Truth among set relations for BOWL / ELBOW / BELLOW: Let A = letters of 'BOWL', B = letters of 'ELBOW', C = letters of 'BELLOW'. Which statement is false?

Difficulty: Easy

Correct Answer: B is a proper subset of C.

Explanation:


Introduction / Context:
We compare sets made from letters of words (duplicates ignored). The task is to identify the false statement among four relations.



Given Data / Assumptions:

  • A = {B, O, W, L}
  • B = {E, L, B, O, W}
  • C = {B, E, L, O, W}


Concept / Approach:
Compute each relation by direct comparison of membership.



Step-by-Step Solution:
A ⊂ B: true (A's four letters are all in B)B = C: true (same five letters)B ⊃ C: true if interpreted as superset allowing equality'B is a proper subset of C': false because B and C are equal



Verification / Alternative check:
Enumerations confirm equality of B and C, invalidating the 'proper subset' claim.



Why Other Options Are Wrong:
They are true under standard set interpretations (A ⊂ B, B ⊃ C, B = C).



Common Pitfalls:
Confusing superset '⊃' with proper superset; here, equality makes 'proper subset' false.



Final Answer:
B is a proper subset of C.

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