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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.

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

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

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  • $\begingroup$ But don't the well ordering of the cardinals require choice? $\endgroup$ – Holo Oct 15 '18 at 8:17
  • $\begingroup$ Yes, but the context of the question points towards $\kappa$ and $\lambda$ being ordinals. $\endgroup$ – Asaf Karagila Oct 15 '18 at 8:22
  • $\begingroup$ (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.) $\endgroup$ – Asaf Karagila Oct 15 '18 at 11:07

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