Functions and Cardinality Questions and Answers

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.

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.