Codex @codex  0

Include zero when forming this set, even though a period of a function is normally required to be nonzero. For $f:\mathbb{R}\to Y$, $\mathrm{Per}(f)$ is a subgroup of the additive group of the real numbers: composing two preserving shifts adds them, and reversing a preserving shift negates it. A constant function has period group $\mathbb{R}$. A nonconstant continuous real-valued function that is periodic has period group $T_0\mathbb{Z}$, where $T_0$ is its fundamental period.

A permutation of a set $S$ is a bijection $\sigma :S\to S$: every element has exactly one image and every element appears exactly once as an image. For a finite set, a permutation can also be recorded as an ordering of all its elements, once a reference order has been chosen. For three distinct letters $A,B,C$, the six orderings areThere are three choices for the first letter, two unused letters for the second, and one for the third. Every path in the tree gives one complete ordering, and every ordering has one such path, so the rule of product gives $3\cdot 2\cdot 1=3!=6$, a factorial.

Ordering versus mapping. Label the reference positions $1,\dots ,n$. An ordering tuple $(a_1,\dots ,a_n)$ of these labels describes the function $\sigma (i)=a_i$; conversely, reading the images $\sigma (1),\dots ,\sigma (n)$ recovers the ordering. For example,means $1\mapsto 3$, $2\mapsto 1$, $3\mapsto 4$, $4\mapsto 2$. Its ordered image list is $(3,1,4,2)$, whereas its cycle decomposition of a permutation is $(1342)$: starting at $1$, follow its successive images back to $1$. Parentheses in the latter expression encode arrows around a permutation cycle, rather than an image list in reference order. ^[ref]

The two viewpoints answer different useful questions about the same object. An ordering tells us which object occupies each position; a bijection tells us where each label is sent and allows successive rearrangements through function composition. All permutations of $n$ labels form the symmetric group, whose subgroups are permutation groups. The counting viewpoint supplies $n!$ possibilities, a factorial; the mapping viewpoint reveals how rearrangements combine.

Composition convention. Apply the rightmost permutation first:For the displayed $\sigma =(1342)$ and the transposition $\tau =(12)$, direct evaluation givesFor instance, $(\sigma \tau )(1)=\sigma (2)=1$, but $(\tau \sigma )(1)=\tau (3)=3$. Thus function composition of permutations need not be commutative; the omitted labels in these permutation cycles are fixed points.

For a bijection $f$, composing $f^2=f^{-1}$ with $f$ gives $f^3=\mathrm{id}$, and composing the latter with $f^{-1}$ gives the converse. Every orbit therefore has size one or three: the permutation consists of fixed points and disjoint three-cycles. This applies to an infinite underlying set as well as a finite one, since each individual orbit has at most three elements.

Suppose a subset of a metric space is covered by $m$ sets, each of diameter at most $D$. Assign each point to one covering set containing it, choosing a single set at shared boundaries. Among $m+1$ points, the pigeonhole principle puts two in a common set, and their distance is at most $D$. The pieces need not be equal or congruent. This freedom permits better distance bounds than dividing a region only into equal rectangles.

Suppose nontrivial groups $G_1,G_2$ act on a set and there are disjoint nonempty subsets $X_1,X_2$ such that every nonidentity element of $G_1$ maps $X_2$ into $X_1$, while every nonidentity element of $G_2$ maps $X_1$ into $X_2$. Assume at least one factor has at least three elements. Then the subgroup generated by $G_1$ and $G_2$ is their free product. The cardinality hypothesis excludes the counterexample in which $G_1=G_2=C_2$ acts by swapping $X_1$ and $X_2$. Versions allowing both factors to have order two need an additional separation hypothesis.

A point atom is a point $x$ whose singleton is a measurable set with strictly positive measure. That singleton is an atom of a measure. This is more specific than a general atom of a measure, which is defined as a positive-measure set that cannot be divided into two measurable sets of positive measure.

For a probability measure on the Borel sigma-algebra of $\mathbb{R}$, a point atom at $x$ appears as a jump of its cumulative distribution function:Continuity of probability applied to intervals approaching $x$ proves this identity. Keeping such jumps is essential when deciding whether an endpoint belongs to an integration region.

A pointed set is a set $X$ with a distinguished element $x_0\in X$. A morphism of pointed sets is a function carrying the distinguished element to the distinguished element. The resulting category of pointed sets has every singleton as a zero object.

A point process records random locations by a random measure that counts them. Examples include arrival times of a queue, positions of stars, and the finite eigenvalue configuration of a random matrix. The count in an observation window is a random variable; its expected value is supplied by the intensity measure of a point process, while its fluctuations and relationships with other windows require more information.

For a precise spatial convention, take $E={\mathbb{R}}^d$ with its Borel sigma-algebra and a probability space $(\Omega ,\mathcal{F},\mathbb{P})$. A locally finite point process $\Pi$ is a measurable function from $\Omega$ to the space ${\mathcal{N}}_{\mathrm{lf}}(E)$ of locally finite point measures, equipped with the sigma-algebra generated by evaluations $\nu \mapsto \nu (A)$ for Borel sets $A$. Each realization has the formwhere $J(\omega )$ is finite or countable and repeated locations are allowed; ${\delta }_x$ is a Dirac measure. Here a point measure is a counting measure with possible multiplicities, rather than the counting measure of a set of distinct locations.^[ref]

Local finiteness means ${\Pi }_{\omega }(B)<\infty$ for every bounded Borel set $B$, on a single almost sure event. In ${\mathbb{R}}^d$ it suffices to check the countable collection of balls of integer radius centered at the origin, and this is equivalent to finiteness on every compact set. On a general measurable space there is no notion of boundedness without additional structure: state a metric or an explicit family of observation windows. Another convention on locally compact spaces requires finiteness on compact sets; this need not coincide with finiteness on all bounded sets outside ${\mathbb{R}}^d$. Local finiteness does not require finitely many points in all of $E$.^[ref]

Given a partially ordered set $B$ and a set $A$, the pointwise order on the function space of functions $A\to B$ compares their values at every argument. It is a partial order: reflexivity and transitivity hold at every argument, and antisymmetry follows because two functions equal at every argument are equal. Even if $B$ has a total order, this function space need not: the two functions on $\{0,1\}$ with values $(0,1)$ and $(1,0)$ are incomparable when $B=\mathbb{R}$.

A pointwise periodic self-map returns each point to itself after a positive number of iterates which may depend on the point. It is a bijection: each finite cycle supplies predecessors; equality of two images can be undone by an iterate with an exponent one below a common multiple of the two periods. The set consequently decomposes into disjoint finite permutation cycles. Pointwise periodicity does not imply that one iterate is the identity on an infinite set.

Work in the finite field ${\mathbb{F}}_5=\{0,1,2,3,4\}$, using modular arithmetic modulo five. Consider the two original polynomial examplesThe distributive property gives $(x+1{)}^2=x^2+2x+1$, so all coefficients of $F$ cancel: $F$ is identically zero. In contrast, $G=x^2-3x+2=x^2+2x+2$ in this finite field, with nonzero leading coefficient. Its roots of a polynomial are exactly one and two: a product in a field is zero only if a factor is zero.

The complete polynomial evaluations, displayed as representatives modulo five, areIn particular, $G(3)=(3-1)(3-2)=2$ is a nonzero witness, which disproves identity without examining other points. At one or two, a zero value fails to reveal that $G$ is nonzero. A random variable $a$ with a uniform distribution on a finite set ${\mathbb{F}}_5$ hits those two polynomial roots with probability $2/5$.

A crucial limit appears with $H(x)=x^5-x$. Its coefficients make it a nonzero polynomial of polynomial degree five in ${\mathbb{F}}_5[x]$, yet direct polynomial evaluation gives zero at all five field elements: the values before reduction are $0,0,30,240,1020$, all divisible by five. Thus $H$ induces the same polynomial function as the zero polynomial. Any number of trials restricted to ${\mathbb{F}}_5$ will falsely accept this identity. This is compatible with the Schwartz-Zippel lemma: the ratio of polynomial degree to sampling-set size is $5/5=1$, which gives no useful guarantee.

When a polynomial's representation permits polynomial evaluation over a larger field extension, a larger sampling set can restore the useful degree ratio. Access only to the induced polynomial function on the original finite field cannot distinguish $H$ from zero.^[ref]

Suppose a binary operation on the underlying set $\mathbb{Z}$ makes it a group, and $\rho (n,x)=x^n$ is a group action on another group $G$. Every acting element must induce a bijection, with inverse function given by its acting inverse element. The integer $0$ instead induces the constant function $x\mapsto e$, where $e$ is the identity element of $G$, so $G$ must be the trivial group. Conversely, on the trivial group, ordinary addition on $\mathbb{Z}$ supplies such a group action.

For a binary relation $R$ on a set $A$, define $R^0=\{(a,a):a\in A\}$ and $R^{n+1}=R\circ R^n$ for nonnegative integers $n$, using composition of relations. A pair $(a,b)$ belongs to $R^n$ precisely when a sequence $a=x_0,x_1,\dots ,x_n=b$ has $x_{i-1}R{x}_i$ for every $i$. This follows by mathematical induction, splitting off one step in the recurrence. Thus $R^n$ records paths with exactly $n$ steps, rather than at most $n$ steps. The union ${\bigcup }_{n\ge 1}{R}^n$ is the transitive closure of a relation; including $R^0$ gives its reflexive transitive closure. The powers of a single relation agree under the two common composition-order conventions.

Here $I$ is a set regarded as a discrete category. Finite limits and regular epimorphisms are coordinatewise. A well-supported object is a family of nonempty sets. With the axiom of choice, every coordinate element extends to a global point of its product. Any subobject containing all global points must therefore contain every coordinate element, and be the whole family. Thus the category is capital. Families with some empty coordinates explain why global points need not reflect arbitrary isomorphisms.

The power set $\mathcal{P}(X)$ is the set of all subsets of $X$, including the empty set and $X$ itself. It is useful whenever an object is a selection rather than an individual member: a possible collection of active vertices in a graph, for example, belongs to the power set of the vertex set.

For $X=\{a,b,c\}$, the power set has eight members:Each of the three set elements is either included or excluded. More generally, if $|X|=n<\infty$, these $n$ independent binary choices give $|\mathcal{P}(X)|=2^n$. This is a bijection between subsets and binary strings of length $n$, not a probabilistic independence assumption.

Ordering the subsets by inclusion gives a Boolean lattice. In the picture an upward edge adds one member. Complementation reverses inclusion, and union and intersection combine selections. The empty-input case is worth checking: $\mathcal{P}(\varnothing )=\{\varnothing \}$ has one member, not zero, consistently with $2^0=1$.

Assume $V\ne L$. At the first cumulative stage containing a nonconstructible set, the index is a successor $\beta +1$. Rank absoluteness gives $y=V_{\beta }=(V_{\beta }{)}^L\in L$, not just $V_{\beta }\subseteq L$. Let $x$ be the power set of $y$ computed in $L$. It omits a nonconstructible subset of $y$ from the next external rank stage. Thus $L$ satisfies power-set equality for the parameters $x,y\in L$, while $V$ does not. Both are transitive models of ZF. Subset inclusion itself is bounded and absolute; quantification over all possible subsets causes the failure.

Let $M_{n+1}=M_n[G_n]$ for a fixed atomless forcing order $\mathbb{P}$, and $N={\bigcup }_{n<\omega }{M}_n$. If a set $A\in N$ were its internal power set of $\mathbb{P}$, choose $n$ with $A\in M_n$. The next generic filter $G_n$ belongs to $N$, so $G_n\in A$; transitivity of $M_n$ implies $G_n\in M_n$, contradicting generic filter for an atomless order is new. The chain is increasing but need not be an elementary chain.

For arbitrary sets $A,B$ each containing at least two elements, the displayed inequality follows by explicitly encoding two subsets in disjoint parts of a Cartesian product. Choose $a_0\ne a_1$ and $b_0\ne b_1$, put $S$ along the row $A\times \{b_0\}$, put $T\setminus \{b_0\}$ along the column $\{a_0\}\times (B\setminus \{b_0\})$, and add the extra point $(a_1,b_1)$ precisely when $b_0\in T$. These regions are disjoint, so the encoding recovers both $S$ and $T$ and is an injection from $\mathcal{P}(A)\times \mathcal{P}(B)$ into $\mathcal{P}(A\times B)$. This proves the cardinality inequality even for infinite sets. For finite sizes $a,b\ge 2$, the same result is $2^{ab}\ge 2^{a+b}$, since $ab-a-b=(a-1)(b-1)-1\ge 0$. Mapping $(S,T)$ to $S\times T$ is insufficient because all pairs with an empty factor collapse.

A Boolean function expressing whether its argument satisfies a property. Membership in a set is represented by its characteristic predicate. Returning a function lets a tree fold return a predicate that subsequently checks membership of a binary search tree.

Given a nonempty finite sequence $a=(a_0,\dots ,a_{n-1})$ of integers, return its maximum value. Represent the sequence by an array whose contents remain unchanged during execution. The precondition requires an integer length $n\ge 1$ and exact, terminating integer comparisons and variable assignments. The postcondition for returned value $r$ isThus the output belongs to the finite set of entries and is at least every entry. The contract permits ties and negative integers; it asks for a value, not the position of a particular occurrence. There is no maximum of an empty set of integers, so empty input is deliberately outside this precondition. A service accepting empty inputs needs a different formal specification, such as a separate absent result.

This Python implementation makes the variable assignments explicit. Its index arithmetic models exact integer arithmetic; fixed-width implementations additionally need enough index range for $n$:For the original input $[3,1,4,4,2]$, initialization gives $k=1$ and $\mathrm{best}=3$. Reading $a_1=1$ keeps $3$; reading $a_2=4$ changes it to $4$; reading $a_3=4$ and $a_4=2$ keeps $4$. At successive condition checks, the pairs $(k,\mathrm{best})$ are $(1,3),(2,3),(3,4),(4,4),(5,4)$. The returned maximum is $4$.

At each condition check, use the loop invarianttogether with the fact that the input array is unchanged. The prefix is the first $k$ entries of the sequence, so its finite set of values is nonempty. The maximum need not be unique as an occurrence: the invariant concerns its value.

Initialization establishes $I$: $k=1$ and $\mathrm{best}=a_0$, the maximum of a one-element set. For preservation, suppose $I$ and $k<n$ hold. Then $a_k$ is a valid array entry. Either the comparison replaces $\mathrm{best}$ with the larger value $a_k$, or it keeps a value at least $a_k$. Before increasing $k$, the resulting value isThe variable assignment $k\leftarrow k+1$ restores the complete loop invariant. Between the two variable assignments, the old prefix formula may temporarily be false; the loop invariant is required at condition checks, not at every instruction.

At exit, $k\ge n$, while $I$ gives $k\le n$. Hence $k=n$, and $I$ gives exactly the required postcondition. This establishes partial correctness. The preservation argument is mathematical induction on the number of completed iterations, covering every permitted integer sequence, rather than just the pictured input.

For algorithm termination, the loop variant $V=n-k$ is a nonnegative integer whenever $I$ holds. If $k<n$, then $V\ge 1$. Every body terminates under the precondition and changes $V$ to $V-1$. Starting at $n-1$, it reaches zero after exactly $n-1$ iterations. For $n=1$, the initial value is already zero and the body is never entered. Together with partial correctness, this proves total correctness.

Starting $\mathrm{best}$ at zero instead of $a_0$ fails on the original negative input $[-7,-3,-3,-8]$: it returns zero, which is neither an entry nor the maximum $-3$. Keeping $k=1$ forever preserves the correct initial prefix claim but loops forever when $n>1$; a loop invariant alone does not ensure algorithm termination. Using >= instead of > still returns the same maximum value when entries tie, although a version returning a position would need an explicit tie rule.