Functions and Cardinality Questions and Answers

Q: \(g(.)\) is a function from \(A\) to \(B\), \(f(.)\) is a function from \(B\) to \(C\), and their composition defined as \(f(g(.))\) is a mapping from \(A\) to \(C\).If \(f(.)\) and \(f(g(.))\) are onto (surjective) functions, which ONE of the following is TRUE about the function \(g(.)\)?

The correct statement is \(g(.)\) is not required to be a one-to-one or onto function.That \(f \circ g\) is onto forces f to be onto, but places no requirement on g. A counterexample shows both properties can fail. Let \(A = \{1, 2\}\), \(B = \{x, y, z\}\), \(C = \{c\}\). Take \(g(1) = g(2) = x\), and let f send everything to c.Then f and \(f(g(.))\) are onto C, yet g is not one-to-one (1 and 2 both go to x) and not onto (nothing maps to y or z). So g need not be injective, surjective or bijective, and options A, B and C, which say it must be, are false.

Q: Let \(N\) be the set of natural numbers. Consider the following sets.\(P\): Set of Rational numbers (positive and negative)\(Q\): Set of functions from {0, 1} to \(N\)\(R\): Set of functions from \(N\) to {0, 1}\(S\): Set of finite subsets of \(N\).Which of the sets above are countable?

The countable sets are P, Q and S.P: the rationals are countable, since each is a pair of integers and the pairs can be listed.Q: a function from {0, 1} to N is just an ordered pair of natural numbers, and N × N is countable.S: the finite subsets of N can be listed by size and largest element, or matched with the natural numbers through binary representation.R: a function from N to {0, 1} is an infinite binary sequence, and Cantor's diagonal argument shows these are uncountable.So any option containing R, or leaving out one of P, Q or S, is wrong.

Q: Consider the set X of all sequences \((x_{n})\) of natural numbers.Let \(Y = \{(x_{n}) \in X : x_{n} = 0\) for infinitely many \(n\}\),\(Z = \{(x_{n}) \in X : x_{n} = 0\) for infinitely many \(n\}\),\(W = \{(x_{n}) \in X : x_{n} \neq 0\) for finitely many \(n\}\).Then which of the following sets is countable ?

The countable set is W.W consists of sequences of natural numbers that are non-zero for only finitely many \(n\), that is, sequences that are eventually 0. Each such sequence is determined by a finite list of natural numbers, and the set of finite lists of natural numbers is a countable union of countable sets, hence countable.The paper prints Y and Z with the same definition (\(x_{n} = 0\) for infinitely many \(n\)), so they are the same set, and it is uncountable: every 0-1 pattern on the odd places with zeros on the even places gives a distinct member, and there are uncountably many such patterns. The complement of Z, sequences with only finitely many zeros, is uncountable too, since it contains all sequences of non-zero natural numbers.

\(g(.)\) is a function from \(A\) to \(B\), \(f(.)\) is a function from \(B\) to \(C\), and their composition defined as \(f(g(.))\) is a mapping from \(A\) to \(C\).

If \(f(.)\) and \(f(g(.))\) are onto (surjective) functions, which ONE of the following is TRUE about the function \(g(.)\)?

A.\(g(.)\) must be an onto (surjective) function.
B.\(g(.)\) must be a one-to-one (injective) function.
C.\(g(.)\) must be a bijective function, that is, both one-to-one and onto.
D.\(g(.)\) is not required to be a one-to-one or onto function.

Exam Year(s)

GATE CS - 2025,

Answer: D

The correct statement is \(g(.)\) is not required to be a one-to-one or onto function.

That \(f \circ g\) is onto forces f to be onto, but places no requirement on g. A counterexample shows both properties can fail. Let \(A = \{1, 2\}\), \(B = \{x, y, z\}\), \(C = \{c\}\). Take \(g(1) = g(2) = x\), and let f send everything to c.

Then f and \(f(g(.))\) are onto C, yet g is not one-to-one (1 and 2 both go to x) and not onto (nothing maps to y or z). So g need not be injective, surjective or bijective, and options A, B and C, which say it must be, are false.

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Let \(N\) be the set of natural numbers. Consider the following sets.

\(P\): Set of Rational numbers (positive and negative)
\(Q\): Set of functions from {0, 1} to \(N\)
\(R\): Set of functions from \(N\) to {0, 1}
\(S\): Set of finite subsets of \(N\).

Which of the sets above are countable?

A.Q and S only
B.P and S only
C.P and R only
D.P, Q and S only

Exam Year(s)

GATE CS - 2018,

Answer: D

The countable sets are P, Q and S.

P: the rationals are countable, since each is a pair of integers and the pairs can be listed.
Q: a function from {0, 1} to N is just an ordered pair of natural numbers, and N × N is countable.
S: the finite subsets of N can be listed by size and largest element, or matched with the natural numbers through binary representation.
R: a function from N to {0, 1} is an infinite binary sequence, and Cantor's diagonal argument shows these are uncountable.

So any option containing R, or leaving out one of P, Q or S, is wrong.

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Consider the set X of all sequences \((x_{n})\) of natural numbers.

Let \(Y = \{(x_{n}) \in X : x_{n} = 0\) for infinitely many \(n\}\),
\(Z = \{(x_{n}) \in X : x_{n} = 0\) for infinitely many \(n\}\),
\(W = \{(x_{n}) \in X : x_{n} \neq 0\) for finitely many \(n\}\).

Then which of the following sets is countable ?

A.Y
B.Z
C.W
D.Complement of Z in X

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: C

The countable set is W.

W consists of sequences of natural numbers that are non-zero for only finitely many \(n\), that is, sequences that are eventually 0. Each such sequence is determined by a finite list of natural numbers, and the set of finite lists of natural numbers is a countable union of countable sets, hence countable.

The paper prints Y and Z with the same definition (\(x_{n} = 0\) for infinitely many \(n\)), so they are the same set, and it is uncountable: every 0-1 pattern on the odd places with zeros on the even places gives a distinct member, and there are uncountably many such patterns. The complement of Z, sequences with only finitely many zeros, is uncountable too, since it contains all sequences of non-zero natural numbers.

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Let f and g be the functions from the set of integers to the set integers defined by \(f(x) = 2x + 3\) and \(g(x) = 3x + 2\) Then the composition of f and g and g and f is given as

A.\(6x + 7\), \(6x + 11\)
B.\(6x + 11\), \(6x + 7\)
C.\(5x + 5\), \(5x + 5\)
D.None of the above

Exam Year(s)

UGC NET - 2013,

Answer: A

The composition f o g means f(g(x)) and g o f means g(f(x)). Substituting gives each in turn.

\[ f(g(x)) = 2(3x + 2) + 3 = 6x + 7 \]\[ g(f(x)) = 3(2x + 3) + 2 = 6x + 11 \]

So the pair is 6x + 7 and 6x + 11 in that order. The second option reverses them, and 5x + 5 results from adding the functions, not composing them.

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