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An absolutely normal number is a normal number in every integer base $K\ge 2$. This requires every finite word in every such positional notation to have its appropriate limiting digit block frequency. Knowing base-ten normality alone does not establish absolute normality; the base is part of the statement.

Borel's 1909 result says that absolute normality holds almost everywhere for Lebesgue measure: within $[0,1)$ the exceptions form a Lebesgue null set. In particular, the set of absolutely normal numbers has Lebesgue measure one there. This is a statement about the size of an exceptional set, not a test for a specified constant. ^{[ref] [ref]}

A modern route uses the Birkhoff ergodic theorem and ergodicity of integer multiplication on the circle. For each fixed base and digit interval, the time proportion equals its Lebesgue measure almost everywhere. The bases and finite words form countable sets, so a countable intersection of these measure-one sets still has Lebesgue measure one. For an absolutely normal number, the word $11$ therefore has frequency $1/4$ in base two, $1/9$ in base three, and $1/100$ in base ten. The same symbols represent different intervals in different bases.

For limit ordinals $\alpha \le \beta$, the structures $V_{\alpha }$ and $V_{\beta }$ agree on whether any shared set is a cardinal number. They agree on being an ordinal. If $\delta <\alpha$ is not a cardinal number, a bijection from some $\gamma <\delta$ onto $\delta$ has rank of a set at most $\delta +3$, hence belongs to $V_{\alpha }$. Its defining properties are bounded formulas in set theory. Both structures therefore see the same failure witness. This full agreement is stronger than downward absoluteness of cardinalhood between arbitrary transitive models.

The rule of sum counts alternatives in disjoint finite sets by adding their sizes. For a disjoint union $A\bigsqcup B$, each element belongs to exactly one alternative, so $|A\bigsqcup B|=|A|+|B|$. Overlapping alternatives require accounting for repeated elements, as in the inclusion-exclusion principle. The groups indexed by first-return position in the Catalan recurrence are disjoint alternatives.

Additive combinatorics studies how the cardinalities and sumset representation functions of sumsets reveal structure in finite sets contained in abelian groups. Write the group operation as addition: the ambient abelian group may be the integers $\mathbb{Z}$, a finite cyclic group, or another abelian group. The same definitions apply, but the ambient group changes which bounds are possible.

The basic experiment is to add every ordered pair of members of a finite set. The sumset records the distinct answers; the sumset representation function records how often each answer occurs; the additive energy counts coincidences between those answers. An arithmetic progression produces many repeated sums, so its sumset grows slowly and its additive energy is relatively large. The doubling constant measures that growth relative to the original cardinality.

The examples below compare the arithmetic progression $A=\{0,1,2,3\}$ with the more spread-out integer set $B=\{0,1,4,10\}$. Their four members give sixteen ordered pairs in each case, but different numbers of distinct sums. This small comparison illustrates the definitions; it does not establish a claim about arbitrarily large sets. For a systematic introduction, see Tao and Vu's author page for Additive Combinatorics, with free sample chapters on sum set estimates and additive geometry.

The additive energy of a finite set $S\subseteq G$ in an abelian group $G$ counts additive quadruples, with order and repeated members allowed:At a fixed sum $s$, independently choosing two of its $r_{S+S}(s)$ ordered pairs gives $r_{S+S}(s{)}^2$ additive quadruples. Thus additive energy weights heavily represented sums more strongly than the cardinality of the sumset does.

For $A=\{0,1,2,3\}$ and $B=\{0,1,4,10\}$, the sumset representation functions giveThe arithmetic progression has additional coincidences such as $0+3=1+2$. The set $B$ still has coincidences forced by commutativity, including $0+1=1+0$, so its additive energy is not merely the sixteen identical-pair comparisons.

The same additive energy can be counted using the difference set. Define $r_{S-S}(t)=|\{(a,b)\in S\times S:a-b=t\}|$. The identity $a+b=c+d$ is equivalent to $a-c=d-b$, and the coordinate rearrangement $(a,b,c,d)\mapsto (a,c,d,b)$ is a bijection. ConsequentlyFor $A$, the difference counts from $-3$ through $3$ are again $1,2,3,4,3,2,1$. For $B$, the zero difference has four representations and each of the twelve nonzero differences has one, giving $4^2+12\cdot 1^2=28$. The sumset representation functions for sums and differences can look different while their squared totals agree.

For every nonempty finite set $S\subseteq G$ in an abelian group $G$,The lower bound follows from the Cauchy-Schwarz inequality and the total sumset representation function count:For the upper bound, any three members $a,b,c$ determine at most one fourth member $d=a+b-c$ of an additive quadruple. See Yufei Zhao's MIT lecture 25, pages 1-3, for representation counts and the energy bounds.

In the two integer examples, the lower bounds are $256/7$ and $256/10$, and the upper bound is $64$. For the subgroup $H=\{0,2,4\}$ of $\mathbb{Z}/6\mathbb{Z}$, each element of $H$ has three sum representations, so $E(H)=3\cdot 3^2=27$ and both bounds are equalities.

For nonempty $S$, dividing additive energy by $|S{|}^3$ gives a relative measure between $1/K(S)$ and one. Small doubling constant therefore forces relatively large additive energy. A converse requires care: the Balog-Szemerédi-Gowers theorem extracts a large subset with small doubling constant, rather than asserting that all of $S$ has small doubling constant.

A bundle-value function for which an agent's value for a set of items is the sum of its item values. Different agents may assign different values to the same item. With nonnegative item values, a bundle containing no positively valued item has value zero. The formula excludes interactions such as a pair of items being valuable only when held together.

A value has cooperative-game additivity when adding two characteristic functions on the same player set adds their allocation vectors. Here $(v+w)(S)=v(S)+w(S)$ for every coalition. The sum of two simple cooperative games need not be simple; the axiom is naturally stated on transferable utility games.

For a forcing atom $p$ in a ground-model forcing order, $G_p$ is a ground-model generic filter. It contains $p$ and is upward closed. For $q,r\in G_p$, choose $q^{\prime }\le p,q$ and $r^{\prime }\le p,r$. The atom property gives $s\le q^{\prime },r^{\prime }$, so $s\in G_p$ is a common strengthening of $q,r$. Thus $G_p$ is a filter in an ordered set. Every dense subset of a forcing order has some $d\le p$, and $d\in G_p$. The defining compatibility predicate is bounded to the ground-model set of conditions, so axiom schema of separation forms $G_p$ inside the ground model. For a nonminimal atom, this filter can strictly contain the principal filter above $p$.

An algebraic structure is a set equipped with specified algebraic operations satisfying selected axioms. An operation's domain is part of the structure: a binary operation on $S$ takes two elements of $S$ and produces another element of $S$. Groups specify one such operation, rings specify addition and multiplication, and vector spaces specify vector addition and scalar multiplication by a separate field.

Different choices of operations can give the same underlying set different structures. The integers form a group under addition, since every integer has an additive inverse. They do not form a group under multiplication: for example, $2$ has no multiplicative inverse among the integers. Even the nonzero integers fail that inverse requirement. In contrast, nonzero rational numbers form a group under multiplication.

Finite multiplication tables make such distinctions visible. Arithmetic modulo $5$ gives a field; arithmetic modulo $6$ gives a commutative ring with zero divisors. In the former, multiplying by any nonzero element permutes all residue classes. In the latter, multiplying by $2$ reaches only $0,2,4$, and $2\cdot 3=0$ even though both factors are nonzero.

The goal is to extract reusable properties rather than memorize isolated tables. A homomorphism preserves the specified operations, and an isomorphism shows that two structures have identical operation patterns after relabeling.

An order is atomless if it has no forcing atom, equivalently every condition has a pair of incompatible forcing conditions below it. For a fixed order belonging to a transitive model, this property is absolute because its quantifiers range over the unchanged set of conditions. Infinite-domain Fn forcing with at least two possible values is atomless: prescribe different values at one fresh coordinate.

An atom is a minimal nonzero Boolean-algebra element. In a Lindenbaum algebra, its nonempty formula class isolates a complete type. A finite Boolean algebra is the power set of its finite atom set. An infinite atomic Boolean algebra need not contain every union of atoms: the finite-cofinite algebra is a simple counterexample.

The supremum of an increasing bounded sequence need not be a maximum of its set of terms. For $a_n=1-1/n$, every term is strictly less than one; the supremum and limit of a sequence are one, but there is no maximum. Every candidate term is beaten by the next. The decreasing sequence $1+1/n$ similarly has infimum one without a minimum.

If an increasing sequence with limit of a sequence $L$ attains $a_k=L$, it must remain there: $L=a_k\le a_n\le L$ for $n\ge k$. The decreasing case is identical with inequalities reversed. Thus attainment of the limit of a sequence is equivalent to an eventually constant tail of a sequence for a convergent monotone sequence. The increasing sequence $0,1/2,1,1,\dots$ has both supremum and maximum one. ^[ref]

An algebra homomorphism onto a semisimple Banach algebra is continuous. To see the spectral mechanism, suppose $a_n\to 0$ and $T(a_n)\to b$. For any $c\in B$, choose $u,v\in A$ with $T(u)=b$ and $T(v)=c$, and set $d=cb$, $d_n=T(v{a}_n)$. Apply the spectral-radius three-circle inequality to $T((1-z)v{a}_n+zvu)$. On the outer circle its norm tends uniformly to $\Vert d\Vert$; on the circle of radius $1/R$ its spectral radius is bounded by $(1+1/R)\Vert v{a}_n\Vert +\Vert vu\Vert /R$. Hence $r(d{)}^2\le \Vert d\Vert \Vert vu\Vert /R$ for every $R>1$, so $r(cb)=0$. The unit criterion for the Jacobson radical gives $b\in J(B)=0$, and the closed graph theorem finishes the proof.

Every set belongs to some level of the constructible hierarchy. The assertion is $\forall x\exists \alpha (x\in L_{\alpha })$, not that each set is definable without parameters. The constructible universe theorem supplies a relative-consistency model of this assertion from ZFC.

The axiom of countable choice allows one to choose an element from each member of a countable family of nonempty sets. It is a restricted form of the axiom of choice. In the usual proof that a countable union of countable sets is countable, it suffices to choose one enumerating injection for each set; if those enumerations are already supplied, that proof requires no further choice.

If a nonempty set carries a relation in which every point has a successor, the axiom of dependent choice supplies an infinite sequence following that relation. Full choice implies this by selecting a successor at every point and applying the recursion theorem on the natural numbers.

Two sets with exactly the same members are equal. In a transitive model, all actual members of its sets are available internally, so ambient extensionality gives the internal axiom as well.

A single class function chooses one member of every nonempty set. In the usual class theory with replacement, it is equivalent to a set-like global well-ordering of the universe. Ordinary choice asserts choice only for each set-sized family separately.

For any two sets there is a set containing exactly those two members. Together with other ZFC axioms it permits ordinary ordered-pair coding. Countable-subset closure implies this axiom for the structure of hereditarily countable sets.

Start with a well-founded universe of sets over an infinite set $A$ of urelements. The full symmetric group of $A$ acts recursively on sets. Retain the objects all of whose descendants have a finite support in a permutation action. This class satisfies set-extensionality, foundation, infinity, pairing, union, separation, replacement and its internal power set axiom. A definable functional image has the finite union of the supports of its domain and parameters as a support; uniqueness makes the image invariant. The family of two-element subsets of $A$ has no choice function: a transposition of two atoms outside the function's finite support fixes their pair and moves either possible choice. For untyped membership with extensionality merely restricted to nonempty objects, atoms are also vacuous subsets; add all atoms to each internal power-set object.