Why is Q dense in R? Consider a topological space $S$ and an arbitrary subset $E$. Then if the closure of $E$ given by $\bar{E} = S$, we say that $E$ is dense in $S$.
How can we prove that $\mathbb{Q}^{n}$ is dense in the vector space $\mathbb{R}^{n}$ with the standard metric topology?
I intended to prove that $\mathbb{Q}^{n}$ is dense in the vector space $\mathbb{R}^{n}$ in the topological sense, not the usual "between two $\mathbb{R}$ there is a $\mathbb{Q}$". Of course, both might be equivalent, but that remains to be proved.

Hint:


*

*$\mathbb{Q}$ is dense in $\mathbb{R}$.

*It suffices to show $\forall x=(x_i)\in \mathbb{R}^n$, and $x\in I=(a_1,b_1)\times (a_2,b_2)\cdots \times (a_n,b_n)\cap \mathbb{Q}^n\neq \emptyset$. That is $\exists q=(q_i)\in \mathbb{Q}^n$, s.t. $q\in I$.

