Use the Element Method for Proving a Set Equals the Empty

Question

Examlex · Accepted Answer

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

A

and

C

such that

( A - C ) \cap ( C - A ) eq \emptyset

.
Thus there is an element, say

x

, in

( A - C ) \cap ( C - A )

.
By definition of intersection,

x \in A - C

and

x \in C - A

.
By definition of set difference applied to both expressions,

x \in A

and

x otin C

and

x \in C

and

x otin A

.
Hence, in particular,

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 - C ) \cap ( C - A ) = \emptyset

.]

$Element Proof: Suppose the given statement is false. [We must show that this supposition leads logically to a contradiction.] Then there exist sets A and C such that ( A - C ) \cap ( C - A ) eq \emptyset . Thus there is an element, say x , in ( A - C ) \cap ( C - A ) . By definition of intersection, x \in A - C and x \in C - A . By definition of set difference applied to both expressions, x \in A and x otin C and x \in C and x otin A . Hence, in particular, x \in A and x otin A , which is a contradiction. [Arriving at this contradiction shows that the given statement is not false; so it is true that for all sets A and B , ( A - C ) \cap ( C - A ) = \emptyset .] Suppose the given statement is false. Then there exist sets \bar { B } and C such that ( B - C ) - B eq \emptyset . Thus there is an element x in (B − C) − B. Applying the definition of set difference twice gives that x \in B and x otin C and x otin B . Hence x \in B and x otin B , which is a contradiction [and so the given statement is true].$ Suppose the given statement is false. Then there exist sets

\bar { B }

and

C

such that

( B - C ) - B eq \emptyset

.
Thus there is an element x in (B − C) − B. Applying the definition of set difference twice gives that

x \in B

and

x otin C

and

x otin B

. Hence

x \in B

and

x otin B

, which is a contradiction [and so the given statement is true].

Use the Element Method for Proving a Set Equals the Empty

Definitions: