Math Problem Statement

For all subsets A, B, and C of some universal set U, prove or disprove the following statements: (a) (A ⊆ B ∧ A ⊆ C) ⇒ A ⊆ B ∩ C (b) A ⊆ B ⇔ A ∩ B = A (c) A^c \ B^c = B \ A (d) A ⊆ B ⇒ A ∪ (B \ A) = B.

Solution

Let’s analyze and determine whether each of the given statements is true or false for subsets A,B,CA, B, C of a universal set UU:


(a) (ABAC)    ABC(A \subseteq B \land A \subseteq C) \implies A \subseteq B \cap C

  • Analysis:

    • ABA \subseteq B means every element of AA is in BB.
    • ACA \subseteq C means every element of AA is in CC.
    • The intersection BCB \cap C consists of elements common to both BB and CC.
    • Therefore, if ABA \subseteq B and ACA \subseteq C, then every element of AA is also in BCB \cap C.
  • Conclusion: True.


(b) AB    AB=AA \subseteq B \iff A \cap B = A

  • Analysis:

    • ABA \subseteq B: Every element of AA is in BB.
    • AB=AA \cap B = A: The intersection of AA and BB contains all elements of AA.
    • If ABA \subseteq B, then all elements of AA are in BB, so AB=AA \cap B = A.
    • Conversely, if AB=AA \cap B = A, then all elements of AA must be in BB, so ABA \subseteq B.
  • Conclusion: True.


(c) AcBc=BAA^c \setminus B^c = B \setminus A

  • Analysis:

    • AcA^c: The complement of AA, which consists of all elements not in AA.
    • BcB^c: The complement of BB, which consists of all elements not in BB.
    • AcBcA^c \setminus B^c: Elements that are in AcA^c but not in BcB^c.
      • This is equivalent to elements not in AA but in BB.
    • BAB \setminus A: Elements that are in BB but not in AA.

    Since both expressions represent the set of elements in BB but not in AA, the equality holds.

  • Conclusion: True.


(d) AB    A(BA)=BA \subseteq B \implies A \cup (B \setminus A) = B

  • Analysis:

    • ABA \subseteq B: Every element of AA is in BB.
    • BAB \setminus A: Elements in BB but not in AA.
    • A(BA)A \cup (B \setminus A): The union of AA and the elements of BAB \setminus A.
      • This covers all elements of AA (which are already in BB) and all elements in BB but not in AA.
    • Thus, A(BA)A \cup (B \setminus A) equals BB.
  • Conclusion: True.


Final Results:

(a) True
(b) True
(c) True
(d) True


Do you need further clarifications on any of these proofs? Let me know if you’d like detailed steps for any specific statement.


Related Questions:

  1. What is the definition of set complement, intersection, and union?
  2. Can you explain the concept of logical equivalence using set theory?
  3. What happens if we replace \subseteq with \subset in statement (b)?
  4. How would the proofs change if the universal set UU is finite?
  5. Can you derive other similar identities involving set difference and complements?

Tip:

Set operations such as union, intersection, and difference follow logical rules that closely align with Boolean algebra. Practice simplifying these operations using Venn diagrams for clarity.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Set Theory
Subsets
Set Operations
Logical Equivalence

Formulas

Subset relation: A ⊆ B means ∀x (x ∈ A ⇒ x ∈ B)
Set complement: A^c = {x | x ∉ A}
Set intersection: A ∩ B = {x | x ∈ A ∧ x ∈ B}
Set difference: A \ B = {x | x ∈ A ∧ x ∉ B}
Logical equivalence ⇔ (if and only if)

Theorems

Properties of subsets
Laws of set complements
Set union, intersection, and difference laws

Suitable Grade Level

Undergraduate (Introductory Discrete Mathematics)