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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.

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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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  • $\begingroup$ I think it works, thank you. However, I actually needed the case where all the $A_m$ are equal. How can I prove it ? $\endgroup$ – Richard Feb 8 '16 at 19:17
  • $\begingroup$ @Richard : I edited my answer. Let me know if anything is unclear (which might be the case…) $\endgroup$ – Watson Feb 8 '16 at 19:25
  • $\begingroup$ I think here it's essential to highlight the fact that $X$ has infinite countably components, if not $X$ is not an element of $A^{\mathbb N}$ in spite of your argument. You proved an inclusion but the other is doable in the same way. If $X$ is $A^{\mathbb N}$ then its first component is in $A_0=A_{n+1}$, the second is in $A_1=A_0=A_{n+2}$ and so on. Therefore $X$ is in $A^{\mathbb N\setminus \{1,...,n\}}$. Same argue about the number of components. Do you agree? $\endgroup$ – Richard Feb 8 '16 at 20:20
  • $\begingroup$ @Richard : you are right, the fact that X has infinite countably components is essential. Yes I agree for the other inclusion. $\endgroup$ – Watson Feb 8 '16 at 20:25

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