Union of cylinder sets in Banach space

Let $$X$$ be a real separable Banach space and $$X^*$$ be its topological dual. We call sets of the form $$C=\{x\in X~|~ (f_1(x),\dots,f_n(x))\in C_0\in B(\mathbb{R}^n)\}$$

a Cylinder set with Base $$C_0$$ and cut points $$f_1\dots f_n$$. I want to show the collection of cylinder sets constitute an algebra but I don't see how to define unions. Denote $$\mathscr{C}$$ the collection of cylinder sets, then choosing $$C_0=\mathbb{R}$$ and $$f\neq 0$$, we have $$C=X\in \mathscr{C}$$.

For $$C_{C_0}=C=\{x\in X~|~ (f_1(x),\dots,f_n(x))\in C_0\in B(\mathbb{R}^n)\}\in \mathscr{C}$$, we have $$C_{C_0^c}=\{x\in X~|~ (f_1(x),\dots,f_n(x))\in C_0^c\in B(\mathbb{R}^n)\}\in \mathscr{C}$$ Now for unions, I don't see how to define unions of sets with different cut off points, namely let for example $$A=\{x| (f_1(x),f_2(x))\in A_0\in B(\mathbb{R}^2)\},\quad B=\{x| (g_1(x),g_2(x),g_3(x))\in B_0\in B(\mathbb{R}^3)\}$$ with $$f_1\neq g_1$$ and $$f_2\neq g_2$$. How is the union $$A\cup B$$ represented ?

If $$A = \{x \in X: (f_1(x), \dots, f_n(x)) \in A_0\} \qquad B = \{x \in X: (g_1(x), \dots, g_m(x)) \in B_0\}$$ then we have that $$A \cap B = \{x \in X: (f_1(x), \dots f_n(x), g_1(x), \dots, g_m(x)) \in A_0 \times B_0\}$$ and so $$A \cap B$$ is also a cylinder set.
It is also possible to just write down a representation of $$A \cup B$$ as a cylinder set directly. We have $$A \cup B = \{ x \in X: (f_1(x), \dots, f_n(x), g_1(x), \dots g_m(x)) \in (\mathbb{R}^n \times B_0) \cup (A_0 \times \mathbb{R}^m) \}.$$