By a D-finite set, we mean a set admitting no injection from the natural numbers (or equivalently, a set not in bijection with any proper subset).
I have encountered a proof that the following are equivalent in ZF:
A. Every D-finite set is finite.
B. D-finite unions of D-finite sets are D-finite.
C. Images of D-finite sets are D-finite.
D. Power sets of D-finite sets are D-finite.
C implies D implies A implies B is achieved without difficulty, but it seems to me that they took a circular approach to showing that B implies C. They proceeded as follows:
Assume $X$ D-finite, $f:X\to Y$ a surjection. Then $Y=\bigcup_{x\in X}\{f(x)\}$, so as a D-finite union of D-finite sets, $Y$ is D-finite.
Of course, singletons are D-finite since finite, and of course $Y$ is the union of said singletons since $f$ is surjective. However, since the set of such singletons is in bijection with $Y$, isn't the D-finite union claim tantamount to claiming the conclusion is true?
Does anyone know of an alternate approach one can take to prove these are all equivalent?