Question about limit point. Let $S$ be an infinite subset of $\mathbb{R}$ such that $S\cap \mathbb{Q}=\varnothing$. Then which of the following statement is true?


*

*$S$ must have a limit point which belongs to $\mathbb{Q}$.

*$S$ must have a limit point which belongs to $\mathbb{R/Q}$.

*$S$ cannot be a closed set in $\mathbb{R}$.

*$\mathbb{R}/S$ must have a limit point which belongs to $S$.
I consider the set {$\sqrt2,\sqrt3,\sqrt5,\sqrt6,\sqrt7,\sqrt8,\sqrt10,..\sqrt15,\sqrt17,....$}
This set satisfies only option 4. So I thought answer should be 4, but answer key says option 3. Am I doing anything wrong?
 A: Only option 4. is true for every infinite set $S\subseteq \mathbb{R}=\emptyset$ such that $S\cap\mathbb{Q}$.  Since $S$ is contained in $\mathbb{R}\setminus\mathbb{Q}$, it follows that $\mathbb{Q}\subseteq \mathbb{R}\setminus S$.  Since every irrational is a limit point of $\mathbb{Q}$, it follows that every member of $S$ is a limit point of $\mathbb{R}\setminus S$.
You example shows that the other three options are, in general, false.
A: It's true that for your example, $\Bbb{R} \setminus S$ has a limit point in $S$. But showing that it's true for one example isn't the same as showing that it must be true for any example.
Nonetheless, I think you're right. Every real number is the limit of a Cauchy sequence of rational numbers, essentially by definition. So if you have some element of $S \subseteq \Bbb{R} \setminus \Bbb{Q}$, then every element of $S$ must be the limit of some sequence in $\Bbb{Q} \subseteq \Bbb{R} \setminus S$, and thus a limit point of $\Bbb{R}\setminus S$.
Your example of $S$ is indeed closed, as is $\{ \pi + n \mid n \in \Bbb{N} \}$, for example. So option 3 cannot be correct.
