# Is it true that for any cardinal $\kappa$ that in injects into $\aleph_{\omega}$, $(\kappa)^{+}$ injects into $\aleph_{\omega}$?

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.

• Doesn't $\aleph_\omega$ inject into $\aleph_\omega$? Nov 24, 2018 at 15:14
• Yes, you are right. Thank you
– MMR
Nov 24, 2018 at 15:30

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).
• @Matthew No cardinal has that property: $\kappa$ always injects into $\kappa$. Nov 24, 2018 at 15:35
• 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?
• @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 $\kappa^+<\aleph_\omega$. Nov 24, 2018 at 16:14