I searched the site as thoroughly as I could, but couldn't find anything that's close enough to what I'm asking.
Let S be some well-ordered set. Let M(S) be the set of all ordinals α, such that there is an injection from α into S. It's easily provable that M(S) is an ordinal itself, and that |S| ≤ |M(S)|. How can one show, without Choice, that |S| < |M(S)|? That is, in the sense that there is no injection from M(S) into S.
I tried to assume that there exists an injection from M(S) into α, the unique ordinal that's isomorphic to S, aiming to somehow reach the conclusion that α = M(S), in contradiction to the fact that α ∈ M(S). But I've been stuck on this for too long without making progress.
Will appreciate any guidance. Thank you!
Suppose there were an injection $M(S)\to S$. Since $M(S)$ is the set of all ordinals injecting into $S$ and $M(S)$ itself is an ordinal we have $M(S)\in M(S)$, a contradiction