Q:

To show that $\mathcal{B}\left(\mathbb{R}^d\right)=\left\{\mathcal{B}\left(\mathbb{R}\right)\right\}^d$

Let $\mathcal{B}\left(\mathbb{R}\right)$ be the Borel $\sigma$-field of $\mathbb{R},$ i.e. the $\sigma$-field generated by the class of open sets in $\mathbb{R}$.
Let $\mathcal{B}\left(\mathbb{R}^d\right)$ be the Borel $\sigma$-field of $\mathbb{R}^d,$ i.e. the $\sigma$-field generated by the class of open sets in $\mathbb{R}^d$.
To show that $\mathcal{B}\left(\mathbb{R}^d\right)=\left\{\mathcal{B}\left(\mathbb{R}\right)\right\}^d,$
where $\left\{\mathcal{B}\left(\mathbb{R}\right)\right\}^d$ 
denotes the cartesian product $\mathcal{B}\left(\mathbb{R}\right)
 \times \cdots \times \mathcal{B}\left(\mathbb{R}\right) (d \text{ times})$.

A:

The claim is not true; just look at an open set of $\mathbb{R}^d$ which is not the product of open sets of $\mathbb{R}$, like the unit ball. It would be true that the Borel subsets of $\mathbb{R}^d$ are generated by $\mathcal{B}(\mathbb{R})^d$.

