The closure of the irrational numbers on $\mathbb{R}$

Question

Exercise: What is the closure in $\mathbb{R}$of each of the following set:

i) the set $\mathbb{P}$ of all irrational numbers.

I have no idea on how to solve this question.

Definition: Let $\tau$ be a topology on a set $X$ and consider $A\subseteq X$. If $x\in X$ is an accumulation point of $A$ if:

$\forall \mathscr{U}\in\tau,x\in \mathscr{U}\implies A\cap\mathscr{U}$ contains another point other than $x$.

However I don not know how to proceed using the definition.

Question:

Can someone show me how to solve this question?

Thanks in advance!

Answer 1

Here $\tau$ represents the standard topology ,namely, the open intervals in $\Bbb{R}$. So take any real $x \in B \subset \tau$

Is $B \cap P $ contains a number other than $x$?

Surely yes! because any open interval $B$ contains both rational as well as irrationals, since they are dense

Answer 2

Hint: The irrationals are dense in $\mathbb{R}$, and they do not form a closed set.

Answer 3

Every rational number is a limit point of the set of irrational numbers. Actually it can be shown that between any two rationals lies an irrational (and vice-versa). You could do something like $x_n=\frac1n\sqrt2+q$, to approximate any rational $q$ by irrationals. Thus the the limit points of $\mathbb P$ consists in all real numbers.

Thus, since $\bar S=S\cup S'$ for any set S, we have for the closure of $\mathbb P$, that $\overline{\mathbb P}=\mathbb P\cup\mathbb R=\mathbb R$.