Can ZF prove the cartesian product of an ordinal by itself to be subnumerous to that ordinal?

31 Views Asked by At

Does ZF (without choice) prove the following?

$\forall \ ordinal \ \alpha \ \exists f( f:\alpha \times \alpha \hookrightarrow \alpha)$

where: $ \alpha\times \alpha $ is the cartesian product of $\alpha$ by itself; and $\hookrightarrow$ signifies injection.