Prove that $\{(a,b): a,b\in \Bbb Q \land aI dont have a clue about that. If I take the definition of base of a topology it says:

A collection of open sets $\cal B$ is a base of a topology $\tau$ if any open set is a union of elements of this collection.

So I dont know how is possible that for any interval $(c,d)$ with $c,d\in\Bbb R$ exists some $\bigcup_{i,j\in I}(a_i,b_j)=(c,d)$ where $a_i,b_j\in\Bbb Q$. To me this is impossible, imagine that $c$ and $d$ are irrational numbers... how I can compose by union of interval defined by rationals an interval as $(c,d)$?
Some hint? Some big mistake in my formulation? Any help is welcome, thank you in advance.
 A: Every real number is a limit of an increasing/decreasing sequence of rational numbers. Let $(x_n)_{n=1}^\infty$ be a sequence of rational numbers $<d$ decreasing to $c$ and let $(y_n)_{n=1}^\infty$ be a sequence of rational numbers $>c$ increasing to $d$. Then
$$(c,d)=\bigcup_{n=1}^\infty (x_n,y_n).$$
A: In fact you just need that $\mathbb Q$ is dense in $\mathbb R$ and that for every open set on $\mathbb R$ there a subset included of the form you stated. 
A: The rationals are dense in $\Bbb R$. 
Given $(c,d)$, there's a rational $s_1\in (c+\frac {d-c} 2, d)$, then a rational $s_2 \in  (\max(s_1, c+\frac {3(d-c)} 4), d)$, ... thus, a strictly increasing sequence of rationals $(s_n)$ with $(c+d)/2 < s_1 < s_2 < ... < d = \lim_n s_n$. (Concluding that the entire sequence $(s_n)$ exists implicitly uses the Axiom of Dependent Choice.)
Similarly, there's a strictly decreasing sequence of rationals $(r_n)$ with $\lim_n r_n = c < ... < r_2 < r_1 < (c+d)/2$. 
Let $I_n = (r_n, s_n)$,  $n\in\Bbb N$. Then $(c,d) = \bigcup_{n} I_n$.
A: The sets $(a,b)$ with $a,b\in\mathbb Q$ form a base of the Euclidean topology on $\mathbb R$.
In general if $\mathcal B$ serves as base for some topology on $X$ and $Y\subseteq X$ then $\{Y\cap B:B\in\mathcal B\}$ serves as a base of the relative topology on $Y$. 
