Computer Science Engineering (CSE) Exam > Computer Science Engineering (CSE) Questions > Let f : A → B be an injective (one-to-on...
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Let f : A → B be an injective (one-to-one) function.
Define g : 2A → 2B as :
g(C) = {f(x) | x ∈ C}, for all subsets C of A.
Define h : 2B → 2A as :
h(D) = {x | x ∈ A, f(x) ∈ D}, for all subsets D of B.
g(C) = {f(x) | x ∈ C}, for all subsets C of A.
Define h : 2B → 2A as :
h(D) = {x | x ∈ A, f(x) ∈ D}, for all subsets D of B.
Q.
Which of the following statements is always true ?
Which of the following statements is always true ?
- a)g(h(D)) ⊆ D
- b)g(h(D)) ⊇ D
- c)g(h(D)) ∩ D = ф
- d)g(h(D)) ∩ (B - D) ≠ ф
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
Let f : A → B be an injective (one-to-one) function.Define g : 2A...
Understanding the Functions
Let’s break down the functions defined:
- f : A → B is an injective function, meaning each element in A maps to a unique element in B.
- g : 2A → 2B is defined as \( g(C) = \{f(x) | x \in C\} \); it maps subsets of A to subsets of B.
- h : 2B → 2A is defined as \( h(D) = \{x | x \in A, f(x) \in D\} \); it maps subsets of B back to subsets of A.
Exploring the Statements
Now, let’s analyze the statements provided:
1. g(h(D)) ⊆ D
2. g(h(D)) ⊇ D
3. g(h(D)) ∩ D = ∅
4. g(h(D)) ∩ (B - D) ≠ ∅
Why is Option A (g(h(D)) ⊆ D) Always True?
- If \( y \in g(h(D)) \), then there exists \( x \in h(D) \) such that \( y = f(x) \).
- Since \( x \in h(D) \), it follows that \( f(x) \in D \) (by the definition of h).
- Thus, \( y \in D \).
- This shows that every element produced by g from h(D) is indeed in D.
Why Other Options Are Not Always True?
- Option B (g(h(D)) ⊇ D): Not true, as g(h(D)) may not cover all elements of D.
- Option C (g(h(D)) ∩ D = ∅): Not true, as some elements in g(h(D)) can be in D.
- Option D (g(h(D)) ∩ (B - D) ≠ ∅): Not guaranteed, as g(h(D)) could be entirely contained in D.
Conclusion
Therefore, the only universally valid statement is g(h(D)) ⊆ D, confirming that option A is correct.
Let’s break down the functions defined:
- f : A → B is an injective function, meaning each element in A maps to a unique element in B.
- g : 2A → 2B is defined as \( g(C) = \{f(x) | x \in C\} \); it maps subsets of A to subsets of B.
- h : 2B → 2A is defined as \( h(D) = \{x | x \in A, f(x) \in D\} \); it maps subsets of B back to subsets of A.
Exploring the Statements
Now, let’s analyze the statements provided:
1. g(h(D)) ⊆ D
2. g(h(D)) ⊇ D
3. g(h(D)) ∩ D = ∅
4. g(h(D)) ∩ (B - D) ≠ ∅
Why is Option A (g(h(D)) ⊆ D) Always True?
- If \( y \in g(h(D)) \), then there exists \( x \in h(D) \) such that \( y = f(x) \).
- Since \( x \in h(D) \), it follows that \( f(x) \in D \) (by the definition of h).
- Thus, \( y \in D \).
- This shows that every element produced by g from h(D) is indeed in D.
Why Other Options Are Not Always True?
- Option B (g(h(D)) ⊇ D): Not true, as g(h(D)) may not cover all elements of D.
- Option C (g(h(D)) ∩ D = ∅): Not true, as some elements in g(h(D)) can be in D.
- Option D (g(h(D)) ∩ (B - D) ≠ ∅): Not guaranteed, as g(h(D)) could be entirely contained in D.
Conclusion
Therefore, the only universally valid statement is g(h(D)) ⊆ D, confirming that option A is correct.
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Let f : A → B be an injective (one-to-one) function.Define g : 2A → 2B as :g(C) = {f(x) | x ∈ C}, for all subsets C of A.Define h : 2B → 2A as :h(D) = {x | x ∈ A, f(x) ∈ D}, for all subsets D of B.Q.Which of the following statements is always true ?a)g(h(D)) ⊆ Db)g(h(D)) ⊇ Dc)g(h(D)) ∩ D = d)g(h(D)) ∩ (B - D) ≠ Correct answer is option 'A'. Can you explain this answer?
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Let f : A → B be an injective (one-to-one) function.Define g : 2A → 2B as :g(C) = {f(x) | x ∈ C}, for all subsets C of A.Define h : 2B → 2A as :h(D) = {x | x ∈ A, f(x) ∈ D}, for all subsets D of B.Q.Which of the following statements is always true ?a)g(h(D)) ⊆ Db)g(h(D)) ⊇ Dc)g(h(D)) ∩ D = d)g(h(D)) ∩ (B - D) ≠ Correct answer is option 'A'. Can you explain this answer? for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about Let f : A → B be an injective (one-to-one) function.Define g : 2A → 2B as :g(C) = {f(x) | x ∈ C}, for all subsets C of A.Define h : 2B → 2A as :h(D) = {x | x ∈ A, f(x) ∈ D}, for all subsets D of B.Q.Which of the following statements is always true ?a)g(h(D)) ⊆ Db)g(h(D)) ⊇ Dc)g(h(D)) ∩ D = d)g(h(D)) ∩ (B - D) ≠ Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let f : A → B be an injective (one-to-one) function.Define g : 2A → 2B as :g(C) = {f(x) | x ∈ C}, for all subsets C of A.Define h : 2B → 2A as :h(D) = {x | x ∈ A, f(x) ∈ D}, for all subsets D of B.Q.Which of the following statements is always true ?a)g(h(D)) ⊆ Db)g(h(D)) ⊇ Dc)g(h(D)) ∩ D = d)g(h(D)) ∩ (B - D) ≠ Correct answer is option 'A'. Can you explain this answer?.
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