Is the axiom of choice independent of the Unrestricted Comprehension Principle?

Question

In Why is the axiom of choice separated from the other axioms?, a comment on the top answer mentions how 'Choice is special in that it isn't a special case of the problematic Unrestricted Comprehension principle', which to me implies that the Axiom of Choice cannot be derived from the Unrestricted Comprehension Principle--can someone explain why that is?

Answer 1

This a problematic question. First of all, unrestricted comprehension is inconsistent, so you can prove anything from it.

But if we try to avoid explosion-style argument, which is often something productive from a pedagogical point of view, we still can't say much.

The axiom of choice can be formulated in many different equivalent ways. One specific way would be that if $R$ is a relation, then there is a function $F\subseteq$ such that $\operatorname{dom}(F)=\operatorname{dom}(R)$.

But now we can ask, can we actually specify this choice function? Well, again, using the unrestricted comprehension schema we can prove everything, which means this is useless to us.

So let's restrict the question and ask, if $R$ is a given relation, can we prove the existence of $F$ as above, without using the axiom of choice, and just the basic axioms of $\sf ZF$?

The answer, of course, is negative. As in the case such thing is provable, we would have that $\sf AC$ is provable from $\sf ZF$, and we know that if $\sf ZF$ is consistent, then the axiom of choice is not provable from $\sf ZF$. So the whole thing would be a proof that $\sf ZFC$ is inconsistent, and a lot of the mathematics we do will require closer re-examining.

More to the point, there is no way to prove the unrestricted schema does not imply choice, as it is inconsistent. However its bounded part does not imply choice.

The intuition for the fact that $\sf ZF$, with restricted comprehension, does not imply the axiom of choice, comes from the fact that in general, having a function $F\subseteq R$ would require us to literally specify the properties of $F$ as a subset of $R$ which is impossible. If you want to argue otherwise, sure, let's try at first with a choice of elements from every set of reals, and remember most sets are not intervals.