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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.

For each unordered pair $\{a,b\}$ with $a\ne b$, antisymmetry allows three choices: neither direction, only $(a,b)$, or only $(b,a)$. This contributes $3^{\left(\genfrac{}{}{0ex}{}{n}{2}\right)}$. The diagonal is unrestricted by antisymmetry, but being not irreflexive requires a nonempty subset of the $n$ diagonal pairs. There are $2^n-1$ such choices. HenceThis allows some diagonal pairs to be absent; “not irreflexive” does not require a reflexive relation.

There are $n^2$ possible ordered pairs and hence $2^{n^2}$ binary relations. A reflexive relation must contain the $n$ diagonal pairs, while its remaining $n^2-n$ pairs are unrestricted. Subtracting the reflexive relations givesbinary relations that are not reflexive.

Use the convention that $R\circ S$ applies $S$ first. Reflexivity inserts an identity step: if $aRb$, use $aSa$ to obtain $(a,b)\in R\circ S$; if $aSb$, use $bRb$ for the same conclusion. Thus $R\cup S\subseteq R\circ S$. By monotonicity of relation composition,For strictness, take $A=\{0,1,2,3\}$ and let $R=S$ consist of the identity pairs together with $(0,1),(1,2),(2,3)$. Both are reflexive relations. The left side permits at most two forward chain steps, so does not contain $(0,3)$. The right side permits four steps, including $0\to 1\to 2\to 3\to 3$, so does contain it. Equality therefore need not hold.

Check each partial order axiom. For every $(a,b)$, $a\le a$ and $b\ge b$, so $S$ is a reflexive relation. If both $(a,b)S(c,d)$ and $(c,d)S(a,b)$, the first coordinates satisfy $a\le c\le a$ and the second coordinates satisfy $b\ge d\ge b$. Therefore $a=c$ and $b=d$, which proves $S$ is an antisymmetric relation.

Finally, if $(a,b)S(c,d)$ and $(c,d)S(e,f)$, then $a\le c\le e$ and $b\ge d\ge f$. Hence $(a,b)S(e,f)$, making $S$ a transitive relation. All three axioms hold, so $S$ is a partial order. It is a product order, with the usual order on the first coordinate and the reversed order on the second.

A binary relation $R$ on $A$ is a reflexive relation precisely when $aRa$ for every $a\in A$.

The reflexive closure of a binary relation $R\subseteq A\times A$ is $R\cup {\Delta }_A$, where ${\Delta }_A=\{(a,a):a\in A\}$. Every reflexive relation containing $R$ must contain the diagonal, so this is the least such extension.

The least reflexive relation and transitive relation containing $R$. Here $R^0$ is the identity relation. Membership means that a finite sequence of zero or more $R$-steps connects the ordered pair. Adding paths can create reverse connections that were absent in $R$, so initial antisymmetry need not be preserved.