Let $A$ be an arbitrary set of nonlimit ordinals less than a given $\kappa$, and let the union $A$ be also less than $\kappa$. Does this union have to be a nonlimit ordinal if
- $\kappa=\omega$;
- $\kappa$ is an uncountable cardinal.
In the first case, the answer is yes? $A$ must be a subset of omega such that $\bigcup A < \omega$. So $A$ must be finite? And then the union will be a limit ordinal?
In the second case, I tried to consider $\aleph_1$. But here it is unclear how to describe its subsets $A$ for which $\bigcup A < \aleph_1$...
For 1, it is trivially true since the first limit ordinal is $\omega$ and you are given that the union is less than $\omega$
For 2, it's easily shown to be false. Just take $A$ to be all the natural numbers. Then their union is a limit ordinal $\omega$