Answer
(a) $A =\cal{R}$ (real numbers), $B =\cal{R}-${$0$} (nonzero real numbers)
(b) $A= (0,1) \cup N$ (all real numbers between 0 and 1 and all nonnegative integers) , $B= (0, 1)$ (all real numbers between 0 and 1)
c) $A=\cal{ R}$ (real numbers) ,$B= R^- \cup \text{{0}}$
Work Step by Step
(a) For example, the set $A =\cal{R}$ (real numbers) is uncountable and the set $B =\cal{R}-${$0$} (nonzero real numbers) is also uncountable.
Using the definition of the difference and the fact that 0 is the only element
of A that is not in B:
$A- B= \cal{R} -(\cal{R}- ${$0$}) = {0}
Since the set {$0$} contains only 1 element, the set {$0$} is finite and thus $A- B$ is finite as well.
(b) For example, the set $A= (0,1) \cup N$ (all real numbers between 0 and 1 and all nonnegative integers) is uncountable and the set $B= (0, 1)$ (all real numbers between 0 and 1) is also uncountable.
Using the definition of the difference and the fact that the nonnegative integers are the only element of $A$ that is not in $B$
$A-B=((0, 1) \cup N)- (0, 1) =N$
The set $N$ of non negative integers is countably infinite.
c) For example, the set $A=\cal{ R}$ (real numbers) is uncountable and the set $B= R^- \cup \text{{0}}$ (negative real numbers and zero) is also uncountable.
Using the definition of the difference and the fact that the positive real numbers are the only element of $A$ that is not in $B$
$A- B= \cal{R}- (\cal{R}^- \cup $ { $0$ } $= \cal{R}^+$
The set $\cal{R}^-$ of positive real numbers is uncountable.