# 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:

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

2. Can this be concluded only from the above?

• You have the first statement off, it means each real is a limit of rationals, so change to "if $a \in \mathbb{R}$." We can find a sequence of irrationals limiting to any real, so question 1 is "yes". – coffeemath Feb 6 '17 at 12:08

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:

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

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 $x_n\to a\in\Bbb R \setminus \Bbb Q$.

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.