I am reading Jech's Axiom of Choice, and I want to prove:
For a non-empty set $I$, if $\{0,1\}^I$, the generalized Cantor space, is non-empty compact, then $\prod_{i\in I}A_i$ where $|A_i|=2$ for all $i\in I$ is non-empty compact where the topologies on both $\{0,1\}$ and $A_i$'s are discrete.
Jech notes that it suffices to show that $\prod_{i\in I} A_i$ is non-empty, and I am wondering if the below argument works formally. (argument for why proving that $\prod_{i\in I} A_i$ is non-empty is sufficient).
My idea right now is: once we know $\prod_{i\in I} A_i$ is non-empty, let $f\in \prod_{i\in I} A_i$. For each $i\in I$, we may identify $f(i)$ with $0$. So can we create a map $F:\{0,1\}^I\to \prod_{i\in I} A_i$ by: given $\alpha\in \{0,1\}^I$, $F$ sends $\alpha$ to $f_\alpha$ where $f_\alpha(i)=f(i)$ if $\alpha(i)=0$ and otherwise, take $f_\alpha(i)$ be the unique element of the set $A_i\setminus \{f(i)\}$. If we can create the such map $F$, then $F$ will give an isomorphism.
Thank you in advance for any help.
This is the right idea for the step you’re asking about. Moreover, it shows that the assumption that the $A_i$‘s have the discrete topology is superfluous. Your $F$ is bijective, and it’s continuous regardless of the topologies of the $A_i$’s (why?). Thus, $\prod_i A_i$ is the continuous image of a compact set.
But I don’t think the result is quite correct how you’ve stated it, since I don’t think you can actually prove the nonemptiness of $\prod_{i\in I} A_i$ from the compactness of $2^I.$ (Or at least I don’t know how and the claim isn’t implicit in what Jech actually wrote in the exercises you got this from.) What you can do is prove the nonemptiness and compactness of any product of sets of size 2 from the general statement that $2^J$ is compact for any set $J$.