Please help me with this problem. Will rate and reward the person I believe has

Question

Please help me with this problem. Will rate and reward the person I believe has the best answer

S = {(X,Y)epsilon B times B x epslion X y epsilon Y (xRy)}. Prove that if R is transitive, then so is S. Why did the empty set have to be excluded form the set B to make this proof work? Suppose R is relation on A, and define a relation S on p (A) as follos: S = {(X,Y) epsilon p (A) times P (A) x epsilon X y epsilon Y (xRy)}. For each part, give either a proof or a counterexample to justify your answer. If R is reflexive, must S be reflexive? If R is symmetric, must S be symmetric? if R is transitive, must S be transitive?

Explanation / Answer

(A) Yes.

Notice that since R is reflexive, then for all x in X, xRx.

Hence (X,X) are in S since for all x in X there exists y in X (we take y=x) such that (xRy).

(B) No.

Let R be a relation on the set {a,b,c,d} such that R= { (a,b), (a,c),(a,d), (b,a),(c,a),(d,a) } . Then R is symmetric, and defining S we get that ( {b,c}, {a,d}) is in in S since for all x in {b,c} there is a y in {a,d} such that xRy (that y is a, this is we can take y = a)}.

However, ({a,d}, {b,c}) is not in S, since there is nothing in {b.c} such that d is related to that. (d is only related to a under R).

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