Limit points of irrationals We know that the set of all limit points of $\Bbb Q$ is $\Bbb R$. This means that if $a \in \Bbb Q$,  we can find a rational number as close to $a$ as we want. We also know that between every two rational numbers there exists an irrational number.
My question is:


*

*Is $\Bbb R$ the set of all limit points of $\Bbb R \setminus \Bbb Q$ (of the irrational numbers)?

*Can this be concluded only from the above?
 A: First observe the following simple fact:

If $a, b\in\mathbb{R} $ with $a<b$ then there are numbers $c, d$ with $c\in \mathbb{Q}, a<c<b$ and $d\in\mathbb{R} \setminus \mathbb{Q}, a<d<b$. 

The following are then immediate consequences of the above statement:


*

*$\mathbb{R} $ is the set of limit points of $\mathbb{Q} $. 

*$\mathbb{R} $ is the set of limit points of $\mathbb{R} \setminus \mathbb{Q} $. 

A: Let $a=0.675356777649\cdots\in\Bbb R \setminus \Bbb Q$
Consider this sequence:
\begin{eqnarray}
x_1 &=& 0.6   \\
x_2 &=& 0.67   \\
x_3 &=& 0.675  \\
x_4 &=& 0.6753   \\
x_5 &=& 0.67535  \\
x_6 &=& 0.675356   \\
x_7 &=& 0.6753567  \\
 &\vdots&
\end{eqnarray}
$x_n\in\mathbb{Q}$ and $x_n\to a\in\Bbb R \setminus \Bbb Q$.
A: From "each real is a limit point of rationals" we can, given a real $c,$ create a sequence $q_1,q_2,\cdots$ of rational numbers converging to $c.$ Then if we multiply each $q_j$ by the irrational $1+(\sqrt{2}/j),$ we get a sequence of irrationals converging to $c.$ 
The point of using $1+\frac{\sqrt{2}}{j}$ is that it gives a sequence of irrationals which converges to $1.$
So statement 2 "almost" follows from the rationals being dense in the reals.
