Question

Show that the closure of a relation \(R\) with respect to a property \({\bf{P}}\), if it exists, is the intersection of all the relations with property \({\bf{P}}\) that contain \(R\).

Short answer

The closure of a relation \(R\) with respect to a property \(P\) is \({R_P} = \bigcap\limits_{i = 1}^k {{R_i}} .\)

Step-by-step solution

Given

The closure of a relation \(R\) with respect to a property \(P\).

Concept of Relation

Relation is a subset of the Cartesian product. Or simply, a bunch of points (ordered pairs). In other words, the relation between the two sets is defined as the collection of the ordered pair, in which the ordered pair is formed by the object from each set.

Find the closure of the Relation

Let\({R_P}\)be the closure of the relation\(R\)with respect to the property\(P\).

If\(\left\{ {{R_1},{R_2}, \ldots ,{R_k}} \right\}\)be the set of relations so that each of them satisfies\(P\)and contains\(R\)then\({R_P} \subseteq \)\({R_i}\forall i \Rightarrow {R_P} \subseteq \bigcap\limits_{i = 1}^k {{R_i}} \).

Obviously\({R_P} = {R_j}\)for some\(j\)as it satisfies\(P\)and\(R \subseteq {R_P}\). Thus\({R_P} = \bigcap\limits_{i = 1}^k {{R_i}} \).

The closure of a relation \(R\) with respect to a property \(P\) is \({R_P} = \bigcap\limits_{i = 1}^k {{R_i}} .\)