Strengthening of the axiom of Dependent Choice

Question

I have in mind the following statement:

For every non empty set $X$ and for every entire binary relation $R$ over $X$, there exists a function $f:X\rightarrow X$ such that, for all $x \in X, \ \ x \ R \ f(x)$

This statement obviously implies $\text{DC}$ in ZF, but how much stronger is it? Is it a proper strengthening? Is it equivalent to other known choice axioms?

Thanks

Answer 1

The axiom you propose is equivalent to the axiom of choice. It can be used to produce choice functions on $V_\alpha\setminus\{\emptyset\}$ for arbitrary $\alpha$:

Let $X=V_\alpha$ and say $x\ R\ y $ iff $y\in x$ or $x=\emptyset=y$. The last clause is only there to make sure that $R$ is entire. If now $f: V_\alpha\rightarrow V_\alpha$ satisfies $x\ R\ f(x)$ for any $x$ then $f(x)\in x$ for $x\neq\emptyset$.

This of course implies that any $X$ (with $\emptyset\notin X$) admits a choice function as $X$ is included in some $V_\alpha$ for $\alpha$ large enough.