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

See similar textbooks

Similar questions

Let S be the universal set, where: S = {1, 2, 3, ..., 18, 19, 20} Let sets A and B be subsets of S, where: Set A {12, 18, 20} Set B = {3, 7, 13, 18, 20} Find the following: LIST the elements in the set (AU B): (AUB) = { } Enter the elements as a list, separated by commas. If the result is the empty set, enter DNE LIST the elements in the set (A n B): (AN B) = { } %3D Enter the elements as a list, separated by commas. If the result is the empty set, enter DNE You may want to draw a Venn Diagram to help answer this question.
typewritten please or handwritten but write legibly
U=(4,5,6,7,8,9,10,11,12,13,14,15) U is the Universal set. A=(x)x is a natural number and 9 ≤x
) Use builder notation and logical equivalences to show that A UB = ANB
i need elaborative and perfect ans
Let A = {5,6, 7}. What is P(A)? Do not merely describe the set, but rather write down all the elements.
Hand written plz asap i'll upvote
do not use sentences to explain it enough with numbers and symbols. sentences should only be for the final conclusion. thank you
Consider the universal set U: {1, 3, 7, 8, 15, 16, 17, 20} Set A: {3, 8, 16, 20} Set B: {1, 15, 17, 20} Find AC ∩ B.
Can you please check my prove and please let me know if I did it correctly. Thank you! One more question , is x is not in (A and B ) by definition of set different or set of intersection? Directions for the problem: use element argument to prove number 11. Assume that all sets are subset of a universal of U.
Please help explain in detail. I want to understand the process involved. Thank you.
Set Theory, Proving