Let $(S,\mathcal{S})$ be a measurable space, $J\subset T$ and denote $$S^T=\{x=(x_t)_{t\in T} \ \lvert \ x_t\in S \}.$$ Let $\mathcal{S}^{\otimes T}$ be the smallest $\sigma$-Algebra that contains all the cylinder sets $\{x\in\mathcal{S}^T \ \lvert \ x_{t_1}\in A_1,\ldots,x_{t_n}\in A_n\},t_i\in T,A_i\in\mathcal{S}.$ Consider the coordinate projections $\pi_{T,J}:S^T\longrightarrow S^J, (x_t)_{t\in T}\mapsto (x_j)_{j\in J}$. Show, that $$ \mathcal{A}=\bigcup_{J\subset T\text{ finite}} \pi_{T,J}^{-1}(\mathcal{S}^{\otimes J})$$ is an algebra over $\mathcal{S}$.
Here’s what I tried. I think $\mathcal{A}$ is non-empty, because it should at least contain the cylinder sets regarding $\mathcal{S}^J$, right? But when it comes to show the other two features of an algebra, namely $$\cdot \ A,B\in\mathcal{A} \Rightarrow A\cup B\in\mathcal{A}$$ $$\cdot A\in\mathcal{A} \Rightarrow A^c\in\mathcal{A}$$ I am completely lost. What does $A\cup B$ even mean in this context? Would really appreciate some help.