# Let $A \subseteq \mathbb R$ such that $A \cap A^d = \emptyset$. Which of the following is true?

Let $A \subseteq \mathbb R$ such that $A \cap A^d = \emptyset$ (where $\mathbb R$ denotes the the set of all real numbers, $A^d$ denotes the derived set of $A$ and $\emptyset$ denotes the empty set).Then which of the following is true?

$(a)$ $A^d$ must be empty.

$(b)$ $A^d$ is at most a singleton set.

$(c)$ $A^d$ is at most a finite set.

$(d)$ $A^d$ may be infinite.

My attempt:

Let us consider a collection $\{A_p\}_{p\ is\ prime}$ where $A_p=\{p^{\frac {1} {2} + \frac {1} {2^2} + \cdots + \frac {1} {2^n}} : n \in \mathbb N \}$ for all $p.$ Then clearly $\{A_p\}^d=\{p\}$ since the sequence $\{p^{\frac {1} {2} + \frac {1} {2^2} + \cdots + \frac {1} {2^n}}\}_n$ converges to $p$. Let $A = \underset {p\text{ is prime}} \cup A_p$ and $\mathbb P$ denotes the set of all prime numbers. Then $A^d=\mathbb P$ and hence $A \cap A^d = \emptyset$. This shows that $A^d$ may be an infinite set. So according to me $(d)$ is the correct option.

Is the above reasoning correct at all? Please verify it. Thank you in advance.

Yes, it is correct, but there are much simpler examples. For instance$$A=\left\{k+\frac1n\,\middle|\,k\in\mathbb{Z}\text{ and }n\in\mathbb{N}\setminus\{1\}\right\}.$$Then $A'$ (that's the standard notation for derived set) is equal to $\mathbb Z$ and therefore $A'\cap A=\emptyset$.
• The choice of $n$ from $\mathbb N \setminus \{1\}$ is excellent. If $n \in \mathbb N$ then we would have $A' \cap A = \mathbb Z$. Thanks for your suggestion @Jose Carlos Santos. – Arnab Chatterjee. Jun 20 '17 at 13:40
Edited:The set $A=${$\frac{1}{2^m}+\frac{1}{3^n}: m,n\in\mathbb N$} serves as counterexample for options $a,b,c$.
• Here in your example @toddler $\frac {1} {2} \in A^d$ but $\frac {1} {2} \in A$ since $\frac {1} {4} + \frac {1} {4} =\frac {1} {2}$.So $A \cap A^d \ne \phi$ here which violates our hypothesis. – Arnab Chatterjee. Jun 20 '17 at 13:33
• Yes...I missed that point...may be raising them to the powers of $2$ will work. – Nitin Uniyal Jun 20 '17 at 13:54