can we construct a countable set $A$, such that the cardinality of $A'$ is $c$, and $A\cap A'=\emptyset$? I came up with a problem when learning real analysis.
We have known that since $Q$ is a countable set and $Q'=R$,  the cardinality of $Q'$ is $c$.
Then, can we construct a countable set $A$, such that the cardinality of $A'$ is $c$, and  $A\cap A'=\emptyset$ ?
Some textbook on real analysis provides this conclusion, but I wonder how to construct it.
Your help will be appreciated.

Note: 
  
  
*
  
*$c$ denotes the cardinality of $\mathbb{R}$;  
  
*$A^\prime$ denotes the derived set of $A$.
  

 A: Here's one way, but it might not be the simplest. Let $C$ be any Cantor set in the reals ($C$ is a set that is nonempty, nowhere dense, and perfect) and let $A$ be the set of all midpoints of the bounded complementary intervals of $C.$ Then $A' = C,$ so $A'$ has cardinality $c.$ Also, $A \cap A' = A \cap C = \emptyset,$ since all points in $A$ belong to the complement of $C.$
ADDED NEXT DAY $\;$You asked (in a comment) for me to explain in detail why $A' = C.$ I'll do this by proving these two inclusions: (1) $\,A' \subseteq C\;$ and$\;$ (2) $\,C \subseteq A'.$
(1) To prove $A' \subseteq C,$ it suffices to prove ${\mathbb R} - C \subseteq {\mathbb R} - A'.$ To this end, choose $x \in {\mathbb R} - C.$ Then $x$ belongs to at most one of the bounded complementary intervals of $C.$ Therefore, all sufficiently small open intervals containing $x$ contain at most one point of $A$ (indeed, no points of $A$ unless $x$ happens to belong to $A).$ It follows that $x$ cannot be a limit point of $A,$ and hence we have $x \in {\mathbb R} - A'.$ This completes the proof that $A' \subseteq C.$
(2) We are to prove $C \subseteq A'.$ To this end, choose $x \in C.$ Since every point of $C$ is a limit point of $C,$ there exists a sequence $\{x_n\}$ of points in $C$ such that $x_n \rightarrow x.$ By passing to a subsequence if necessary, we can assume that $\{x_n\}$ approaches $x$ monotonically from one side. That is, we have $x_n \nearrow x$ or we have $x_n \searrow x.$ Between $x_1$ and $x_2,$ there will be at least one bounded complementary interval of $C.$ Pick one of them and let $a_1$ be the point we chose from that bounded complementary interval when we formed the set $A.$ Between $x_2$ and $x_3,$ there will be at least one bounded complementary interval of $C.$ Pick one of them and let $a_2$ be the point we chose from that bounded complementary interval when we formed the set $A.$ By continuing in this manner, we get a sequence $\{a_n\}$ of points in $A$ such that $a_n \rightarrow x.$ Hence, $x \in A',$ which completes the proof that $C \subseteq A'.$
