Why a cylinder set with base in $\mathbb{R}^n$ can have a base in $\mathbb{R}^{n+1}$ as well?

Question

Let define a cylinder set as:

$$\mathcal{C}(B_1 \times \dots \times B_n)=\{x \in \mathbb{R}^{\infty}|x_1 \in B_1 , \dots , x_n \in B_n \}$$

where $B_k \in \mathcal{B}(\mathbb{R})$ for $k=1,2,...,n$ with base in $\mathbb{R}^n$.

But, we can write also:

$$\mathcal{C}(B_1 \times \dots \times B_n)=\mathcal{C}(B_1 \times \dots \times B_n\times \mathbb{R})$$

with base in $\mathbb{R}^{n+1}$.

Why is this statement true?

Answer 1

Hint:

If $x\in A$ implies that $x\in B$ and $x\in B$ implies that $x\in A$ then we are allowed to conclude that $A=B$.

This because (according to the axiom of extensionality) sets are completely determined by their elements.