# logicsets

• Nov 17th 2008, 09:56 AM
srk619
logicsets
can anyone do this question

The sets A, B and C are given by
A = {1, 2, 3, 4},
• Nov 17th 2008, 10:18 AM
clic-clac
A diagram is easy but with a pen :)

Let see b)

$\displaystyle f (1) = y, f (2) = x, f (3) = z$ and $\displaystyle f:A\rightarrow B=\{x,y,z\}$, so $\displaystyle f$ is surjective. But since $\displaystyle A$ has more elements than $\displaystyle B$, $\displaystyle f$ can't be injective, so can't be bijective.

$\displaystyle g(x) = a, g(y) = b$ and $\displaystyle g:B\rightarrow C=\{a,b\}$, so $\displaystyle f$ is surjective. But since $\displaystyle B$ has more elements than $\displaystyle C$, $\displaystyle g$ can't be injective, so can't be bijective. (yeah the proof is quite similar)

$\displaystyle h$ is the identity map on $\displaystyle C$, it is bijective (i.e. injective and surjective).

$\displaystyle k(a) = 2, k(b) = 2$ That means $\displaystyle k:C\rightarrow A$ is not injective. Furthermore, as $\displaystyle A$ has more elements than $\displaystyle C$, $\displaystyle Range(k) \neq A$ so $\displaystyle k$ is not surjective, in conclusion, $\displaystyle k$ is "none of these".