Let $A^{\mathbb N}=\prod_{m\in\mathbb N} A_m$ be the infinite countably cartesian product of the sets $A_m$. Let $A_i'$ be a subset of $A_i$ for $i=1,...,n$. Is it true that $A_1'\times A_2'\times...\times A_n'\times A^{\mathbb N\setminus\{1,...,n\}}$ is equal to $A_1'\times A_2'\times...\times A_n'\times A^{\mathbb N}$ ?
For me the answer is yes because an infinite countably set minus a finite set is still infinite countably but I don't know how formally using this to solve the problem. Thank you very much.
Answer
Hint : consider $A_m = \{0,1,…,m\}$, $A_m' = \{0\}$.
Then, you can find $(\underbrace{0,0,…,0}_{n \text{ times}},n+1,n+1,…)$ in the first product, but not in the second one (because the $n+1$-th component in the second product is $A_0 = \{0\}$).
However, if all the $A_m$ are equal, then it is true : it is sufficient to prove that $$A^{\mathbb N\setminus\{1,...,n\}} = A^{\mathbb N}$$
To do this, pick $X=(x_{n+1},x_{n+2},…) \in A^{\mathbb N\setminus\{1,...,n\}}$, that means $x_j \in A_j := A_0$ for all $j ≥ n+1$.
Then, you want to prove that $X \in A^{\mathbb N}$. This means : does the $k$-th component of $X$ ($k≥1$) belong to $A_k=A_0$ ? Does every component of $X$ belong to $A_0$ ?
Then, $X$ belongs to $A^{\mathbb N}$ simply because all its components belong to $A_0$... Even if I denoted the first component of $X$ by $x_{n+1}$, this doesn't really matter because what is important is that $x_{n+1} \in A_{n+1} = A_0$.
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