# 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.

• You don't need choice to well-order the image of a well-ordered set. – Asaf Karagila Oct 15 '18 at 7:01

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.

• You actually don't need Cantor–Bernstein on this one. You need the well-order comparability theorem. – Asaf Karagila Oct 15 '18 at 7:00

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 $$\kappa=\lambda$$.

• But don't the well ordering of the cardinals require choice? – Holo Oct 15 '18 at 8:17
• Yes, but the context of the question points towards $\kappa$ and $\lambda$ being ordinals. – Asaf Karagila Oct 15 '18 at 8:22
• (I agree that in general the term "cardinals" should apply more broadly in choiceless contexts. But this question doesn't even make sense for non-well ordered cardinals.) – Asaf Karagila Oct 15 '18 at 11:07