perhaps my title isn't very clear, I'll try to be more precise: The definition I'm using is that (for the "successor" step of the recursion) $\aleph_{\alpha+1}$ is the smallest cardinal which is strictly greater that $\aleph_{\alpha}$. I take this to mean that $\aleph_{\alpha+1} = \inf \{$cardinals $C$ such that there exists a one sided injection, but not a bijection between $C$ and $ \aleph_{\alpha} \}$
What I can't figure out is how to show that with this definition, $\aleph_{\alpha+1}$ indeed remains strictly greater than $\aleph_{\alpha}$, ie there is no bijection between $\aleph_{\alpha}, \aleph_{\alpha+1}$. I know that there exists the generalized continuity hypothesis which would settle this question, but I'm thinking it is probably overkill and that I should be able to show this some other way, no? If someone could help me with what is probably a simple reasoning that would be great.
It's more accurate to say $\aleph_{\alpha+1} = \min\{\kappa \text{ cardinal } \mid (\exists f \text{ injection }: f:\kappa \to \aleph_\alpha) \land \kappa \not\simeq \aleph_\alpha \}$ so a minimum not an infimum (which is also true) because all sets/classes of cardinals (which are sets of ordinals too) have a minimum. In particular $\aleph_{\alpha+1}$ is itself a member of that set of cardinals, so is strictly greater than $\aleph_\alpha$ by definition.