Answer:
The function is neither injective nor surjective
Step-by-step 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
[tex]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 \}[/tex]
[tex]f(\varnothing )=0[/tex]
[tex]f(\left \{ 1 \right \})=f(\left \{ 2 \right \})=f(\left \{ 3 \right \})=1 [/tex]
[tex]f(\left \{ 1,2 \right \})=f(\left \{ 1,3 \right \})=f(\left \{ 2,3 \right \})=2 [/tex]
[tex]f(\left \{ 1,2,3 \right \})=3[/tex]
The function is not injective, for
[tex]P(\left \{ 1 \right \})=P(\left \{ 2 \right \})[/tex]
but
[tex]\left \{ 1 \right \}\neq \left \{ 2 \right \}[/tex]
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