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Archangel Macsika

11. Show that if A and B are sets, then (a) A − B = A ∩ B (b) (A ∩ B) ∪ (A ∩ B) = A

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a) A-B=A\cap B^C

We must show A-B\sube A\cap B^C and A\cap B^C\sube A-B

Show that A-B\sube A\cap B^C

Let x\in A-B. By definition of set difference, x\in A and x\notin B. By definition of complement, x\notin B implies that x\in B^C. Hence, it is true that both, x\in A and x\in B^C. By definition of intersection, x\in A\cap B^C.

Show that A\cap B^C\sube A-B.

Let x\in A\cap B^C. By definition of intersection, x\in A and x\in B^C. By definition of complement, x\in B^C implies that x\notin B. Hence, x\in A and x\notin B. By definition of set difference, x\in A-B.

Thus, A-B=A\cap B^C.

b) (A-B)\cup(A\cap B)=A

Let x\in A. There are two cases, x\in A and x\notin B or x\in A and x\in B.

In the first case x\in A, x\notin B, so by definition set difference, x\in A-B.

In the second case x\in A and x\in B, so by definition of intersection x\in A\cap B.

By definition of union x\in (A-B)\cup(A\cap B).

Thus if x\in A, then x\in (A-B)\cup(A\cap B).

Let x\in (A-B)\cup(A\cap B). This means that either x\in A-B or x\in A\cap B.

In the first case x\in A, x\notin B, in the second case x\in A and x\in B.

Then in either case x\in A.

Two sets are equal, since they have the same elements.Therefore

(A-B)\cup(A\cap B)=A

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