Can I recieve help with part c. I understand part a and b. Thank you. Mainly nee

Question

Can I recieve help with part c. I understand part a and b. Thank you. Mainly need help with part c number 2 and 3 (ii and iii)

Problem 3. Let A {1,2, 3, 4, 5,6,7,8 a). Does there exist an equivalence relation F on set A such that 11 If yes, then give an example of such a relation otherwise justify why it cannot exist. b). Suppose T is an equivalence relation on set A which induces the following partition of A 6 11, 2, 3, 4, 5, 6, 7, 8) Find IT c). We define a relation g on the power set P (A) as follows: let X and Y be subsets of Athen i). Prove that g is an equivalence relation on P (A). ii). Find the equivalence class [11,3,4la. such that Ynf1, 3,5,7) -0 Prove that for every X ESP (A)there is exactly one set YEDXl

Explanation / Answer

c) i) XRY <=> (X-Y) U (Y-X) {1,3,5,7}

We have (X-X) U (X-X) = U =    {1,3,5,7}

Thus R is reflexive.

If XRY then (X-Y) U (Y-X) {1,3,5,7} => (Y-X) U (X-Y) {1,3,5,7} => YRX

Thus R is symmetric.

If XRY and YRZ ,

(X-Y) U (Y-X) {1,3,5,7} and (Y-Z) U (Z-Y) {1,3,5,7}

=> (X-Y) U (Y-X) U (Y-Z) U (Z-Y) {1,3,5,7}

=> (X-Y) U (Z-X) {1,3,5,7}

Thus R is transitive and hence an equivalence relation.

ii) (X-Y) U (Y-X) is the set containing elements in X and Y but not both.

Since this is a subset of {1,3,5,7}, the elements 1,3,5 and 7 can be elements of X or Y but any other element must be common to both.

If X = {1,3,4}, then Y must contain 4 and no element apart from 1,3,5 and 7. 1,3,5 and 7 are in P(A)

Thus [{1,3,4}] = P(A)U{4}

iii) The above result can be generalised for any set Y as

[X] = P(A)U{Y} where [X] is the equivalence class of XA.

We already saw that Y may contain elements 1,3,5 and 7 and some unique elements whose set can be taken as S.

Since 1,3,5,7 are already in P(A),

[X] = P(A)US

Since S contains unique elements and not 1,3,5,7,

=> S(A){1,3,5,7} = Ø

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