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If for every family $(X_i)_{i\in I}$ of sets, there exists a categorical product in the category $\mathbf{Set}$ of sets, does this imply that the set-theoretic construction $\left\{(x_i)_{i\in I}\in\left(\bigcup_{i\in I}X_i\right)^I\middle|x_i\in X_i\text{ for all $i$}\right\}$ is non-empty, i.e. that the Axiom of Choice holds?

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No -- in a world without choice it may simply be that $\varnothing$ satisfies the conditions for being the product object.

It would do so almost vacuously: In order to apply the universal property of products you would need to have an object $Y$ and a family of morphisms $Y\to X_i$; but if $(X_i)_i$ doesn't have a choice function, then such a family of morphisms can't exist unless $Y$ happens to be the empty set.

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  • $\begingroup$ Oh right, if I try to prove that $\{(x_i)\in\bigcup X_i\mid x_i\in X_i\text{ for all $i$}\}$ is a categorical product, I do not need Choice, so the existence of categorical products is always true and thus does not have much to do with the question, if Choice holds. Is that correct? $\endgroup$ – user114885 Dec 23 '13 at 14:25
  • $\begingroup$ Yes this is correct. Roughly, this is because universal objects deal with unique choices, whereas AC deals with random choices. $\endgroup$ – Martin Brandenburg Dec 23 '13 at 14:29
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    $\begingroup$ @user114885: Correct. You don't need Choice to prove that the set you describe exists, only if you want to prove it is non-empty. $\endgroup$ – Henning Makholm Dec 23 '13 at 14:30
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    $\begingroup$ @MartinBrandenburg: "Random" is a rather misleading word here. It does have informal meanings that are not wrong in this context, but a reader who don't already know the various subtleties would be very likely to read it as "stochastic", which is definitely not what AC is about. $\endgroup$ – Henning Makholm Dec 23 '13 at 14:33
  • $\begingroup$ Read: "roughly" $\endgroup$ – Martin Brandenburg Dec 23 '13 at 14:42

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