Let R be a relation on set A that is reflexive and symmetric

Let R be a relation on set A that is reflexive and symmetric but not transitive. Let R[x] be the set of all yA such that (x,y)R. Show the set P{R[x],xA} does not form a partition for set A. (Thus, it is only the equivalence relations that form partitions.)

Solution

A relation on a set S is reflexive- if a a for all a S.

A relation on a set S is symmetric- if a b implies b a for all a, b S.

A relation on a set S is transitive – if a b and b c implies a c for all a, b, c S.

Consider that R be a relation on set A that is reflexive and symmetric but not transitive.

                       R(x)={y:xRy}

The set A={ R(x):x A }is not always a partition on A.

Normally partition of a set A is a collection of non-empty subsets of A which cover all of A and don’t overlap.

XP

(2) [X] = [P]

[a] [Y] =

Union AF A= S .

A relation on a set S is reflexive- if a a for all a S.

A relation on a set S is symmetric- if a b implies b a for all a, b S.

A relation on a set S is transitive – if a b and b c implies a c for all a, b, c S.

Let R be a relation on set A that is reflexive and symmetric but not transitive. Let R[x] be the set of all yA such that (x,y)R. Show the set P{R[x],xA} does no

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