Can you please check my prove and please let me know if i did it correctly in the answer. Thank you! Directions: Use element argument to prove number 29. Assume that all sets are subsets of a universal set of U. **Theorem**: For all sets \( A, B, \) and \( C \), \[(A - C) \cap (B - C) \cap (A - B) = \emptyset\]**Proof (by contradiction)**: Let \( A, B, \) and \( C \) be any sets.Suppose not. That is, suppose there exists \( x \in (A - C) \cap (B - C) \cap (A - B) \neq \emptyset \). Then there is an element \( x \) in \( (A - C) \cap (B - C) \cap (A - B) \).In particular, \( x \in (B - C) \cap (A - B) \) by definition of intersection.- \( x \in (B - C) \Rightarrow x \in B \) and \( x \notin C \)- \( x \in (A - B) \Rightarrow x \in A \) and \( x \notin B \)Thus, in particular, \( x \in B \) and \( x \notin B \).\[x \in (B - C) \cap (A - B) \Rightarrow (B - C) \cap (A - B) = \emptyset\]Therefore, \( (A - C) \cap \emptyset = \emptyset \) because the intersection of an empty set is an empty set.Therefore, the assumption is wrong. \((X)\)So, \[(A - C) \cap (B - C) \cap (A - B) = \emptyset\]\(\square\)

Answered: Image /qna-images/answer/ccd0c6b1-3a8d-488d-886b-29da11da5fec.jpg

29. For all sets A,B and C (A-c) n (B-C) N(A-B) = Ø Pronf (by comtadicton) Let A,B, and c be any sets. Suppose not . That is Suppose 3x E (A-c) n (8-C) n (A-B) ± Ø . Then (A-C) n (B-C) n (A-B) .. there is an elemment x in In particular z t'LB-C) nCA-B) by definiton of Intersecton . X E (B-c) > 'a t B and x¢C and x¢ B X E (A-B) > KE A Thus , in partiaular 1 E B. and x¢ B XE (B-C) n(A-B) (8-C) n(A-B)- Ø (A-C) nd = d because set Therefore intergecton of an empty set is an empty There fore the assumption O (B-C) O (A -B)- e (x). Buesn sI

29. For all sets A,B and C (A-c) n (B-C) N(A-B) = Ø Pronf (by comtadicton) Let A,B, and c be any sets. Suppose not . That is Suppose 3x E (A-c) n (8-C) n (A-B) ± Ø . Then (A-C) n (B-C) n (A-B) .. there is an elemment x in In particular z t'LB-C) nCA-B) by definiton of Intersecton . X E (B-c) > 'a t B and x¢C and x¢ B X E (A-B) > KE A Thus , in partiaular 1 E B. and x¢ B XE (B-C) n(A-B) (8-C) n(A-B)- Ø (A-C) nd = d because set Therefore intergecton of an empty set is an empty There fore the assumption O (B-C) O (A -B)- e (x). Buesn sI

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Can you please check my prove and please let me know if i did it correctly in the answer. Thank you! Directions: Use element argument to prove number 29. Assume that all sets are subsets of a universal set of U.

**Theorem**: For all sets \( A, B, \) and \( C \),
\[
(A - C) \cap (B - C) \cap (A - B) = \emptyset
\]

**Proof (by contradiction)**: Let \( A, B, \) and \( C \) be any sets.

Suppose not. That is, suppose there exists \( x \in (A - C) \cap (B - C) \cap (A - B) \neq \emptyset \). Then there is an element \( x \) in \( (A - C) \cap (B - C) \cap (A - B) \).

In particular, \( x \in (B - C) \cap (A - B) \) by definition of intersection.

- \( x \in (B - C) \Rightarrow x \in B \) and \( x \notin C \)
- \( x \in (A - B) \Rightarrow x \in A \) and \( x \notin B \)

Thus, in particular, \( x \in B \) and \( x \notin B \).

\[
x \in (B - C) \cap (A - B) \Rightarrow (B - C) \cap (A - B) = \emptyset
\]

Therefore, \( (A - C) \cap \emptyset = \emptyset \) because the intersection of an empty set is an empty set.

Therefore, the assumption is wrong. \((X)\)

So,
\[
(A - C) \cap (B - C) \cap (A - B) = \emptyset
\]

\(\square\)

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