Show each interval (a,b) is open in the basis B

Question

I am using Topology Without Tears by Morris

Let B={[a,b)$\in R$:a<b} B is a basis for some topology $\tau$ on R,but not the Euclidean one.

Regardless show for each interval (a,b)is open in (R $\tau$)

Attempted proof

Let B be a base for $\tau$.

If for each x$\in$ (a,b)$\subseteq$ U

Let $I_n=[a,b-1/n) $

Then the union of all intervals (a,b)=$\bigcup_{n\in N}I_n$

for some index n$\in $N and $I_n\in$ B

But x $\in $U implies x$\in I_n$ for n$ \in N$ then [x,b)$\in $(a,b) and [x,b)$\in $B

Any help would be appreciated?

Answer 1

Since $a<b$, there is some $N\in\Bbb N$ such that $a+\frac1N<b$. Then$$\bigcup_{n\geqslant N}\left[a+\frac1n,b\right)=(a,b).$$

Answer 2

Why don't you just say that

$$(a,b)=\bigcup_{n=1}^{\infty}\left[a+\frac{b-a}{n+1}, b\right)$$