Closure in a metric space

Question

Let $P \subset \mathbb{R}$ be an open set and define the metric space $(P, d)$ with $d$ the usual metric on the set of real numbers.

Show that $\overline{\mathbb{Q} \cap P} = P$ where the closure is in the above metric space.

I am thinking of the following rough argument:

$P \cap Q$ is the set of rationals in the open set $P$. Taking the closure of $P \cap Q$ in the metric space $(P, d)$ gives the set of all limit points of the rationals in the set $P$. These limit points are the set of reals in the open set $P$. Since $P \subset \mathbb{R}$, $\overline{P \cap Q} = P$.

How to mathematically show this?

Answer 1

The inclusion $Cl_P(\mathbb{Q} \cap P) \subset P$ is obvious. Now, take $x \in P$. Then there is a sequence ${x_n}$ of rational numbers such that $x_n \rightarrow x$. Since $P$ is open in $\mathbb{R}$, there must be an open ball $B(x; \epsilon) \subset P$ and an integer $N$ such that $n>N$ implies $x_n \in B(x; \epsilon)$. So $x_n \in \mathbb{Q} \cap P$. Therefore, $x \in Cl_P(\mathbb{Q} \cap P)$.

Answer 2

The better statement is $\overline{\Bbb Q \cap P}=\overline{P}$. (Closure in $\Bbb R$. The fact for closure in $P$ is an immediate consequence).

The left to right inclusion just follows from $\Bbb Q \cap P \subseteq P$.

OTOH, if $x \in \overline{P}$, let $B(x,r)$ be an open ball around $x$, and we have some $p \in B(x,r) \cap P$. As $B(x,r)$ and $P$ are open we find $r'>0$ such that

$$p \in B(p,r') \subseteq B(x,r) \cap P$$

and then we have some $q \in B(p,r') \cap \Bbb Q$ as $\Bbb Q$ is dense.

This $q \in (P \cap \Bbb Q) \cap B(x,r)$ by the previous and so $x \in \overline{\Bbb Q \cap P}$ as required.