closure of a relation with respect to a property
Introduction
Fix a set $A$. A property $\mathcal{P}$ of $n$ary relations on a set $A$ may be thought of as some subset of the set of all $n$ary relations on $A$. Since an $n$ary relation^{} is just a subset of ${A}^{n}$, $\mathcal{P}\subseteq P({A}^{n})$, the powerset of ${A}^{n}$. An $n$ary relation is said to have property $\mathcal{P}$ if $R\in \mathcal{P}$.
For example, the transitive property is a property of binary relations on $A$; it consists of all transitive^{} binary relations on $A$. Reflexive^{} and symmetric properties are sets of reflexive and symmetric binary relations on $A$ correspondingly.
Let $R$ be an $n$ary relation on $A$. By the closure^{} of an $n$ary relation $R$ with respect to property $\mathcal{P}$, or the $\mathcal{P}$closure of $R$ for short, we mean the smallest relation $S\in \mathcal{P}$ such that $R\subseteq S$. In other words, if $T\in \mathcal{P}$ and $R\subseteq T$, then $S\subseteq T$. We write ${\mathrm{Cl}}_{\mathcal{P}}(R)$ for the $\mathcal{P}$closure of $R$.
Given an $n$ary relation $R$ on $A$, and a property $\mathcal{P}$ on $n$ary relations on $A$, does ${\mathrm{Cl}}_{\mathcal{P}}(R)$ always exist? The answer is no. For example, let $\mathcal{P}$ be the antisymmetric property of binary relations on $A$, and $R={A}^{2}$. For another example, take $\mathcal{P}$ to be the irreflexive^{} property, and $R=\mathrm{\Delta}$, the diagonal relation on $A$.
However, if ${A}^{n}\in \mathcal{P}$ and $\mathcal{P}$ is closed under^{} arbitrary intersections^{}, then $\mathcal{P}$ is a complete lattice^{} according to this fact (http://planetmath.org/CriteriaForAPosetToBeACompleteLattice), and, as a result, ${\mathrm{Cl}}_{\mathcal{P}}(R)$ exists for any $R\subseteq {A}^{n}$.
Reflexive, Symmetric, and Transitive Closures
From now on, we concentrate on binary relations on a set $A$. In particular, we fix a binary relation $R$ on $A$, and let $\mathcal{X}$ the reflexive property, $\mathcal{S}$ the symmetric property, and $\mathcal{T}$ be the transitive property on the binary relations on $A$.
Proposition 1.
Arbitrary intersections are closed in $\mathrm{X}$, $\mathrm{S}$, and $\mathrm{T}$. Furthermore, if $R$ is any binary relation on $A$, then

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${R}^{=}:={\mathrm{Cl}}_{\mathcal{X}}(R)=R\cup \mathrm{\Delta}$, where $\mathrm{\Delta}$ is the diagonal relation on $A$,

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${R}^{\leftrightarrow}:={\mathrm{Cl}}_{\mathcal{S}}(R)=R\cup {R}^{1}$, where ${R}^{1}$ is the converse^{} of $R$, and

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${R}^{+}:={\mathrm{Cl}}_{\mathcal{T}}(R)$ is given by
$$\bigcup _{n\in \mathbb{N}}{R}^{n}=R\cup (R\circ R)\cup \mathrm{\cdots}\cup \underset{n\mathit{\text{fold power}}}{\underset{\u23df}{(R\circ \mathrm{\cdots}\circ R)}}\cup \mathrm{\cdots},$$ where $\circ $ is the relational composition^{} operator.

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${R}^{*}:={R}^{=+}={R}^{+=}$.
${R}^{=}$, ${R}^{\leftrightarrow}$, ${R}^{+}$, and ${R}^{*}$ are called the reflexive closure^{}, the symmetric closure, the transitive closure^{}, and the reflexive transitive closure of $R$ respectively. The last item in the proposition^{} permits us to call ${R}^{*}$ the transitive reflexive closure of $R$ as well (there is no difference^{} to the order of taking closures). This is true because $\mathrm{\Delta}$ is transitive.
Remark. In general, however, the order of taking closures of a relation is important. For example, let $A=\{a,b\}$, and $R=\{(a,b)\}$. Then ${R}^{\leftrightarrow +}={A}^{2}\ne \{(a,b),(b,a)\}={R}^{+\leftrightarrow}$.
Title  closure of a relation with respect to a property 

Canonical name  ClosureOfARelationWithRespectToAProperty 
Date of creation  20130322 17:36:26 
Last modified on  20130322 17:36:26 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  15 
Author  CWoo (3771) 
Entry type  Definition 
Classification  msc 08A02 
Related topic  Property2 
Defines  reflexive closure 
Defines  symmetric closure 
Defines  transitive closure 
Defines  reflexive transitive closure 