1.2.7: Operations on Relations

Sets, Logic, Computation · LibreTexts

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    From LibreTexts, under CC BY, as the source states it; its equations are written in TeX, which the PDF shows typeset. This page at LibreTexts.

    1.2.7: Operations on Relations

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    It is often useful to modify or combine relations. In Proposition 2.5.1, we considered the union of relations, which is just the union of two relations considered as sets of pairs. Similarly, in Proposition 2.5.2, we considered the relative difference of relations. Here are some other operations we can perform on relations.

    Definition \(\PageIndex{1}\)

    Let \(R\), \(S\) be relations, and \(A\) be any set.

    The inverse of \(R\) is \(R^{-1} = \Setabs{\tuple{y, x}}{\tuple{x, y} \in R}\).

    The relative product of \(R\) and \(S\) is \((R \mid S) = \{\tuple{x, z} : \exists y(Rxy \land Syz)\}\).

    The restriction of \(R\) to \(A\) is \(\funrestrictionto{R}{A}= R \cap A^2\).

    The application of \(R\) to \(A\) is \(\funimage{R}{A} = \{y : (\exists x \in A)Rxy\}\)

    Example \(\PageIndex{1}\)

    Let \(S \subseteq \Int^2\) be the successor relation on \(\Int\), i.e., \(S = \Setabs{\tuple{x, y} \in \Int^2}{x + 1 = y}\), so that \(Sxy\) iff \(x + 1 = y\).

    \(S^{-1}\) is the predecessor relation on \(\Int\), i.e., \(\Setabs{\tuple{x,y}\in\Int^2}{x -1 =y}\).

    \(S\mid S\) is \(\Setabs{\tuple{x,y}\in\Int^2}{x + 2 =y}\)

    \(\funrestrictionto{S}{\Nat}\) is the successor relation on \(\Nat\).

    \(\funimage{S}{\{1,2,3\}}\) is \(\{2, 3, 4\}\).

    Definition \(\PageIndex{2}\): Transitive closure

    Let \(R \subseteq A^2\) be a binary relation.

    The transitive closure of \(R\) is \(R^+ = \bigcup_{0 < n \in \Nat} R^n\), where we recursively define \(R^1 = R\) and \(R^{n+1} = R^n \mid R\).

    The reflexive transitive closure of \(R\) is \(R^* = R^+ \cup \Id{X}\).

    Example \(\PageIndex{2}\)

    Take the successor relation \(S \subseteq \Int^2\). \(S^2xy\) iff \(x + 2 = y\), \(S^3xy\) iff \(x + 3 = y\), etc. So \(S^+xy\) iff \(x + n = y\) for some \(n > 1\). In other words, \(S^+xy\) iff \(x < y\), and \(S^*xy\) iff \(x \le y\).

    Problem \(\PageIndex{1}\)

    Show that the transitive closure of \(R\) is in fact transitive.


    This page titled 1.2.7: Operations on Relations was last modified on Fri, 08 Mar 2024 05:56:12 GMT and is shared under a CC BY license and was authored, remixed, and/or curated by Richard Zach et al. (Open Logic Project) .