finite sets, their union could not be the original infinite set.

The assumption of finitely many parts is essential. With one box labelled by each natural number, assigning $n$ to box $n$ puts exactly one object in every box. Infinitely many objects and infinitely many boxes therefore need not produce even a shared box. The infinite pigeonhole principle is useful whenever finitely many labels organize an endless collection: one label must keep occurring infinitely often, providing an infinite set on which that label is constant.

For $0<\epsilon <1$ and any $n$ points in a Hilbert space, there is a linear map into Euclidean dimension $O({\epsilon }^{-2}\log n)$ that multiplies every pairwise distance by a factor between $1-\epsilon$ and $1+\epsilon$.

The striking feature is that the target dimension depends on the number of points and the allowed error, independently of the ambient dimension. Fix a finite set $S=\{x_1,\dots ,x_n\}\subset {\mathbb{R}}^D$, with $n\ge 2$, and $0<\epsilon <1$. An equivalent sufficient statement uses squared Euclidean distances: a linear map $T:{\mathbb{R}}^D\to {\mathbb{R}}^k$ exists, with $k=O({\epsilon }^{-2}\log n)$, such thatThe finite-set theorem cited here supplies an explicit sufficient target dimension for this squared-distance formulation. ^[ref]

Taking square roots gives distance factors $\sqrt{1-\epsilon }$ and $\sqrt{1+\epsilon }$, which lie inside $[1-\epsilon ,1+\epsilon ]$. Conversely, bounds $1\pm \epsilon$ on distances only give squared-distance bounds $(1\pm \epsilon {)}^2$; the same numerical tolerance should not be silently used for both. Under the squared-distance bounds, the bi-Lipschitz distortion is at most $\sqrt{(1+\epsilon )/(1-\epsilon )}$.

The Johnson–Lindenstrauss lemma preserves this fixed finite set, rather than every vector in the ambient Euclidean space. A linear map with $k<D$ has a nonzero kernel by the rank-nullity theorem, so some nonzero vectors collapse completely. For finitely many points in a real Hilbert space, apply the finite-dimensional argument on their linear span and extend by orthogonal projection onto that linear span. If the sufficient target dimension exceeds the ambient dimension, the identity map already preserves every Euclidean distance exactly, with zero padding if a larger target is requested.

A combinatorial structure whose elements carry distinct labels from a prescribed finite set. For example, a rooted tree can have labelled vertices; assigning labels is part of the structure counted by an exponential generating function.

A tree whose vertices carry distinct labels from a fixed finite set, as a labelled combinatorial structure. Two labelled trees on the same vertex set are equal when their sets of edges are equal: any comparison must keep every label fixed. Cayley's formula counts these trees.

The complete bipartite graph $K_{3,3}$ has vertex sets $A=\{a_1,a_2,a_3\}$ and $B=\{b_1,b_2,b_3\}$, with every edge between $A$ and $B$ and none inside either part. Its six vertices support $3\cdot 3=9$ edges and no triangles in a graph: among any three vertices, two belong to the same independent set and have no joining edge.

Add the single edge $a_1{a}_2$. Each $b_j$ is a common neighbour of its endpoints, producing the triangle in a graph $a_1{a}_2{b}_j$. These are exactly the three new triangles in a graph, because every triangle in a graph must use the only edge inside a part. The new count is ten edges; nine was the last possible count without a triangle in a graph on six vertices.

In general, every simple graph with a finite set of $n\ge 0$ vertices that is triangle-free has at mostedges. This is the $r=2$ case of Turán's theorem, with forbidden clique $K_3$. Equality holds exactly for the balanced complete bipartite graph $K_{\lfloor n/2\rfloor ,\lceil n/2\rceil }$, up to graph isomorphism; empty parts are allowed for $n<2$.^[ref]

The floor function gives the exact integer bound. Splitting $n$ vertices into two independent sets of sizes $a$ and $n-a$, and putting in every cross edge, givesThe count is largest when $a$ is an integer nearest $n/2$. For $n=2k$ it is $k^2$; for $n=2k+1$ it is $k(k+1)$. Both equal $\lfloor n^2/4\rfloor$. This proves attainability for every $n$, including the empty graph.

A short proof of the upper bound uses degrees of vertices. In a triangle-free graph, the endpoints $u,v$ of an edge have disjoint vertex neighbourhoods: a common neighbour would complete a triangle in a graph. Thus $d(u)+d(v)\le n$. Summing over edges counts $d(v)$ once for each of the $d(v)$ edges incident to $v$, givingFor $n>0$, the Cauchy-Schwarz inequality and handshaking lemma yieldIf $m>0$, division gives $m\le n^2/4$; $m=0$ is immediate. Since $m$ is an integer, apply the floor function. The case $n=0$ is immediate too. This proves the bound; the full equality characterization follows from Turán's theorem.

The useful reversal is that more than $\lfloor n^2/4\rfloor$ edges force a triangle in a graph, whatever their placement. A global count certifies a local pattern. The extremizer is bipartite, but a triangle-free graph need not be bipartite: a five-vertex graph cycle has five edges, below the bound of six. Historically, Willem Mantel posed the triangle-exclusion question in 1907.^[ref]

Mark and recapture estimates the cardinality of a finite set of individuals when counting everybody is impractical. The useful information is the overlap between observations: with fixed sample sizes, finding many of the same individuals again suggests a smaller population, while finding few suggests a larger one. More elaborate capture-recapture models also estimate survival and changing probabilities of detection.

In a closed-population capture-recapture model, let $N$ be the unknown population cardinality. Capture $M$ distinct individuals, mark them, release them all, and allow them to mix with the rest. Next take a second sample of $C$ distinct individuals using sampling without replacement, and count the $R$ individuals carrying a mark. The observed marked fraction $R/C$ estimates the population's marked fraction $M/N$, giving the Lincoln-Petersen estimatorRelease happens between the two samples; sampling without replacement means that an individual is counted at most once within the second sample. See Richard J. Barker, Open Population Mark-Recapture Models, §1.1.1.

For a constructed teaching example, mark $M=20$ individuals and later capture $C=25$, including $R=5$ marked individuals. One fifth of the second sample is marked, suggesting that the twenty marked individuals are one fifth of the population:Let $A$ and $B$ be the sets of individuals in the first and second samples. Their intersection has cardinality five. By the inclusion-exclusion principle, the set union containsThus $N\ge 40$ is an observed lower bound, whereas $100$ is the output of an estimator based on the capture-recapture model. These are toy counts, not measurements from a medical or ecological study.

The diagram shows only observed individuals. It does not draw an assumed population of one hundred. The capture-recapture model uses the overlap to infer how many individuals may have escaped both samples; its likelihood function shows why one estimator output cannot describe all that uncertainty.

Let $E$ be a finite set and $\mathcal{I}\subseteq \mathcal{P}(E)$ a family of subsets, where $\mathcal{P}(E)$ is the power set. The matroid independence axioms areThe second condition is heredity; the third is exchange, also called augmentation. The set difference $J\setminus I$ matters: the new element must actually increase the cardinality of $I$. Exchange adds one element without requiring any deletion from $I$. Nonemptiness together with heredity is equivalent to the first condition. ^[ref]