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(a) Let A, B be sets. Prove that A\(An B) = A\B. You may assume all sets in the question are non-empty. (b) Let A, B be non-empty set such that AC B. Prove that P(A) C P(B).

(a) Let A, B be sets. Prove that A\(An B) = A\B. You may assume all sets in the question are non-empty. (b) Let A, B be non-empty set such that AC B. Prove that P(A) C P(B).

Advanced Engineering Mathematics
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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**Problem (a)** Let \( A, B \) be sets. Prove that \( A \setminus (A \cap B) = A \setminus B \). You may assume all sets in the question are non-empty. **Problem (b)** Let \( A, B \) be non-empty sets such that \( A \subseteq B \). Prove that \( P(A) \subseteq P(B) \). **Explanation** - In problem (a), we are dealing with the difference and intersection of sets, and we are tasked to show the equality of two different ways of expressing the difference of sets involving intersections. - In problem (b), subset notation \( A \subseteq B \) means every element of \( A \) is also in \( B \), and we want to prove a similar inclusion for their power sets, \( P(A) \) and \( P(B) \). There are no graphs or diagrams included in this image.
Transcribed Image Text:**Problem (a)** Let \( A, B \) be sets. Prove that \( A \setminus (A \cap B) = A \setminus B \). You may assume all sets in the question are non-empty. **Problem (b)** Let \( A, B \) be non-empty sets such that \( A \subseteq B \). Prove that \( P(A) \subseteq P(B) \). **Explanation** - In problem (a), we are dealing with the difference and intersection of sets, and we are tasked to show the equality of two different ways of expressing the difference of sets involving intersections. - In problem (b), subset notation \( A \subseteq B \) means every element of \( A \) is also in \( B \), and we want to prove a similar inclusion for their power sets, \( P(A) \) and \( P(B) \). There are no graphs or diagrams included in this image.
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