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Suppose $$A=A_1\times A_2\space\times\space...\space=\prod_{i\in\mathbb{Z_+}} A_i$$ and $$B=B_1\times B_2\space\times\space...\space=\prod_{i\in\mathbb{Z_+}} B_i\neq\emptyset$$ and that $B\subset A$. I'm curious whether the proof that $B_i\subset A_i$ for all $i\in\mathbb{Z_+}$, requires the Axiom of Choice.

First since $B$ is nonempty, $B_i\neq\emptyset$ for all $i\in\mathbb{Z_+}$ so let $b_i\in B_i$ for some arbitrary $i\in\mathbb{Z_+}$. Since $B\subset A$, by definition of the cartesian product any function $$f:\mathbb{Z_+}\to\bigcup_{i\in\mathbb{Z_+}}B_i$$ such that $f(i)\in B_i$ f0r all $i\in\mathbb{Z_+}$ is actually a function $$g:\mathbb{Z_+}\to\bigcup_{i\in\mathbb{Z_+}}A_i$$ such that $g(i)\in A_i$ for all $i\in\mathbb{Z_+}$, hence $f(i)=g(i)$ for all $i\in\mathbb{Z_+}$.

There then exists a function $f$ such that $b_i=f(i)$

so that since $f(i)=g(i)$, we have $b_i=g(i)\in A_i$. Thus $B_i\subset A_i$ since $i$ was arbitrary. $\blacksquare$

The highlighted portion requires the $AC$, right?

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    $\begingroup$ At worst you'd need countable choice. $\endgroup$
    – J.G.
    Commented Aug 20, 2023 at 8:36

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No, you don't need any form of choice. You are already assuming $B \neq \emptyset$, which is the only place some form of choice might be needed.

Fix $f_0 \in B$. Then for any $i$ and $b_i \in B_i$, you can define the function $$f(k) = \begin{cases}b_i, & \text{ if } k=i \\ f_0(k), & \text{ otherwise}\end{cases}$$

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  • $\begingroup$ I think however, that the OP is right that the way that the original proof is phrased it uses choice. But I find it somewhat confusing anyway. $\endgroup$
    – Carsten S
    Commented Aug 21, 2023 at 8:12

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