Let $B=\{(a,b]:a,b\in R,a<b\}$ is basis of topology $T$ on $R$ show that each interval$(a,b)$ in open in $(R,T)$

Question

please check my proof

Suppose $B$ is basis of topology on $R$

Let $U$ is collection of union of all interval $(a_{n},b_{n})$

There exist some interval $\{I:(a_{n},b_{n}):a,b\in R\}$

for some interval $(a_{n},b_{n})$

such that

$(a_{n},b_{n})\in I\subseteq U$

therefore U is open and each interval (a,b) is open

Answer 1

That it is a base for a topology I just showed here.

Every set $(a,b)$ is open, because if $x \in (a,b)$, then $x \in [x,b) \subseteq (a,b)$ and $[x,b) \in \mathcal{B}$, so $(a,b)$ is open in the topology generated by $\mathcal{B}$.