Let $B=\{[a,b)\subset\Bbb R\mid a,b\in\Bbb Q\text{ and }a
Let $B=\{[a,b)\subset\Bbb R\mid a,b\in\Bbb Q\text{ and }a<b\}$ Show that $B$ is basis for topology on $\Bbb R$.
please check my proof 
1.Let $\bigcup B=\Bbb R$ and $x\in\Bbb Q $ then $x\in [x,x+1)$ and $[x,x+1)\in B$
2.Let $B_{1}=[a,b)a,b\in\Bbb Q$ and $B_{2}=[c,d)c,d\in\Bbb Q$
and
$a\leq c\leq d\leq b$ and $x\in B_{1}\cap B_{2}$
It must exist interval $B_{3}=[e,f) e,f\in\Bbb Q$
that lie between $a\leq c\leq d\leq b$ 
therefore $x\in B_{3}\subseteq B_{1}\cap B_{2}$ and $B$ is basis for a topology on $\Bbb R$
 A: Your proof is not a proof at all. If you wrote down this proof in an exam and I was correcting it, I would give you a point or two out of 25 for having some correct ideas, but no more. It is entirely unclear (1) what you are proving, (2) what you are assuming and (3) what statement follows from what other statement.
For example, you begin your proof by saying

Let $\bigcup B=\mathbb R$ and $x\in \mathbb Q$.

But you cannot just say "let this thing be true", if "this thing" is exactly what you want to prove! You have to prove that $\bigcup B=\mathbb R$. To do that, you need to prove that every element $x\in\mathbb R$ is an element of $\bigcup B$. Therefore, your proof should start with the words:

Let $x\in\mathbb R$.

and conclude with 

Therefore, $x\in\bigcup B$.


Similarly, for point $2$, your proof should start with two arbitrary elements $B_1,B_2\in B$, and an element $x\in B_1\cap B_2$. Your proof should conclude with the discovery of $B^*\in B$ (or whatever else you want to call the set) such that $x\in B^*\subseteq B_1\cap B_2$
A: To give an idea of a more valid proof:
Let $\mathcal{B} = \{[a,b): a,b \in \mathbb{Q}, a < b\}$ be our candidate base.


*

*Let $x \in \mathbb{R}$. As the rationals are order dense in the reals we can find some $q \in \mathbb{Q}$ such that $x-1 < q < x$. But then $x \in [q,q+1)$ and $[q,q+1) \in \mathcal{B}$, so $\bigcup \mathcal{B} = \mathbb{R}$.

*Let $B_1= [a,b) \in \mathcal{B}$ and $B_2 = [c,d) \in \mathcal{B}$.
Let $x \in B_1 \cap B_2$ be arbitrary. Then we know that $a \le x < b$ and $c \le x < d$ so that $\max(a,c) \le x$ and $x < \min(b,d)$. As all of $a,b,c,d$ are in $\mathbb{Q}$, $[\max(a,c),\min(b,d)) \in \mathcal{B}$ as well and clearly $$x \in [\max(a,c),\min(b,d)) \subseteq B_1 \cap B_2$$ as required.
So $\mathcal{B}$ obeys the two conditions to generate a topology.
