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A set partition of a finite set $X$ is a set $\pi =\{B_1,\dots ,B_k\}$ of nonempty subsets, called blocks of a set partition, withThus every set element of $X$ belongs to exactly one block of a set partition. This is the finite counting version of a partition of a set. The blocks of a set partition are unlabelled: their subscripts help describe them, but give them no roles or order. Stanley uses this definition in his classification of counting problems.^[ref]

All possibilities for three elements. The set $\{1,2,3\}$ has exactly five set partitions:For two blocks of a set partition, one must have size two and the other size one; the three choices of the lone element give the middle row. One block of a set partition forces the whole set, and three blocks of a set partition force three singleton sets. This proves the enumeration complete, with counts $1,3,1$ by number of blocks of a set partition. Their sum is the Bell number $B_3=5$.

What information is kept? Exchanging the two blocks of a set partition in $\{\{1,2\},\{3\}\}$ leaves the same set partition; writing the tuple $(\{1,2\},\{3\})$ instead makes an ordered list, different from $(\{3\},\{1,2\})$. Exchanging elements within a block of a set partition also changes nothing. In contrast, $\{\{1,3\},\{2\}\}$ is a different set partition, because the actual memberships differ. Both have the same integer partition $2+1$ of sizes. An integer partition, or its Ferrers diagram, forgets which distinct objects were grouped together.

For four named people, $\{\{1,2\},\{3,4\}\}$ and $\{\{1,3\},\{2,4\}\}$ therefore describe different ways of making two unnamed teams, although both have sizes $2+2$. If every team must have exactly two people, only three set partitions qualify; if any nonempty team sizes are permitted, the Stirling number of the second kind $S(4,2)=7$ counts them. Naming the teams changes the counting problem again.

A set partition is useful whenever membership matters but category names do not. Define $x\sim y$ when $x$ and $y$ belong to the same block of a set partition. This is an equivalence relation, and its equivalence classes recover the set partition. Conversely, an equivalence relation supplies its set partition of equivalence classes. The beautiful simplification is that one grouping can be described by whole subsets or by a rule comparing two elements. Additional distinctions are organized by refinement of a set partition.

The empty set has one set partition, the empty collection $\pi =\varnothing$, with zero blocks of a set partition. This differs from $\{\varnothing \}$, which is forbidden because its proposed block of a set partition is empty.

For a sequence of sets, the limit superior of sets consists of points belonging to infinitely many members, while the lower limit consists of points belonging to every sufficiently late member. If the two sets coincide, their common value is the set-theoretic limit.

Set theory studies sets, their set elements, and the constructions and sizes possible for them. It supplies a shared vocabulary for mathematics: a function connects two sets, a binary relation specifies which pairs of objects are related, and a probability space starts with a set of possible outcomes. Giving these objects a common language lets a construction from one subject be used in another without repeating its underlying bookkeeping.

A particularly beautiful question is how to compare sizes without first assigning a number. Two sets have the same cardinality when a bijection pairs every member of one with exactly one member of the other. For finite sets, this agrees with ordinary counting. For infinite sets, it produces surprises: $n\mapsto 2n$ bijects the natural numbers with the nonnegative even integers, and the Cantor pairing function bijects the Cartesian product ${\mathbb{N}}^2$ with $\mathbb{N}$. A proper subset can have the same cardinality as its containing set.

This does not make all infinite sets equally large. Cantor's diagonal argument shows that all infinite binary sequences cannot be listed by the natural numbers. Cantor's theorem gives the wider principle: the power set of any set has strictly greater cardinality. The contrast between an explicit enumeration and an impossibility argument makes the meaning of infinity much sharper than a picture of something merely continuing forever.

Set formation itself requires care. Russell's paradox shows why not every condition defines a set of all objects satisfying it. Zermelo–Fraenkel set theory instead uses axioms specifying permitted constructions, including set comprehension within a previously given set. The restrictions are useful: they let us form the collections ordinary arguments need while blocking the unrestricted construction that produces the paradox.

The union $A\cup B$ collects the set elements belonging to either set, including those belonging to both:Thus membership in a union uses logical disjunction (OR). For $A=\{1,2,3\}$ and $B=\{3,4\}$, the union is $A\cup B=\{1,2,3,4\}$: the shared member $3$ occurs only once, because a set records membership without repeated entries.

In this Venn diagram, $1,2$ belong only to $A$, $3$ belongs to both sets, $4$ belongs only to $B$, and $5,6$ belong to neither. The surrounding universal set is $U=\{1,2,3,4,5,6\}$. Numbers are placed by membership; circle areas do not represent cardinalities. Each panel shades the region selected by its operation.

For a database illustration, let $A$ collect records matching one condition and $B$ collect records matching another. Their union selects records matching at least one condition, with a record satisfying both still appearing once. This interpretation turns an abstract operation into a reusable rule for combining selections. For an authored treatment of union, intersection, set difference, and complement, see Lehman, Leighton and Meyer, Mathematics for Computer Science (2015), Definition 4.1.1, pp. 82–83.

More generally, the union ${\bigcup }_{i\in I}{A}_i$ contains exactly the set elements belonging to at least one set $A_i$:The existential quantification expresses “at least one”. With no indexing members, the union is the empty set. Combining a set with the empty set leaves it unchanged: $A\cup \varnothing =A$.

An uncountable subset $S$ of the real numbers whose intersection with every null set for Lebesgue measure is countable. Testing only Borel sets suffices because every Lebesgue-null set has a Borel null superset. No measurability of $S$ is assumed.

A sigma-algebra on a set $\Omega$ is a collection $\mathcal{F}$ of subsets such that $\Omega \in \mathcal{F}$, $A\in \mathcal{F}$ implies $\Omega \setminus A\in \mathcal{F}$, and $A_n\in \mathcal{F}$ for all $n$ implies ${\bigcup }_n{A}_n\in \mathcal{F}$. Thus it contains the empty set and is closed under countable intersections as well. This is precisely the closure needed to apply countable additivity without leaving the domain of a measure.

For $\Omega =\{a,b,c\}$,is a sigma-algebra. It describes observations that distinguish $a$ from the pair $\{b,c\}$, but cannot distinguish $b$ from $c$. The power set provides a finer sigma-algebra with all eight subsets. Both are valid; choosing a sigma-algebra records which questions the model can answer.

On the real numbers, the Borel sigma-algebra contains all open sets and everything generated from them by these operations. The Lebesgue measurable sets form a larger sigma-algebra, including all subsets of Borel null sets. Closure under countable unions accommodates constructions involving an entire sequence of observations or approximations.

A simple function is a measurable function whose image is a finite set. Its value sets $A_j=\{x:s(x)=a_j\}$ form a finite measurable partition. For nonnegative $a_j$, its Lebesgue integral is ${\sum }_j{a}_j\mu (A_j)$, with the convention $0\cdot \infty =0$. The pieces need not be intervals, or even connected sets. See MIT lecture 10, Definitions 100 and 106.

For example, on $[0,1]$ let $s$ be $2$ on $[0,1/4)$, $1$ on $[1/4,3/4)$, and $0$ on $[3/4,1]$. Its Lebesgue integral is $2(1/4)+1(1/2)=1$. Its superlevel sets have Lebesgue measure $3/4$ for $0\le t<1$, $1/4$ for $1\le t<2$, and $0$ thereafter. The layer cake representation gives the same answer: $(3/4)(1)+(1/4)(1)=1$. These two calculations are the building blocks of the general Lebesgue integral.

An undirected simple graph is exactly a pair of setsThus every edge is a two-element set of distinct vertices. The unordered pair $\{u,v\}$ equals $\{v,u\}$, so there is no direction. The distinctness condition excludes self-loops, and using a set of edges excludes parallel copies. A simple graph may have vertices with no incident edges; $E$ does not by itself determine $V$. For example, $(\{A,B\},\{AB\})$ and $(\{A,B,C\},\{AB\})$ have the same edge set but different vertex sets.^[ref]

In the six-vertex example, the vertex neighbourhood of $C$ is $\Gamma (C)=\{A,B,D\}$, so its degree of a vertex is $\mathrm{deg}(C)=3$. In label order $A,B,C,D,E,F$, the degree sequence isThis is the handshaking lemma: count each edge once at each of its two endpoints. The average degree of a graph here is $14/6=7/3$. The degree sequence is a compact consistency check on an edge list; an odd sum could not describe any finite undirected graph. It also forces the number of odd degrees of vertices to be even, since the even terms do not affect the parity of the sum.

The adjacency matrix in the figure uses label order $A,B,C,D,E,F$. Its entry $M_{uv}$ is one exactly when $\{u,v\}\in E$, and zero otherwise. It is a symmetric matrix with zero diagonal, and each row sum gives the corresponding vertex degree. For example, row $C$ is $(1,1,0,1,0,0)$. The fourteen ones count both directions of the seven undirected edges; they do not indicate fourteen different edges. This translates a picture into a representation suitable for calculation.

The restrictions matter. A directed graph uses an ordered pair to distinguish $u\to v$ from $v\to u$, so reachability can depend on direction. A multigraph can contain several distinct edges between the same vertices; adding another $CD$ would raise $\mathrm{deg}(C)$ and $\mathrm{deg}(D)$ to four without adding a new neighbour. A self-loop at $A$ contributes two to its undirected vertex degree, making it four, because both endpoints lie at $A$. Specify whether self-loops are allowed in a multigraph. Finally, a weighted graph attaches numerical information to edges: if links carry travel times, minimizing the number of edges need not minimize total time. These are changes to the model, rather than changes to how the same simple graph is drawn.

An abstract simplicial complex is a pair $K=(V,\Sigma )$, where $V$ is a set of vertices of a simplicial complex and $\Sigma$ is a collection of finite nonempty subsets, called simplexes. Every singleton $\{v\}$ belongs to $\Sigma$, and every nonempty subset of a simplex is a simplex. Its dimension is the supremum of $|\sigma |-1$ over $\sigma \in \Sigma$. Some conventions also include an empty simplex; it contributes no points to the geometric realization.

A group action on a nonempty set $X$ is simply transitive when exactly one group element sends any chosen point to any other:Equivalently it is both a transitive group action and a free group action; this is also a regular group action. Transitivity supplies a transport, and freeness makes it unique. Conversely, putting $y=x$ in the uniqueness condition shows that every stabilizer subgroup is $\{e\}$, since the identity element already fixes $x$. In a transitive group action, checking that just one stabilizer subgroup is trivial suffices: all the others are conjugate subgroups.

The cyclic group $C_3$ in the diagram under transitive group action is simply transitive. More generally, the left regular action of any group $G$ on itself is simply transitive: the equation $gx=y$ has the unique solution $g=y{x}^{-1}$, using the inverse element of $x$. This shows why the property is useful: every position has a unique label relative to a chosen starting position.

Choose a point $x_0\in X$. Its orbit mapis a bijection and an equivariant map from the left regular action: ${\varphi }_{x_0}(hg)=h\cdot {\varphi }_{x_0}(g)$. This identifies $X$ with a copy of $G$, but $X$ does not come with a preferred identity element. If $x_1=a\cdot x_0$, then ${\varphi }_{x_1}(g)={\varphi }_{x_0}(ga)$: changing the chosen point composes this identification with right multiplication by $a$. An old label $g$ therefore becomes the new label $g{a}^{-1}$, which also specifies the side of multiplication for a noncommutative group.

The transport from $x$ to $y$ remains independent of the chosen point. Writing $x=b\cdot x_0$ and $y=c\cdot x_0$, it is $c{b}^{-1}$. With $x_1=a\cdot x_0$, the new labels are $b{a}^{-1}$ and $c{a}^{-1}$, andFor a finite group, simple transitivity implies $|X|=|G|$. Conversely, a transitive group action with $|X|=|G|<\infty$ is simply transitive by the orbit-stabilizer theorem. Equality of cardinalities alone is insufficient without transitivity: the cyclic group $C_2$ can fix both points of a two-point set, giving two group orbits and two possible self-transports at each point.

A probability distribution is singular continuous if it has no point masses and assigns probability one to a set of zero Lebesgue measure. Its cumulative distribution function is continuous, but the distribution has no probability density function with respect to Lebesgue measure. This distinguishes continuity of the cumulative distribution function from an absolutely continuous probability distribution, whose probabilities are integrals of a probability density function.

The solution set of a mathematical condition $P$ in a specified domain $D$ collects every allowed value that satisfies it:It describes the answer without committing to the steps used to find it. For the equation $x^2=1$ over the real numbers, the solution set is $\{-1,1\}$; restricting to positive real numbers leaves $\{1\}$. For $2x=1$, the solution set is $\{1/2\}$ over the rational numbers and the empty set over the integers. The domain is part of the question.

An equation can have one solution, several solutions, no solutions, or an entire domain of solutions. Over the real numbers, $2x+3=9$ gives the singleton set $\{3\}$, $x^2+1=0$ gives the empty set, and the mathematical identity $(x+1{)}^2=x^2+2x+1$ gives the whole real line. Changing that second equation's domain to the complex numbers gives $\{-i,i\}$. The elementary definition by values that satisfy an equation and the classification by solution sets appear in College Algebra 2e, §2.2: Solving Linear Equations in One Variable.

Requiring several conditions simultaneously takes the intersection of their solution sets. The squared equation $x=(x-2{)}^2$ has solution set $\{1,4\}$. Adding the restriction $x\ge 2$ giveswhich is also the solution set of $\sqrt{x}=x-2$. The extraneous solution example explains why the sign restriction is needed. Requiring at least one of several conditions instead takes their union. This turns logical combinations of constraints into reusable operations on sets.

For several variables, a solution is an ordered assignment, usually represented by a tuple. The real equation $x+y=1$ has solution set $\{(t,1-t):t\in \mathbb{R}\}$, a straight line in the Euclidean plane with coordinates $(x,y)$. Adding $x-y=0$ intersects that straight line with another, leaving the single tuple $(1/2,1/2)$. A system of linear inequalities has no solution exactly when the intersection of its individual solution sets is the empty set. For optimization constraints, this common solution set is the feasible region.

Because a solution set is a set, the order in which its members are listed is irrelevant. The multiple root of $(x-1{)}^2=0$ is still the one solution $1$; its multiplicity of a root is additional polynomial information, rather than a second occurrence inside the solution set. Equivalent equations preserve the solution set on the stated domain, even if their algebraic forms differ.

For $U:\mathcal{C}\to \mathcal{D}$, the left-adjoint solution-set condition requires a weakly initial set in every $(D\downarrow U)$. Equivalently, there is a set of arrows $d_i:D\to U{C}_i$ through which every $d:D\to UC$ factors as $U(h)d_i$. The dual right-adjoint condition asks for weakly terminal solution sets in $(U\downarrow D)$.

The output of a sorting algorithm must be a permutation of the complete input sequence. For equal lengths $n$, require a permutation $\pi$ of $\{0,\dots ,n-1\}$ such thatEquivalently, the input and output have the same multiset of complete entries: each entry has the same multiplicity. Equality of the underlying sets is insufficient; $(1,1,2)$ and $(1,2,2)$ have the same set of values and the same length but different multisets. When tracking occurrences rather than contents, retain original identity labels to check that each one is moved exactly once. A swap preserves this multiset automatically; an overwrite need not.

Define a sequence by $a_n=1$ when $n$ is a power of two, including $1=2^0$, and $a_n=0$ otherwise. Every tail of a sequence contains both a later power of two and a later nonpower. Therefore $\mathrm{liminf}{a}_n=0$ and $\mathrm{limsup}{a}_n=1$, and the sequence does not converge. Its peak indices become increasingly sparse, yet never stop appearing.

Through index $N\ge 1$, the number of peaks is $1+\lfloor {\log }_{2}N\rfloor$, so their fraction isThe set of powers of two has natural density zero. The same fraction is the Cesàro mean of the first $N$ terms, which tends to zero even though the limit superior stays one. At $N=1024$, eleven peaks give mean $11/1024\approx 0.0107422$. An average and a long-run upper edge answer different questions about the same sequence.

Take $X=\{0,1,2,\dots \}$, the nonnegative integers, and $Y=\{1,2,3,\dots \}$, the positive integers. Choose the injectionsHere $g$ is an inclusion map, while $f$ reaches only the positive even integers. In particular, $f$ is not a bijection: it misses every positive odd integer. Simply using the inverse function of $g$ on its image misses input zero.

The remedy is to follow the alternating injections starting at the missing input zero:The subscripts distinguish an input in $X$ from an output in $Y$, even when their numerical values agree. The resulting subset of $X$ is $A=\{0,2,6,14,30,\dots \}$. Its sequence of members satisfies $a_0=0$ and $a_{k+1}=2a_k+2$, giving $a_k=2^{k+1}-2$ by mathematical induction. Define the function

The image $h(A)=\{2,6,14,30,\dots \}=A\setminus \{0\}$ covers all the positive members of $A$ exactly once. The other inputs cover $Y\setminus A$ exactly once by staying fixed. These output sets are disjoint and their union is $Y$, proving that $h$ is a bijection. In particular, $h(0)=2$, $h(2)=6$, $h(6)=14$, while $h(1)=1$ and $h(4)=4$.

For completeness, its inverse function isThe familiar bijection $n\mapsto n+1$ also works, but this example shows something more specific: the two chosen injections can repair the missing zero by moving only a sparse sequence of inputs. The repair never has to shift the entire set.

A finite set $A$ of positive integers such that nonempty disjoint sets contained in $A$ have different sums, and the larger subset has the larger sum when their sizes differ. For $a_1<\cdots <a_n$, the size condition is equivalent to $a_1+\cdots +a_{k+1}>a_{n-k+1}+\cdots +a_n$ for $1\le k\le \lfloor (n-1)/2\rfloor$: the left side is the smallest possible sum of $k+1$ elements and the right side the largest sum of $k$ elements.

For $B=\overline{\mathbb{C}[1,x]}$ inside $A$, the displayed set is the full spectrum of $x$ in $A$. On a bounded complementary component, the maximum modulus principle gives $|p(\lambda )|\le {\max }_{{\sigma }_A(x)}|p|\le \Vert p(x)\Vert$. Evaluation $p(x)\mapsto p(\lambda )$ therefore extends to a character of an algebra on $B$, showing $\lambda \in {\sigma }_B(x)$. Conversely the unbounded resolvent component survives by resolvent components in a closed unital subalgebra. In particular, if zero lies in that component, $x^{-1}$ belongs to the norm closure of polynomials in $x$.

Suppose $f,g:X\to \mathbb{R}$ are bounded functions on the same nonempty set, and the supremum norm of their difference is at most $\delta$. Then their optimal values differ by at most the same amount:Indeed, $f(x)\le g(x)+\delta \le {\sup }_{X}g+\delta$ for every $x$, so ${\sup }_{X}f\le {\sup }_{X}g+\delta$; exchange $f$ and $g$ for the other direction. Adding a constant shifts a supremum by that constant, a basic compatibility between upper bounds and addition.^[ref]

For a chosen example on $[-1,1]$, take $f(x)=-x^2$ and $g(x)=-x^2+0.2x$. Their supremum error is $0.2$, since $|g(x)-f(x)|=0.2|x|$. Completing the square givesThus $f$ has maximum zero at $x=0$, while $g$ has maximum $0.01$ at $x=0.1$. The change in optimal value, $0.01$, is smaller than the certified bound $0.2$. For the vertical translation $h(x)=f(x)+0.2$, the bound is exact: its supremum error and optimal-value change are both $0.2$.

A stable optimal value does not guarantee a stable maximizing input. On $[-1,1]$, the two linear functions $f_{\epsilon }(x)=\epsilon x$ and $g_{\epsilon }(x)=-\epsilon x$ satisfyFor every positive $\epsilon$, their unique maximizing inputs remain two units apart, at $1$ and $-1$. Both functions approach the flat zero function, which has many maximizing inputs. Consequently an accurate objective function value is not by itself a certificate that the optimizing decision is accurate.

The estimate also gives a convenient consequence of uniform convergence. If $f_n$ converges uniformly to a bounded function $f$ on a nonempty $X$, then ${\sup }_X{f}_n\to {\sup }_{X}f$ for the bounded tail of the sequence. Uniform control handles the changing input that defines the best value. Pointwise convergence alone supplies no such control; the existing examples under uniform convergence illustrate that distinction.

For a group action of a group $G$ on a set $X$, the stabilizer subgroup of $x\in X$ consists of the transformations that leave $x$ fixed:It is a subgroup: the identity element fixes $x$, products of transformations fixing $x$ fix $x$, and the inverse element of a transformation fixing $x$ also fixes $x$. A stabilizer subgroup is a subset of $G$, whereas the orbit of a group action is a subset of $X$. See Ackerman's algebra notes, definition 0.4.1.2.

For the square group actions defined under group action, direct calculation givesThe reflection $s$ fixes $A$ and $C$ individually. The reflection $rs$ preserves the unoriented side of a polygon $AB$ by exchanging its two vertices of a polygon. The half-turn rotation $r^2$ exchanges the two ends of the unoriented diagonal of a polygon $AC$ while preserving that diagonal of a polygon. Thus fixing a side of a polygon or diagonal of a polygon as an object does not mean fixing every point on it; the latter condition uses a pointwise stabilizer. The four stabilizer subgroups have group orders $2,2,4,8$, matching group orbit sizes $4,4,2,1$ through the orbit-stabilizer theorem.

A stabilizer subgroup records the transformations still available after imposing a constraint. Fixing a vertex of a polygon of the square leaves two choices; fixing its centroid leaves all eight. In a faithful group action, a stabilizer subgroup can still contain a nonidentity transformation, as $s\in (D_4{)}_A$ shows: a faithful group action requires that only the identity element fix every point of $X$ simultaneously, not that every individual stabilizer subgroup be trivial. The two-point cyclic group example under group action has both stabilizer subgroups equal to the group kernel $\{0,2\}$ of its associated group homomorphism.