Let $A=\{\frac{1}{n}:n\in \mathbb N\}$ and $\mathscr{B}=\{(a,b):a<b\}\cup \{(a,b)\setminus A:a<b\}.$ Prove that $\mathscr B$ is a basis for topology $\mathscr T$ on $\mathbb R$. Is $\mathscr T$ is a usual topology on $\mathbb R$?
Obviously,
(1)$\bigcup_{a,b\in \mathbb R}(a,b)=\mathbb R$
For the second condition of basis,
(2)Case 1:-
Let $x\in B_1,B_2\in \{(a,b):a<b\}$ , then obviously there exists $x\in B_3$ such that $B_3\subset B_1 \cap B_3$
Case2:-$x\in B_1\in \{(a,b):a<b\}$and $B_2 \in\{(a,b)\setminus A:a<b\}$ I am not able to prove existance of $B_3$ such that $B_3\subset B_1 \cap B_3$
Case 3 $B_1,B_2 \in\{(a,b)\setminus A:a<b\}$. How do I prove formally the existance of $B_3$?
to check whether the given basis generate usual topology or not?
any open set can be generated by using $\{(a,b):a<b\}$.
Please check my answers.
Best Answer
The assertion “there exists $x\in B_3$ such that $B_3\subset B_1 \cap B_3$” doesn't make sense because no $B_3$ has been defined at this point. What you should prove is that for each $x\in B_1\cap B_2$, there exists some $B_3\in\mathcal{B}$ such that $x\in B_3$ and that $B_3\subset B_1\cap B_2$.
And $\mathcal B$ doesn't generate the usual topology, since $A^\complement\in\mathcal B$ and $A^\complement$ doesn't belong to the usual topology. The topology generated by $\mathcal B$ is strictly finer that the usual one.