Let N be the set of natural numbers. Consider the following statements:
(A) If A is a finite set and \(f : A \to A\) is one-one, then f is onto.
(B) If \(n(A) = 5\) and \(n(B) = 3\), then there is no one-one function from A to B.
(C) Every one-one function \(f : N \to N\) is onto.
(D) Every onto function \(f : N \to N\) is one-one.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) and (B) only
If A has n elements and \(f : A \to A\) is one-one, the n images are all different, so they use up all n elements of A, and f is onto.
A one-one function from A to B needs a different image for each element of A, so it needs \(n(B) \ge n(A)\).
For the infinite set N, these counting arguments fail. Two examples show this:
\(f(n) = n + 1\): one-one, but 1 is not an image
\(g(1) = 1\), \(g(n) = n - 1\) for \(n \ge 2\): onto, but \(g(1) = g(2) = 1\)
So on a finite set a one-one self-map is onto, a 5-element set has no one-one map into a 3-element set, and on N one-one does not force onto, nor onto force one-one.
Checking the statements
(A) A one-one \(f : A \to A\) on a finite set is onto. — Correct. n different images fill all n elements of A.
(B) No one-one function from a 5-element set to a 3-element set. — Correct. 5 different images cannot fit in 3 elements.
(C) Every one-one \(f : N \to N\) is onto. — Incorrect. \(f(n) = n + 1\) is one-one, but 1 is not an image.
(D) Every onto \(f : N \to N\) is one-one. — Incorrect. \(g(1) = 1\), \(g(n) = n - 1\) for \(n \ge 2\) is onto, but \(g(1) = g(2)\).