max of limit cardinals smaller than a successor cardinal bigger than $\aleph_\omega$

Question

This is the statment i want to proove : if $k$ is a successor cardinal bigger than $\aleph_{\omega}$ then there is a max of $\{\lambda < k | \lambda \text{ is a limit cardinal}\}$. Assuming (AC), if $k \geq \omega$ then $k^{+}$ is regular, so we can assume that $k$ is regular. I've found this argument but i'm not sure that it's correct : By contradiction, if there is no max, then the union of all $\lambda$ limit smaller than $k$ is a successor cardinal $\leq k$. But a successor cardinal is regular, so it cannot be reach by this union : a contradiction. So there must be a max. Am I right?

Answer 1

Hint: Let $\rho = \bigcup \{ \lambda < \kappa \mid \lambda \text{ is a limit cardinal } \}$. $\rho$ is a nonempty union of cardinals and hence itself a cardinal. If you can show that $\rho$ is a limit cardinal, you're done.

Answer 2

Choice has nothing to do with this. The $\aleph$ numbers are defined the same. Of course, for non-$\aleph$ cardinals there is an issue for what would be a successor, or limit, and so on. So let's ignore that issue and focus on $\aleph$ cardinals.

You can easily prove this by induction, noting that successor cardinals are exactly those whose $\aleph$ index is a successor ordinal. So the question is reduced to limit ordinals below infinite successor ordinals, and this question is easier to grasp.