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Here, $(\kappa)^+$ is the Hartog's Number, the least cardinal $\lambda$ so that there is no injection from $\lambda$ to $\kappa$.

My feeling is that this is true, since for any ordinal $\alpha < \omega $, $\ \alpha+1<\omega$, so for each infinite cardinal $\aleph_\alpha<\aleph_\omega$, $\ \aleph_{\alpha+1}=(\aleph_{\alpha})^+<\aleph_\omega$, but I feel I am making some leaps in logic here or have missed something and this doesn't constitute any kind of a proof.

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  • $\begingroup$ Doesn't $\aleph_\omega$ inject into $\aleph_\omega$? $\endgroup$ – Andrés E. Caicedo Nov 24 '18 at 15:14
  • $\begingroup$ Yes, you are right. Thank you $\endgroup$ – MMR Nov 24 '18 at 15:30
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It is true that $\aleph_\alpha<\aleph_\omega\implies\aleph_{\alpha+1}<\aleph_\omega$ - this is true simply because $\omega$ is a limit ordinal - but "$\aleph_\alpha$ injects into $\aleph_\omega$" does not imply $\aleph_\alpha<\aleph_\omega$ (e.g. $\aleph_\omega$ injects into $\aleph_\omega$, obviously).

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  • $\begingroup$ Ah yes, I see. How would you find an example of a cardinal with this property? $\endgroup$ – MMR Nov 24 '18 at 15:32
  • $\begingroup$ @Matthew No cardinal has that property: $\kappa$ always injects into $\kappa$. $\endgroup$ – Noah Schweber Nov 24 '18 at 15:35
  • $\begingroup$ Would it be true instead for the well ordering on ordinals i.e. If $\kappa<\aleph_\omega$, then $(\kappa)^+<\aleph_\omega$? Since in this case $\aleph_\omega$ is not strictly less than itself? $\endgroup$ – MMR Nov 24 '18 at 15:49
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    $\begingroup$ @Matthew Yes. You can also phrase it in terms of cardinalities the same way. $\endgroup$ – Andrés E. Caicedo Nov 24 '18 at 15:53
  • $\begingroup$ @Matthew And note that you can turn that into an embedding-phrased condition: if $\kappa$ injects as a proper initial segment of $\aleph_\omega$, then $\kappa^+<\aleph_\omega$. $\endgroup$ – Noah Schweber Nov 24 '18 at 16:14

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