I'm currently in a measure theory class and I need to show $B(\mathbb{R^2}) \subseteq B(\mathbb{R}) \otimes B(\mathbb{R})$ (here $B(\mathbb{R})$ is the borel $\sigma$-algebra on $\mathbb{R}$, $B(\mathbb{R}) \otimes B(\mathbb{R})$ is the product $\sigma$-algebra of the product space $\mathbb{R} \times \mathbb{R}$) but I don't really know how.
I've read some answers using basis of topology but I haven't learned topology before(I only know a topology space consist of a universal space and a class of open sets).
Is there a solution without using basis of topology? Thank you.
If $U$ is any open set in $\mathbb R^{2}$ and $(x,y) \in U$ then there is an open rectangle $R(x,y)$ containing $(x,y)$ having vertices with rational coordinates contained in $U$. The union of all these rectangles is equal to $U$. Since open rectangles are of the type $(a,b)\times (c,d)$ they belong to $B(\mathbb R) \otimes B(\mathbb R))$. Hence, every open set belongs to $B(\mathbb R) \otimes B(\mathbb R))$ and this implies that $B(\mathbb R^{2})$ is contained in $B(\mathbb R) \otimes B(\mathbb R))$. I hoep this proof is acceptable to you.