Let $A$ be a set and $R$ be an $n$ -ary relation on $A$ , $n\ge 1$ . A subset $B$ of $A$ is said to be closed under $R$ , or $R$ -closed, if, whenever $b_1,\ldots,b_{n-1}\in B$ , and $(b_1,\ldots,b_{n-1},b_n)\in R$ , then $b_n\in B$ .

Note that if $R$ is unary, then $B$ is $R$ -closed iff $R\subseteq B$ .

More generally, let $A$ be a set and $\mathcal{R}$ a set of (finitary) relations on $A$ . A subset $B$ is said to be $\mathcal{R}$ -closed if $B$ is $R$ -closed for each $R\in\mathcal{R}$ .

Given $B\subseteq A$ with a set of relations $\mathcal{R}$ on $A$ , we say that $C\subseteq A$ is an $\mathcal{R}$ -closure of $B$ if

1. $B\subseteq C$ ,
2. $C$ is $\mathcal{R}$ -closed, and
3. if $D\subseteq A$ satisfies both 1 and 2, then $C\subseteq D$ .

By condition 3, $C$ , if exists, must be unique. Let us call it the $\mathcal{R}$ -closure of $B$ , and denote it by $\operatorname{Cl}_{\mathcal{R}}(B)$ . If $\mathcal{R}=\lbrace R\rbrace$ , then we call it the $R$ -closure of $B$ , and denote it by $\operatorname{Cl}_R(B)$ correspondingly.

Here are some examples.

1. Let $A=\mathbb{Z}$ , $B=\lbrace 5\rbrace$ , and $R$ be the relation that $mRn$ whenever $m$ divides $n$ . Clearly $B$ is not closed under $R$ (for example, $(5,10)\in R$ but $10\notin B$ ). Then $\operatorname{Cl}_{R}(B) = 5\mathbb{Z}$ . If $R$ is instead the relation $\le$ , then $\operatorname{Cl}_{\le}(B)=\lbrace n\in A\mid n\ge 5\rbrace$ .
2. This is an example where $R$ is in fact a function (operation). Suppose $A=\mathbb{Z}$ and $R$ is the binary operation subtraction $-$ . Suppose $B=\lbrace 3,5\rbrace$ . Then $\operatorname{Cl}_{-}(B)=A$ . To see this, set $C=\operatorname{Cl}_{-}(B)$ . Note that $2=5-3\in C$ so $1=3-2$ as well as $-1=2-3\in C$ . This means that if $n\in C$ , both $n-1$ and $n+1\in C$ . By induction, $C=\mathbb{Z}$ . In general, if $B=\lbrace p,q\rbrace$ , where $p,q$ are coprime, then $\operatorname{Cl}_{-} (B) =\mathbb{Z}$ . This is essentially the result of the Chinese Remainder Theorem.
3. If $R$ is unary, then the $R$ -closure of $B\subseteq A$ is just $B\cup R$ . When every $R\in \mathcal{R}$ is unary, then the $\mathcal{R}$ -closure of $B$ in $A$ is $(\bigcup \mathcal{R}) \cup B$ .

Proof. Let $S_{\mathcal{R}}(B)$ be the set of subsets of $A$ satisfying the defining conditions 1 and 2 of $\mathcal{R}$ -closures above, partially ordered by $\subseteq$ . If $\mathcal{C} \subseteq S_{\mathcal{R}}(B)$ , then $\bigcap \mathcal{C} \in S_{\mathcal{R}}(B)$ . To see this, we break the statement down into cases:

• In the case when $\mathcal{C}=\varnothing$ , we have $\bigcap \mathcal{C} = A\in S_{\mathcal{R}}(B)$ .
• When $C:=\bigcap \mathcal{C}\ne \varnothing$ , pick any $n$ -ary relation $R\in \mathcal{R}$ .
  1. If $n=1$ , then, since aach $D\in \mathcal{C}$ is $R$ -closed, $R\subseteq D$ . Therefore, $R\subseteq \bigcap \mathcal{C} = \bigcap \lbrace D \mid D \in \mathcal{C}\rbrace = C$ . So $C$ is $R$ -closed.
  2. If $n>1$ , pick elements $c_1,\ldots, c_{n-1}\in C$ such that $(c_1,\ldots, c_n)\in R$ . As each $c_i\in D$ for $i=1,\ldots, n-1$ , and $D$ is $R$ -closed, $c_n\in D$ . Since $c_n\in D$ for every $D\in \mathcal{C}$ , $c_n\in C$ as well. This shows that $C$ is $R$ -closed.
  In both cases, $B\subseteq C$ since $B\subseteq D$ for every $D\in \mathcal{C}$ . Therefore, $C\in S_{\mathcal{R}}(B)$ .

Hence, $S_{\mathcal{R}}(B)$ is a complete lattice by virtue of this fact, which means that $S_{\mathcal{R}}(B)$ has a minimal element, which is none other than the $\mathcal{R}$ -closure $\operatorname{Cl}_{\mathcal{R}}(B)$ of $B$ .

Remark. It is not hard to see $\operatorname{Cl}_{\mathcal{R}}$ has the following properties:

1. $B\subseteq \operatorname{Cl}_{\mathcal{R}}(B)$ ,
2. $\operatorname{Cl}_{\mathcal{R}}(\operatorname{Cl}_{\mathcal{R}}(B))= \operatorname{Cl}_{\mathcal{R}}(B)$ , and
3. if $B\subseteq C$ , then $\operatorname{Cl}_{\mathcal{R}}(B)\subseteq \operatorname{Cl}_{\mathcal{R}}(C)$ .

Next, assume that $\mathcal{S}$ is another set of finitary relations on $A$ . Then

1. if $\mathcal{R}\subseteq \mathcal{S}$ , then $\operatorname{Cl}_{\mathcal{S}}(B) \subseteq \operatorname{Cl}_{\mathcal{R}}(B)$ ,
2. $\operatorname{Cl}_{\mathcal{S}}(\operatorname{Cl}_{\mathcal{R}}(B)) \subseteq \operatorname{Cl}_{\mathcal{R}\cap\mathcal{S}}(B)$ , and
3. $\operatorname{Cl}_{\mathcal{S}}(\operatorname{Cl}_{\mathcal{R}}(B))= \operatorname{Cl}_{\mathcal{R}\cap\mathcal{S}}(B)$ if $\mathcal{R}\subseteq \mathcal{S}$ or $\mathcal{S}\subseteq \mathcal{R}$ .