# Axiom of choice hiding in seemingly trivial proof?

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?

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

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 $$f(k) = \begin{cases}b_i, & \text{ if } k=i \\ f_0(k), & \text{ otherwise}\end{cases}$$