just wondering, say you have two sets A and B of real numbers, and k a positive real number. Letting $kA= \{$ka: a $\in A$}\, does it follow that k(A\B)=(kA\kB), where \ is set difference? I think it does, since the function kx is a one to map and $kA\setminus kB$= k($A \cap B^c$) which is just k times a set.
Thanks!
The statement is correct. If you are unsure, the safest way to prove that two sets are equal is to show that one is a subset of the other and vice versa.
To show that $kA \setminus kB \subseteq k(A \setminus B)$, let $x \in kA \setminus kB$. By definition, there exists $a \in A$ such that $x = ka$ and there is no $b \in B$ with $x = kb$. Hence, $a$ cannot be in $B$, which implies $a \in A \setminus B$ and thus $x \in k(A \setminus B)$.
To prove $k(A \setminus B) \subseteq kA \setminus kB$, let $x \in k(A \setminus B)$. This means that there is an $a \in A \setminus B$ such that $x = ka$. We thus have $x \in kA$. It remains to show that $x \notin kB$. Assume toward a contraction that $x \in kB$. Then, there is a $b \in B$ with $x = kb$, which implies $ka = x = kb$. Since $k$ has an inverse in $\mathbb{R}$, this implies $b = a$, and therefore $a \in B$, contradicting $a \in A \setminus B$.