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.
 A: 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. $q\in I$.

A: Something to think about (not a proof):
The closure of a subset is the minimal closed superset. What must a superset of $\mathbb{Q}$ look like?
It cannot have a maximal or minimal element, certainly.
Maybe you could make one out of closed intervals around each point of $\mathbb{Q}$, but it would be impossible to make each of this intervals disjoint (think about the space between each rational).
Trying to construct a closed superset always leads us to the whole real line, and hence $\mathbb{Q}$ is dense in $\mathbb{R}$. 
A: You can construct $\mathbf{R}$ is many ways, see this for instance. The first one (by Cauchy sequences) is the usual one. In it (I encourage you to find it somewhere and to sudy it, as it is very enlightening), $\mathbf{Q}$ is dense in $\mathbf{R}$ by contruction. It is also the case in the second construction, from Dedekind cuts.
A: $Cl(A)=AUAd$
here $A=Q$
so, $Cl(Q)=QUQd$
Now,
  as $Qd=R$ 
so, $Cl(Q)=QUR$
and $QUR=R$
Hence $Cl(Q)=R$
So $Q$ is dense in $R$ as from definition of dense set " In a topological space , $A$ is subset of $X$ Then $A$ is dense in $X$ iff $Cl(A)=X$
A: Since x < y, we have y−x>0. from the Archimedian property for a  positive integer n such that
n(y−x)>1...(classics Proof) between two reals we can find irrationnal
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Proof verification that Q is dense in R
