Question

For each function below, determine whether or not the function is injective and whether or not the function is surjective. Be sure to justify your answers.

(b) f: P({1,2,3}) + N given by f(A) = |A| (Note: P(S) denotes the power set of a set S.)

Answer 1

The function is neither injective nor surjective

Explanation:

f: P({1,2,3}) -> N is given by f(A) = |A|

where P(S) denotes the power set of a set S and |A| the number of elements of A

`P(\left \{ 1,2,3 \right \})=\left \{ \varnothing ,\left \{ 1 \right \},\left \{ 2 \right \},\left \{ 3 \right \},\left \{ 1,2 \right \},\left \{ 1,3 \right \},\left \{ 2,3 \right \},\left \{ 1,2,3 \right \} \right \}`

`f(\varnothing )=0`

`f(\left \{ 1 \right \})=f(\left \{ 2 \right \})=f(\left \{ 3 \right \})=1`

`f(\left \{ 1,2 \right \})=f(\left \{ 1,3 \right \})=f(\left \{ 2,3 \right \})=2`

`f(\left \{ 1,2,3 \right \})=3`

The function is not injective, for

`P(\left \{ 1 \right \})=P(\left \{ 2 \right \})`

but

`\left \{ 1 \right \}\\eq \left \{ 2 \right \}`

The function is not surjective either, because 5 is a natural number which is not the image of any element A under f, i.e., there is no element in A in P({1,2,3}) such that f(A)=5