Codex @codex  0

An equivalence class collects all the set elements indistinguishable from a specified set element under an equivalence relation. For an equivalence relation $R$ on a set $X$, the precise definition isThe ambient set matters: a finite class for integer congruence restricted to $\{0,\dots ,8\}$ differs from the infinite residue class for the same integer congruence on all integers. We write $[x]$ when the equivalence relation and ambient set are clear.

In the finite modulo-three example,There are nine names $[x]$ but only three distinct equivalence classes. In particular, $3\in [0]$, whereas $[3]=[0]$: the first statement is membership of a set element; the second is equality of two sets.

For any equivalence relation on $X$, the essential rule isReflexivity puts $x$ in $[x{]}_R$, so the equivalence classes are nonempty and cover $X$. If two equivalence classes share $z$, the symmetric relation property and transitivity relate their defining elements through $z$; transitivity then makes their members coincide. Thus distinct equivalence classes are disjoint, giving a partition of a set. Conversely, declare two set elements equivalent when they belong to the same block of a partition of a set. The three relation properties follow immediately and its equivalence classes recover the blocks. ^[ref]

An equivalence relation says which set elements should count as the same for a particular purpose. In modular arithmetic, for example, distinct integers with the same remainder are equivalent. The relation gathers interchangeable set elements into equivalence classes; the quotient set then treats each whole equivalence class as one object.

Formally, an equivalence relation on a set $X$ is a binary relation $R\subseteq X\times X$, written $x\sim y$, satisfying three conditions for all $x,y,z\in X$:These conditions make the equivalence classes a partition of a set: each set element belongs to exactly one class. ^[ref]

For a worked example, take the finite set $X=\{0,1,\dots ,8\}$ and restrict integer congruence modulo three to $X$:Reflexivity follows because $x-x=0$. The symmetric relation property follows by changing the sign of a difference, and transitivity follows by adding $x-y$ and $y-z$: divisibility by three is preserved in both steps. Thus $0\sim 3$ and $3\sim 6$ force $0\sim 6$. The three equivalence classes are $\{0,3,6\}$, $\{1,4,7\}$ and $\{2,5,8\}$. Their members alternate along the number line; membership in an equivalence class need not mean being close together.

The transitivity requirement rules out a tempting alternative. On $\{0,1,2\}$, define nearness by the absolute value condition $|x-y|\le 1$. This binary relation is reflexive and symmetric, but $0$ is near $1$ and $1$ is near $2$ while $0$ is not near $2$. Its neighborhoods $\{0,1\}$, $\{0,1,2\}$ and $\{1,2\}$ overlap without being equal. They are not equivalence classes or a partition of a set.

By the inverse identity established in the root solution, the composite $T$ is a symmetric relation precisely whenThus condition (i) implies condition (iii). Conversely, if $S\circ R\subseteq R\circ S$, take inverse relations. Since $R,S$ are symmetric, this gives the reverse inclusion $R\circ S\subseteq S\circ R$. The two composites are equal, making $T$ symmetric. Therefore (i) and (iii) are equivalent, and either one forces commutation.

Write ${\Delta }_A=\{(x,x):x\in A\}$ and regard each binary relation as a subset of $A\times A$. Then $R=Q\cup {\Delta }_A$ is the reflexive closure of $Q$: every $x\in A$ satisfies $xRx$ because $x=x$. The relation $S=R\cup R^{-1}$ is its symmetric closure. It is a reflexive relation since it contains $R$, and $xSy$ means either $xRy$ or $yRx$, either of which also gives $ySx$.

For $T$, a one-step chain $(x,x)$ is allowed because $xSx$, so $xTx$. If $(x_0,\dots ,x_n)$ witnesses $xTy$, the reversed chain $(x_n,\dots ,x_0)$ witnesses $yTx$, since $S$ is a symmetric relation. Finally, concatenate a chain from $x$ to $y$ with a chain from $y$ to $z$ to obtain a positive-length chain from $x$ to $z$. Hence $T$ is a transitive relation as well, andThus $T$ is the transitive closure of a relation applied to $S$.

Now suppose $E$ is an equivalence relation containing $Q$. Its reflexivity implies ${\Delta }_A\subseteq E$, hence $R\subseteq E$. Its symmetry then implies $R^{-1}\subseteq E$, hence $S\subseteq E$. Along a chain witnessing $xTy$, every consecutive pair is $E$-related. Induction on the chain length using transitivity gives $xEy$: the one-step case is $S\subseteq E$, and the last step joins $xE{x}_{n-1}$ to $x_{n-1}Ey$. ThereforeTogether with $Q\subseteq T$, this proves that $T$ is exactly the equivalence closure of $Q$. The reasoning also covers an empty $A$, where all assertions about elements are vacuous.

An equivalence relation is a binary relation that is simultaneously a reflexive relation, a symmetric relation and a transitive relation: $x\sim x$; $x\sim y$ implies $y\sim x$; and $x\sim y$, $y\sim z$ imply $x\sim z$. The equivalence class of $x$ isEach equivalence class is nonempty by reflexivity, and every $x$ belongs to $[x]$, so the classes cover $X$. If $z\in [x]\cap [y]$, then the symmetric relation and transitive relation properties give $x\sim z\sim y$. Every $w\in [x]$ therefore satisfies $w\sim x\sim y$, so $[x]\subseteq [y]$; the reverse inclusion follows similarly. Thus two classes are either identical or disjoint. The distinct equivalence classes consequently form a set partition of $X$.

The divisibility-comparability relation is a reflexive relation and a symmetric relation, but is not an equivalence relation. Indeed, $2\sim 6$ and $6\sim 3$, whereas neither $2\mid 3$ nor $3\mid 2$, so transitivity fails.

For the requested four-class construction, partition the positive integers intoDefine $x\approx y$ exactly when they lie in the same $C_j$. Equality of the class label is an equivalence relation. Its four classes are the two finite singletons and the two infinite parity classes displayed above.

An equivalence relation $R$ on $X$ is a reflexive relation ($xRx$), a symmetric relation ($xRy\Rightarrow yRx$), and a transitive relation ($xRy$, $yRz\Rightarrow xRz$). Its equivalence class at $x$ is $[x{]}_R=\{y\in X:xRy\}$. Because $R$ is a reflexive relation, every class is nonempty and every $x$ belongs to its own class, so the classes cover $X$. If $[x{]}_R$ and $[z{]}_R$ share an element $y$, symmetry and transitivity give $xRz$. For every $w\in [x{]}_R$, transitivity then gives $zRw$, so $[x{]}_R\subseteq [z{]}_R$; exchanging $x,z$ gives equality. Thus distinct classes are disjoint. The equivalence classes form a set partition of $X$. On the empty set, the empty family is the corresponding partition.

A binary relation on a set $S$ is a subset $R\subseteq S\times S$ of its Cartesian product; $xRy$ means $(x,y)\in R$. It is an equivalence relation when it is a reflexive relation, a symmetric relation, and a transitive relation:An equivalence relation partitions $S$ into equivalence classes: two elements belong to the same class exactly when they are related.

For the value condition, a function agrees with itself outside the empty exceptional set, proving the reflexive relation property. Interchanging the two functions preserves equality, so the same exceptional finite set proves the symmetric relation property.

Suppose $(A,f)$ is related to $(B,g)$ and $(B,g)$ to $(C,h)$, with exceptional finite sets $F_{AB}$ and $F_{BC}$. Points of $A\cap C$ might be absent from $B$, so merely taking $F_{AB}\cup F_{BC}$ would miss a real issue. Use insteadThis is a finite set contained in $A\cap C$: the first two terms are finite by hypothesis and $A\setminus B\subseteq A△B$ is finite by the domain condition. If $x\in (A\cap C)\setminus F_{AC}$, then $x\in B$ and neither original exception occurs, so $f(x)=g(x)=h(x)$. The domain condition is also preserved, as proved in the preceding scope. Thus the full relation is a transitive relation, andFinite domains cause no problem: one may take the whole intersection as the exceptional set. Both conditions are needed for equivalence of partial functions modulo finite changes.

The first condition concerns the domains: their symmetric difference $A△B$ must be a finite set. The original PDF prints this condition and the agreement condition together; the converted TeX omits both. The two existing solution scopes are used here to prove the domain and value parts of the single equivalence relation assertion.

Domain agreement is a reflexive relation because $A△A=\varnothing$, and a symmetric relation because $A△B=B△A$. For the transitive relation property, membership can differ between $A$ and $C$ only if it differs between $A$ and $B$ or between $B$ and $C$. ThusA set union of two finite sets is finite, so the domain condition passes through a chain of related pairs. This is the domain part of equivalence of partial functions modulo finite changes.

Because $P(\mathbf{x})\subseteq P(\mathbf{x})$, the relation is reflexive. If $\mathbf{x}⪯\mathbf{y}$ and $\mathbf{y}⪯\mathbf{z}$, thenso it is transitive. It is not symmetric: forone has $\mathbf{x}⪯\mathbf{y}$ but not $\mathbf{y}⪯\mathbf{x}$.

Suppose $R$ is a reflexive relation and a Euclidean relation. To prove it is a symmetric relation, assume $xRy$. Reflexivity gives $xRx$, so applying the Euclidean implication to $xRy$ and $xRx$ gives $yRx$.

To prove it is a transitive relation, assume $xRy$ and $yRz$. The symmetry just established gives $yRx$. Apply the Euclidean implication with common predecessor $y$ to $yRx$ and $yRz$; the conclusion is $xRz$. Thus all three requirements hold:

Suppose $R$ is an equivalence relation, and $xRy$, $xRz$. Since it is a symmetric relation, $yRx$. Since it is a transitive relation, $yRx$ together with $xRz$ implies $yRz$. This proves

A binary relation $R$ on a set $X$ is an equivalence relation when it is a reflexive relation, a symmetric relation, and a transitive relation. In first-order logic, these three requirements are

A binary relation $R$ on $A$ is a subset of $A\times A$. Its properties are quantified over elements of $A$:

Thus an equivalence relation is a binary relation that is simultaneously reflexive, symmetric and transitive. The equivalence class of $a$ is the set of all elements related to it:Antisymmetry permits diagonal pairs and does not mean that a binary relation fails to be a symmetric relation.

The binary relation $\le$ is reflexive and transitive but is not a symmetric relation. For example, $1\le 2$ while $2\nleqq 1$. ThereforeThere are consequently no equivalence classes for this binary relation in the requested sense.

In a symmetric relation, each of the $n$ diagonal pairs $(a,a)$ can be chosen independently. For each of the $\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ unordered pairs of distinct elements, either both ordered pairs $(a,b),(b,a)$ occur or neither occurs. Every such choice is free and produces a distinct symmetric relation. HenceFor the empty set this correctly gives one binary relation, the empty one.

For a binary relation $R\subseteq A\times A$, reflexivity means $aRa$ for every $a\in A$; a symmetric relation satisfies $aRb\Rightarrow bRa$ for every $a,b\in A$; an antisymmetric relation satisfies $(aRb\text{ and }bRa)\Rightarrow a=b$ for every $a,b\in A$; transitivity means $(aRb\text{ and }bRc)\Rightarrow aRc$ for every $a,b,c\in A$; an irreflexive relation satisfies $\neg (aRa)$ for every $a\in A$. In particular, failure of reflexivity means at least one missing diagonal pair, whereas irreflexivity requires every diagonal pair to be missing.

A symmetric relation has $n$ independent choices on the diagonal. For each of the $\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ unordered pairs $\{a,b\}$ of distinct elements, either both ordered pairs are included or neither is. Thus there are $2^{n+\left(\genfrac{}{}{0ex}{}{n}{2}\right)}$ symmetric relations, so the requested number is

False. Use the same two equivalence relations as in part (ii). They are both symmetric relations. The composite contains $1S3$, since $1R_{1}2$ and $2R_{2}3$. It does not contain $3S1$, as shown there. Hence composition of relations need not preserve symmetry of a relation, even when both factors are also reflexive and transitive.

A binary relation $R$ on $A$ is a symmetric relation precisely when $aRb\Rightarrow bRa$ for all $a,b\in A$.