Injection from cardinal $\lambda$ to cardinal $\kappa$ implies $\lambda\leq\kappa$ I'm trying to prove that if there is an injection $f:\lambda\to\kappa$ (for $\lambda$,$\kappa$ cardinal numbers) then $\lambda\leq\kappa$. This is not true if they are just ordinal numbers, for example it is easy to build an injection from 
 $\omega+1$ to $\omega$, however $\omega<\omega+1$.
I think the proof should be quite straightforward but I'm not getting it.
  I want to arrive to a contradiction by assuming $\kappa<\lambda$, so $f|_\kappa:\kappa\to\kappa$ is an injection and $f(\kappa)$ is propper subset of $\kappa$ (not necessarly an ordinal number) but I don't realize how this can be problematic or how to move from here.
I thought maybe I should well order $f(\kappa)$ (and for this I think I need AC) and do something with its order type, but again I'm not sure how to proceed.
 A: Hint: The statement you are trying to prove is more or less just a disguised version of the Schroder-Bernstein theorem.
A full proof is hidden below.

 Suppose there is an injection $f:\lambda\to\kappa$ but $\kappa<\lambda$.  Then the inclusion map is an injection $i:\kappa\to\lambda$.  Since there are injections in both directions between $\kappa$ and $\lambda$, by Schroder-Bernstein there is a bijection between them.  But this is a contradiction, since $\lambda$ is a cardinal so it cannot be in bijection with any smaller ordinal.

A: If you already know that given two well-ordered sets one is isomorphic to an initial segment of the other, then this becomes borderline trivial using the fact both are cardinals.

 By the comparability theorem either $\lambda$ is an initial segment of $\kappa$, in which case we are done. Otherwise $\kappa$ is an initial segment of $\lambda$. By the assumption that $\lambda$ is a cardinal it is not equipotent with any of its proper initial segments, so it has to be that $\kappa=\lambda$.

