(a) Prove the following statement using the element method for proving that a set equals the empty set: For all sets $A$ and $B , A \cap ( B - A ) = \emptyset$.
(b) Use the properties in Theorem $6.2 .2$ to prove the statement in part (a). Be sure to give a reason for every step.

Question

Quizplus · Accepted Answer

The Answer is a. Proof: Suppose the given statement is false.
[We must show that this supposition leads logically to a contradiction.]
Then there exist sets

A

and

B

such that

A \cap ( B - A ) eq \emptyset

.
Thus there is an element, say

x

, in

A \cap ( B - A )

.
By definition of intersection,

x \in A

and

x \in B - A

.
By definition of set difference (since

x \in B - A

),

x \in B

and

x otin A

.
Hence

x \in A

and

x otin A

, which is a contradiction.
[Arriving at this contradiction shows that the given statement is not false; so it is true that for all sets

A

and

B , A \cap ( B - A ) = \emptyset . ]

b. $a. Proof: Suppose the given statement is false. [We must show that this supposition leads logically to a contradiction.] Then there exist sets A and B such that A \cap ( B - A ) eq \emptyset . Thus there is an element, say x , in A \cap ( B - A ) . By definition of intersection, x \in A and x \in B - A . By definition of set difference (since x \in B - A ), x \in B and x otin A . Hence x \in A and x otin A , which is a contradiction. [Arriving at this contradiction shows that the given statement is not false; so it is true that for all sets A and B , A \cap ( B - A ) = \emptyset . ] b. 20 Let A and B be any sets. Then the left-hand side of the equation to be shown is . which is the right-hand side of the equation to be shown. [Hence the given statement is true.]$
20 Let

A

and

B

be any sets. Then the left-hand side of the equation to be shown is $a. Proof: Suppose the given statement is false. [We must show that this supposition leads logically to a contradiction.] Then there exist sets A and B such that A \cap ( B - A ) eq \emptyset . Thus there is an element, say x , in A \cap ( B - A ) . By definition of intersection, x \in A and x \in B - A . By definition of set difference (since x \in B - A ), x \in B and x otin A . Hence x \in A and x otin A , which is a contradiction. [Arriving at this contradiction shows that the given statement is not false; so it is true that for all sets A and B , A \cap ( B - A ) = \emptyset . ] b. 20 Let A and B be any sets. Then the left-hand side of the equation to be shown is . which is the right-hand side of the equation to be shown. [Hence the given statement is true.]$
. which is the right-hand side of the equation to be shown. [Hence the given statement is true.]

(A) Prove the Following Statement Using the Element Method for Proving